Applied statistics and probability for engineers 7th edition montgomery solutions manual

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Applied Statistics and Probability for Engineers, 7h edition CHAPTER 9 Section 9.1 9.1.1

a) H 0 :  = 25, H1 :   25 Yes, because the hypothesis is stated in terms of the parameter of interest, inequality is in the alternative hypothesis, and the value in the null and alternative hypotheses matches. b) H 0 :   10, H1 :  = 10 No, because the inequality is in the null hypothesis. c) H 0 : x = 50, H1 : x  50 No, because the hypothesis is stated in terms of the statistic rather than the parameter. d) H 0 : p = 0.1, H1 : p = 0.3 No, the values in the null and alternative hypotheses do not match and both of the hypotheses are equality statements. e) H 0 : s = 30, H1 : s  30 No, because the hypothesis is stated in terms of the statistic rather than the

parameter.

9.1.2 The conclusion does not provide strong evidence that the critical dimension mean equals 100nm. There is not sufficient evidence to reject the null hypothesis.

9.1.3

a)  = P(reject H0 when H0 is true)

 X −

= P( X  11.5 when  = 12) = P

 / n



11.5 − 12   = P(Z  −2) = 0.02275.

0.5 / 4 

The probability of rejecting the null hypothesis when it is true is 0.02275.

(

b)  = P(accept H0 when  = 11.25) = P X  11 .5 |  = 11 .25

 X −

= P

 / n



)

11.5 − 11.25   = P(Z > 1.0) = 1 − P(Z  1.0) = 1 − 0.84134 = 0.15866 0.5 / 4 

The probability of failing to reject the null hypothesis when it is false is 0.15866 c)  = P(accept H0 when  = 11.25) =

(

 X −  11.5 − 11.5    0.5 / 4   / n

)

= P X  11 .5 |  = 11 .5 = P

= P(Z > 0) = 1 − P(Z  0) = 1 − 0.5 = 0.5 The probability of failing to reject the null hypothesis when it is false is 0.5

9.1.4

 X −

a)  = P( X  11.5 |  = 12) = P

 / n



11.5 − 12   = P(Z  −4) = 0.

0.5 / 16 

The probability of rejecting the null hypothesis when it is true is approximately 0 with a sample size of 16.

 X −

b)  = P( X > 11.5 |  =11.25) = P

 / n



11.5 − 11.25   0.5 / 16 

= P(Z > 2) = 1 − P(Z  2)= 1− 0.97725 = 0.02275. c) The probability of failing to reject the null hypothesis when it is false is 0.02275.

9-1

Applied Statistics and Probability for Engineers, 7h edition

 X −  11.5 − 11.5    = P(Z > 0) = 1 − P(Z  0) = 1− 0.5 = 0.5 0.5 / 16   / n

 = P( X > 11.5 |  =11.5) = P

The probability of failing to reject the null hypothesis when it is false is 0.5.

9.1.5



a) x =11.25, then P-value= P Z 





b) x =11.0, then P-value= P Z 





11.0 − 12   = P( Z  −4)  0.000033 0.5 / 4 

c) x =11.75, then P-value= P Z 



9.1.6

11.25 − 12   = P( Z  −3) = 0.00135 0.5 / 4 

11.75 − 12   = P( Z  −1) = 0.158655 0.5 / 4 

a) β = P( X >11.59|µ=11.5) = P(Z>0.36) = 1-0.6406 = 0.3594 b) β = P( X >11.59|µ=11.5) = P(Z>0.72) = 1-0.7642 = 0.2358 c) Notice that the value of  decreases as n increases

9.1.7

a)  = P( X > 185 when  = 175)

 X − 175

= P

 20 / 10



185 − 175   20 / 10 

= P(Z > 1.58) = 1 − P(Z  1.58) = 1 − 0.94295 = 0.05705 b)  = P( X  185 when  = 185)

 X − 185 185 − 185    = P(Z  0) = 0.5  20 / 10 20 / 10 

= P

c)  = P( X  185 when  = 195)

 X − 195

= P

 20 / 10

9.1.8



185 − 195   = P(Z  −1.58) = 0.05705 20 / 10 

a) α = 0.05, n =10, then the critical value 185.37 (from the previous exercise part (b))  = P( X  185.37 when  = 185) = 𝑃(

𝑋−185 20⁄√10

≤

185.37−185 20⁄√10

)= P(Z  0.05849) = 0.5199

b) α = 0.05, n =16, then the critical value 183.2 (from the previous exercise part (d)), then  = P( X  183.2 when  = 185)

9-2

Applied Statistics and Probability for Engineers, 7h edition

 X − 185 183.2 − 185    = P(Z  -0.36) = 0.3594 20 / 16   20 / 16

= P

c) As n increases,  decreases

9.1.9

 20  X  175 + Z     n a) α = 0.01, n = 10, then z  =2.32 and critical value is 189.67 b) α = 0.05, n = 10, then z  =1.64 and critical value is 185.37 c) α = 0.01, n = 16, then z  = 2.32 and critical value is 186.6 d) α = 0.05, n = 16, then z  =1.64 and critical value is 183.2

9.1.10

a)  = P( X  4.85 when  = 5) + P( X > 5.15 when  = 5)

 X −5 4.85 − 5   X − 5 5.15 − 5   + P     0.25 / 8 0.25 / 8   0.25 / 8 0.25 / 8 

= P

= P(Z  −1.697) + P(Z > 1.697) = P(Z  −1.697) + (1 − P(Z  1.697) = 0.045514 + (1 − 0.954486) = 0.0910 b) Power = 1 −   = P(4.85  X  5.15 when  = 5.1)

 4.85 − 5.1

= P

 0.25 / 8



X − 5.1 5.15 − 5.1    0.25 / 8 0.25 / 8 

= P(−2.828  Z  0.5657) = P(Z  0.5657) − P(Z  −2.828) = 0.712260 − 0.002401 = 0.70986 1 −  = 0.290141

9.1.11

X ~ bin(15, 0.4) H0: p = 0.4 and H1: p  0.4 p1= 4/15 = 0.267

p2 = 8/15 = 0.533

ˆ  0.533 Accept Region: 0.267  p Reject Region: pˆ  0.267 or pˆ  0.533 Use the normal approximation for parts a) and b)

ˆ  0.267 ) + P ( pˆ  0.533 ) a) When p = 0.4,  = P ( p         0.267 − 0.4  0.533 − 0.4  = P Z  + P Z    0.4(0.6)  0.4(0.6)      15 15     = P( Z  −1.05) + P( Z  1.05) = P( Z  −1.05) + (1 − P( Z  1.05)) = 0.14686 + 0.14686 = 0.29372

9-3

Applied Statistics and Probability for Engineers, 7h edition b) When p = 0.2     0.267 − 0.2 0.533 − 0.2   = P(0.267  pˆ  0.533) = P Z 0.2(0.8)   0.2(0.8)  15 15   = P(0.65  Z  3.22) = P( Z  3.22) − P( Z  0.65) = 0.99936 − 0.74215 = 0.2572

9.1.12

X ~ Bin(10, 0.3) Implicitly, H0: p = 0.3 and H1: p < 0.3 n = 10 Accept region: pˆ  0.1

ˆ  0 .1 Reject region: p Use the normal approximation for parts a), b) and c):

    0. 1 − 0 . 3   a) When p =0.3  = P ( pˆ  0.1) = P Z   0.3(0.7)    10   = P( Z  −1.38) = 0.08379     0 .1 − 0 .2   b) When p = 0.2  = P ( pˆ  0.1) = P Z   0.2(0.8)    10   = P( Z  −0.79 ) = 1 − P( Z  −0.79 ) = 0.78524 c) Power = 1 −  = 1 − 0.78524 = 0.21476

9.1.13

a) Operating characteristic curve: x = 185

  x−  185 −    = P Z   = P Z      20 / 10 20 / 10 



 185 −   P Z  =  20 / 10 



1−

178 181 184 187 190 193 196 199

P(Z  1.11) = P(Z  0.63) = P(Z  0.16) = P(Z  −0.32) = P(Z  −0.79) = P(Z  −1.26) = P(Z  −1.74) = P(Z  −2.21) =

0.8665 0.7357 0.5636 0.3745 0.2148 0.1038 0.0409 0.0136

0.1335 0.2643 0.4364 0.6255 0.7852 0.8962 0.9591 0.9864

9-4

Applied Statistics and Probability for Engineers, 7h edition Operating Characteristic Curve 1 0.8



0.6 0.4 0.2 0 175

180

185

190

195

200

195

200



b)

Power Function Curve 1 0.8

1−

0.6 0.4 0.2 0 175

180

185

190



Section 9-2 9.2.1

a) H 0 :  = 10, H 1 :   10 b) H 0 :  = 7, H 1 :   7 c) H 0 :  = 5, H 1 :   5

9.2.2

a) α = 0.01, then a = z1−  -2.33 b) α = 0.05, then a = z1−  -1.64 c) α = 0.1, then a = z1−  -1.29

9.2.3

N SE Mean = 0.7495 19.889 − 20 z0 = = −0.468 0.75 / 10 P-value = 1 −  ( Z 0 ) = 1 −  ( −0.468 ) = 1 − 0.3199 = 0.6801 a) StDev =

9-5

Applied Statistics and Probability for Engineers, 7h edition Because the P-value >  = 0.05, we fail to reject the null hypothesis that  = 20 at the 0.05 level of significance. b) A one-sided test because the alternative hypothesis is  > 20 c) 95% CI of the mean is x − z 0.025     x + z 0.025  n n

0.75 0.75    19.889 + (1.96) 10 10 19 .4242    20 .3539 d) P-value = 21 −  ( Z 0 ) = 21 −  (0.468 ) = 21 − 0.6801  = 0.6398 19.889 − (1.96)

9.2.4

a) P-value = 1-  ( Z 0 ) = 1-  ( 2.05 )  0.02 b) P-value = 1-  ( Z 0 ) = 1-  (−1.84 )  0.97 c) P-value = 1-  ( Z 0 ) = 1-  (0.4)  0.34

9.2.5

a) SE Mean from the sample standard deviation =

s 2.365 = = 0.6827 N 12

b) A one-sided test because the alternative hypothesis is  > 99.

100.039 − 98 = 2.8253 2.5 / 12 Because  ( 2.8253 ) is close to 1, the P-value = 1 −  ( 2.8253 ) = 0.002 is very small and close to 0. c) If the null hypothesis is changed to the  = 98, z0 =

Thus, the P-value <  = 0.05, and we reject the null hypothesis at the 0.05 level of significance. d) 95% lower CI of the mean is x − z 0.05    n

100.039 − (1.645)

98 .8518  

2.5  12

100.039 − 99 = 1.4397 2.5 / 12 P-value = 2[1 −  ( Z 0 )] = 2[1 −  (1.4397 )] = 2[1 − 0.9250 ] = 0.15 e) If the alternative hypothesis is changed to the mu ≠ 99, z0 =

Because the P-value >  = 0.05, we fail to reject the null hypothesis at the 0.05 level of significance.

9.2.6

a) 1) The parameter of interest is the true mean crankshaft wear, .

2) H0 :  = 3 3) H1 :   3 4) z0 =

x−

/ n

5) Reject H0 if z0 < −z /2 where  = 0.05 and −z0.025 = −1.96 or z0 > z/2 where  = 0.05 and z0.025 = 1.96 6) x = 2.78,  = 0.9

9-6

Applied Statistics and Probability for Engineers, 7h edition z0 =

2.78 − 3 0.9 / 15

= −0.95

7) Because –0.95 > -1.96, fail to reject the null hypothesis. There is not sufficient evidence to support the claim the mean crankshaft wear differs from 3 at  = 0.05.



b)  =  z 0.025 +



 3 − 3.25  3 − 3.25   −  − z 0.025 +  0.9 / 15  0.9 / 15  

= (1.96 + −1.08) − (−1.96 + −1.08) = (0.88) − (-3.04) = 0.81057 − (0.00118) = 0.80939

c)

9.2.7

n=

(z

+ z )  2 2

 /2

2

=

(z 0.025 + z 0.10 )2  2 (3.75 − 3) 2

=

(1.96 + 1.29) 2 (0.9) 2 = 15.21, n  16 (0.75) 2

a) 1) The parameter of interest is the true mean battery life in hours, .

2) H0 :  = 40 3) H1 :  > 40 4) z0 =

x−

/ n

5) Reject H0 if z0 > z where  = 0.05 and z0.05 = 1.65 6) x = 40 .5 ,  = 1.25

z0 =

40.5 − 40 1.25 / 10

= 1.26

7) Because 1.26 < 1.65, fail to reject H0. There is not sufficient evidence to conclude that the mean battery life exceeds 40 at  = 0.05. b) P-value = 1 −  (1.26 ) = 1 − 0.8962 = 0.1038



40 − 42   = (1.65 + −5.06) = (-3.41)  0.000325 1.25 / 10   (z + z  )2  2 (z0.05 + z 0.10 )2  2 (1.65 + 1.29) 2 (1.25) 2 d) n = = = = 0.844, n  1 2 (40 − 44) 2 (4) 2 c)  =  z 0.05 +

e) 95% Confidence Interval x − z 0.05 / n   40.5 − 1.65(1.25) / 10   39.85   The lower bound of the 90 % confidence interval must be greater than 40 to verify that the true mean exceeds 40 hours.

9.2.8

a) 1) The parameter of interest is the true mean speed, .

2) H0 :  = 100 3) H1 :  < 100

9-7

Applied Statistics and Probability for Engineers, 7h edition 4) z0 =

x− / n

5) Reject H0 if z0 < −z where  = 0.05 and −z0.05 = −1.65 6) x = 102 .2 ,  = 4

z0 =

102.2 − 100 4/ 8

= 1.56

7) Because 1.56 > −1.65 fail to reject the null hypothesis. There is insufficient evidence to conclude that the true mean speed is less than 100 at  = 0.05. b) z 0 = 1.56 , then P-value=  ( z 0 )  0.94

  

c)  = 1 −  − z 0.05 −

(95 − 100) 8   = 1-(-1.65 - −3.54) = 1-(1.89) = 0.02938  4 

Power = 1- = 1-0.0294 = 0.9706

(z + z )  = (z d) n = 2





2

2

+ z 0.15 )  2 (1.65 + 1.03) 2 (4) 2 = = 4.60, n  5 (95 − 100) 2 (5) 2 2

0.05

e) 95% Confidence Interval

     n  4    102.2 + 1.65   8   104.53

  x + z 0.05 

Because 100 is included in the CI, there is not sufficient evidence to reject the null hypothesis.

9.2.9

a) 1) The parameter of interest is the true average battery life, .

2) H0 :  = 4 3) H1 :  > 4 4) z0 =

x− / n

5) Reject H0 if z0 > z where  = 0.05 and z0.05 = 1.65 6) x = 4.05 ,  = 0.2

z0 =

4.05 − 4 = 1.77 0.2 / 50

7) Because 1.77 > 1.65, reject the null hypothesis. Conclude that the true average battery life exceeds 4 hours at  = 0.05. b) P-value=1-  ( Z 0 ) =1-  (1.77 )  0.04

  

c)  =  z0.05 −

(4.5 − 4) 50   = (1.65 – 17.68) = (-16.03) = 0  0.2 

Power = 1- = 1-0 = 1

9-8

Applied Statistics and Probability for Engineers, 7h edition

(z + z )  2

d) n =





2

2

=

(z 0.05 + z 0.1 )2  2 (4.5 − 4) 2

=

(1.65 + 1.29) 2 (0.2) 2 = 1.38, (0.5) 2

n2 e) 95% Confidence Interval

   x − z0.05    n  0 .2  4.05 − 1.65   50  4.003   Because the lower limit of the CI is greater than 4, we conclude that average life is greater than 4 hours at  = 0.05.

9.2.10 a) The alternative hypothesis should be one-sided because the scientists are interested in high adhesion. b) 1) The parameter of interest is the true mean adhesion,  .

2) H0 :  = 2.5 3) H1 :

 > 2.5

4) z0 =

x − 0 / n

5) Reject H0 if z0  z where α = 0.05 and z0.05 = 1.65 6) x = 3.372 ,  = 0.66

3.372 − 2.5 = 2.95 0.66 / 5 7) Because 2.95 > 1.65, reject H0. The true mean adhesion is greater than 2.5 at α = 0.05. c) P-value = 1 −  (2.95) = 1 − 0.998 = 0.002 z0 =

Section 9-3

9.3.1

a) α = 0.01, n = 20, the critical value = 2.539 b) α = 0.05, n = 12, the critical value = 1.796 c) α = 0.1, n = 15, the critical value = 1.345

9.3.2

a) SE Mean =

S 0.717 = = 0.1603 N 20 92.379 − 91 t0 = = 8.6012 0.717 / 20 t 0 = 8.6012 with df = 20 – 1 = 19, so the P-value < 0.0005. Because the P-value <  = 0.05 we reject the

null hypothesis that  = 91at the 0.05 level of significance.

9-9

Applied Statistics and Probability for Engineers, 7h edition 95% lower CI of the mean is x − t 0.05,19 S   n

92.379 − (1.729)

92 .1018  

0.717  20

b) A one-sided test because the alternative hypothesis is  > 91. c) If the alternative hypothesis is changed to  > 90, then t 0 =

92.379 − 90 0.717 / 20

= 14.8385

t 0 = 14 .8385 with df = 20 – 1 = 19, so the P-value < 0.0005. The P-value <  = 0.05 and we reject the null hypothesis at the 0.05 level of significance. Hence the conclusion remains the same as in part a.

9.3.3

a) 2 * 0.025  p  2 * 0.05 then 0.05  p  0.1 b) 2 * 0.025  p  2 * 0.05 then 0.05  p  0.1 c) 2 * 0.25  p  2 * 0.4 then 0.5  p  0.8

9.3.4

a) degrees of freedom = N – 1 = 16 – 1 = 15

S 1.783 = = 0.4458 N 16 35.274 − 34 t0 = = 2.8581 1.783 / 16

b) SE Mean =

c) 0.01<p-value < 0.02.

We reject the null hypothesis if the P-value < . Thus, we can reject the null hypothesis at significance levels greater than p-value. d) If the alternative hypothesis is changed to the one-sided alternative  > 34, the 0.005<p-value<0.01. e) If the null hypothesis is changed to  = 34.5 versus the alternative hypothesis ( ≠ 34.5) the t statistic is reduced. In particular, t 0 =

35.274 − 34.5 1.783 / 16

= 1.7364 and t 0.025,15 = 2.131

Because t 0 = 1.7364  t 0.025,15 , we fail to reject the null hypothesis at the 0.05 level of significance

9.3.5

a) 1 − 0.05  p  1 − 0.025 then 0.95  p  0.975 b) 0.025  p  0.05 c) 1 − 0.4  p  1 − 0.25 then 0.6  p  0.75

9.3.6

a) 1) The parameter of interest is the true mean interior temperature life, .

2) H0 :  = 22.5 3) H1 :   22.5

9-10

Applied Statistics and Probability for Engineers, 7h edition 4) t0 =

x− s/ n

5) Reject H0 if |t0| > t/2,n-1 where  = 0.05 and t/2,n-1 = 2.776 for n = 5 6) x = 22.496 , s = 0.378, n = 5 (To=-0.02366) 22.496 − 22.5 𝑡0 = = −0.02366 0.378⁄√5 7) Because –0.02366 > –2.776, we fail to reject the null hypothesis. There is not sufficient evidence to conclude that the true mean interior temperature differs from 22.5 C at  = 0.05.

Also, 2(0.4) < P-value < 2(0.5). That is, 0.8 < P-value < 1.0. b) The points on the normal probability plot fall along a line. Therefore, the normality assumption is reasonable.

Normal Probability Plot for temp ML Estimates - 95% CI

99

ML Estimates

95

Mean

22.496

StDev

0.338384

90

Percent

80 70 60 50 40 30 20 10 5 1 21.5

22.5

23.5

Data

c) d =

 |  −  0 | | 22.75 − 22.5 | = = = 0.66   0.378

Using the OC curve, Chart VII e) for  = 0.05, d = 0.66, and n = 5, we obtain   0.8 and power of 1−0.8 = 0.2 d) d =

 |  −  0 | | 22.75 − 22.5 | = = = 0.66   0.378

Using the OC curve, Chart VII e) for  = 0.05, d = 0.66, and   0.1 (Power=0.9), n = 40 e) 95% two sided confidence interval

 s   s  x − t 0.025, 4      x + t 0.025, 4    n  n  0.378   0.378  22.496 − 2.776     22.496 + 2.776   5   5  22.027    22.965 We cannot conclude that the mean interior temperature differs from 22.5 at  = 0.05 because the value is included in the confidence interval.

9-11

Applied Statistics and Probability for Engineers, 7h edition 9.3.7

a) 1) The parameter of interest is the true mean of body weight, .

2) H0:  = 300 3) H1:   300 4) t0 =

x− s/ n

5) Reject H0 if |t0| > t/2,n-1 where  = 0.05 and t/2,n-1 = 2.056 for n = 27 6) x = 325.496 , s = 198.786, n = 27

t0 =

325.496 − 300 = 0.6665 198.786 / 27

7) Because 0.6665 < 2.056 we fail to reject the null hypothesis. There is not sufficient evidence to conclude that the true mean body weight differs from 300 at  = 0.05. We have 2(0.25) < P-value < 2(0.4). That is, 0.5 < P-value < 0.8 b) We reject the null hypothesis if P-value < . The P-value = 2(0.2554) = 0.5108. Therefore, the smallest level of significance at which we can reject the null hypothesis is approximately 0.51. c) 95% two sided confidence interval

 s   s  x − t0.025, 26      x + t0.025, 26    n  n  198.786   198.786  325.496 − 2.056     325.496 + 2.056  27  27    246.8409    404.1511 We fail to reject the null hypothesis because the hypothesized value of 300 is included within the confidence interval.

9.3.8

a) 1) The parameter of interest is the true mean rainfall, .

2) H0 :  = 25 3) H1 :   25 4) t0 =

x−

s/ n 5) Reject H0 if t0 > t,n-1 where  = 0.01 and t0.01,19 = 2.539 for n = 20 6) x = 26.04 s = 4.78 n = 20 t0 =

26.04 − 25 4.78 / 20

= 0.97

7) Because 0.97 < 2.539 fail to reject the null hypothesis. There is insufficient evidence to conclude that the true mean rainfall is greater than 25 acre-feet at  = 0.01. The 0.10 < P-value < 0.25. b) The data on the normal probability plot falls along a line. Therefore, the normality assumption is reasonable.

9-12

Applied Statistics and Probability for Engineers, 7h edition

c) d =

 |  −  0 | | 27 − 25 | = = = 0.42   4.78

Using the OC curve, Chart VII h) for  = 0.01, d = 0.42, and n = 20, obtain   0.7 and power of 1 − 0.7 = 0.3.

d) d =

 |  −  0 | | 27.5 − 25 | = = = 0.52   4.78

Using the OC curve, Chart VII h) for  = 0.01, d = 0.52, and   0.1 (Power=0.9), n = 75 e) 99% lower confidence bound on the mean diameter

 s  x − t0.01,19    n  4.78  26.04 − 2.539    20  23.326   Because the lower limit of the CI is less than 25 there is insufficient evidence to conclude that the true mean rainfall is greater than 25 acre-feet at  = 0.01.

9.3.9

a) 1) The parameter of interest is the true mean female body temperature, .

2) H0 :  = 98.6 3) H1 :   98.6 4) t0 =

x− s/ n

5) Reject H0 if |t0| > t/2,n-1 where  = 0.05 and t/2,n-1 = 2.064 for n = 25 6) x = 98 .264 , s = 0.4821, n = 25

t0 =

98.264 − 98.6 = −3.48 0.4821 / 25

9-13

Applied Statistics and Probability for Engineers, 7h edition 7) Because 3.48 > 2.064, reject the null hypothesis. Conclude that the true mean female body temperature differs from 98.6 F at  = 0.05.

P-value = 2(0.001) = 0.002 b) The data on the normal probability plot falls along a line. The normality assumption is reasonable.

Probability Plot of Ex9-59 Normal - 95% CI 99

Mean 98.26 StDev 0.4821 N 25 AD 0.238 P-Value 0.759

95 90

Percent

80 70 60 50 40 30 20 10 5 1

97.0

c) d =

97.5

98.0

98.5 Ex9-59

99.0

99.5

100.0

 |  − 0 | | 98 − 98.6 | = = = 1.24   0.4821

Using the OC curve, Chart VIIe for  = 0.05, d = 1.24, and n = 25, obtain   0 and power of 1 − 0  1 d) d =

 |  − 0 | | 98.2 − 98.6 | = = = 0.83   0.4821

Using the OC curve, Chart VIIe for  = 0.05, d = 0.83, and   0.1 (Power=0.9), n =20 e) 95% two sided confidence interval

 s   s  x − t0.025, 24      x + t0.025, 24    n  n  0.4821   0.4821  98.264 − 2.064     98.264 + 2.064   25   25  98.065    98.463 We conclude that the mean female body temperature differs from 98.6 at  = 0.05 because the value is not included inside the confidence interval.

9.3.10

a) 1) The parameter of interest is the true mean sodium content, .

2) H0 :  = 300 3) H1 :  > 300

9-14

Applied Statistics and Probability for Engineers, 7h edition 4) t0 =

x− s/ n

5) Reject H0 if t0 > t,n-1 where  = 0.05 and t,n-1 = 1.943 for n = 7 6) x = 315 , s = 16 n=7

t0 =

315 − 300 = 2.48 16 / 7

7) Because 2.48 > 1.943 reject the null hypothesis and conclude that the leg strength exceeds 300 watts at  = 0.05.

The P-value is between 0.01 and 0.025 b) d =

 |  − 0 | | 305 − 300 | = = = 0.3125   16

Using the OC curve, Chart VII g) for  = 0.05, d = 0.3125, and n = 7,   0.9 and power = 1−0.9 = 0.1. c) If 1 -  > 0.9 then  < 0.1 and n is approximately 100

 s   = 303.2 <   n

d) Lower confidence bound is x − t ,n−1 

Because 300 is not include in the interval, reject the null hypothesis 9.3.11

a) 1) The parameter of interest is the true mean height of female engineering students, .

2) H0 :  = 65 3) H1 :  > 65 4) t0 =

x−

s/ n 5) Reject H0 if t0 > t,n-1 where  = 0.05 and t0.05,36 =1.68 for n = 37 6) x = 65.811 inches s = 2.106 inches n = 37 t0 =

65.811 − 65 2.11 / 37

= 2.34

7) Because 2.34 > 1.68 reject the null hypothesis. There is sufficient evidence to conclude that the true mean height of female engineering students is greater than 65 at  = 0.05. We obtain 0.01 < P-value < 0.025. b) From the normal probability plot, the normality assumption seems reasonable:

9-15

Applied Statistics and Probability for Engineers, 7h edition

Probability Plot of Female heights Normal

99

95 90

Percent

80 70 60 50 40 30 20 10 5

1

60

c) d =

62

68 − 65 2.11

64

66 Female heights

68

70

72

= 1.42 , n=37

From the OC Chart VII g) for  = 0.05, we obtain   0. Therefore, the power  1. d) d =

66 − 65 2.11

= 0.47

From the OC Chart VIIg) for  = 0.05 and   0.2 (power = 0.8). Therefore, n  30 .

9.3.12

a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true mean concentration of suspended solids, .

2) H0 :  = 55 3) H1 :   55 4) t0 =

x−

s/ n 5) Reject H0 if |t0 | > t/2,n-1 where  = 0.05 and t0.025,59 =2.000 for n = 60 6) x = 59.87 s = 12.50 n = 60 t0 =

59.87 − 55

12.50 / 60

= 3.018

7) Because 3.018 > 2.000, reject the null hypothesis. There is sufficient evidence to conclude that the true mean concentration of suspended solids is not equal to 55 at  = 0.05.

From Table V the t0 value is between the values of 0.001 and 0.0025 with 59 degrees of freedom.

Therefore, 2(0.001) < P-value < 2( 0.0025) and 0.002 < P-value < 0.005. Computer software generates a Pvalue = 0.0038. b) The data tend to fall along a line. The normality assumption seems reasonable.

9-16

Applied Statistics and Probability for Engineers, 7h edition

Probability Plot of Concentration of solids Normal

99.9 99 95

Percent

90 80 70 60 50 40 30 20 10 5 1 0.1

c) d =

20

30

50 − 55 12.50

40

50 60 70 Concentration of solids

80

90

100

= 0.4 , n = 60 so, from the OC Chart VII e) for  = 0.05, d= 0.4 and n = 60 obtain  

0.2. Therefore, the power = 1 - 0.2 = 0.8 d) From the same OC chart, and for the specified power, we would need approximately 75 observations.

d=

50 − 55 12.50

= 0.4

Using the OC Chart VII e) for  = 0.05, d = 0.4, and   0.10 so that the power = 0.90, n = 75 9.3.13 a) 95% two sided confidence interval

 s   s  x − t0.025,99      x + t0.025,99    n  n

 79.01   79.01  299,852.4 − 1.984     299,852.4 + 1.984   100   100  299,836.7244    299,868.0756 b) Michelson's measurements do not seem to be accurate because the true value of 299,734.5 is not included in the confidence interval at α = 0.05.

9-17

Applied Statistics and Probability for Engineers, 7h edition Section 9-4 9.4.1

a) α = 0.01, n = 20, from Table V we find the following critical values 6.84 and 38.58 b) α = 0.05, n = 12, from Table V we find the following critical values 3.82 and 21.92 c) α = 0.10, n = 15, from Table V we find the following critical values 6.57 and 23.68

9.4.2

a) 0.1 < 1-P < 0.5 then 0.5 < P-value < 0.9 b) 0.1 < 1-P< 0.5 then 0.5 < P-value < 0.9 c) 0.99 <1-P <0.995 then 0.005 < P-value < 0.01

9.4.3 a) In order to use the 2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true standard deviation of performance time . However, the solution can be found by performing a hypothesis test on 2.

2) H0 : 2 = 0.752 3) H1 : 2 > 0.752 4)  20 =

( n − 1)s2 2

5) Reject H0 if  0   ,n−1 where  = 0.05 and  0.05,16 = 26 .30 2

2

2

6) n = 17, s = 0.09  20 =

(n − 1) s 2

2

=

16(0.09) 2 = 0.23 .75 2

7) Because 0.23 < 26.30, fail to reject H0. There is insufficient evidence to conclude that the true variance of performance time content exceeds 0.752 at  = 0.05. Because  20 = 0.23, the P-value > 0.995 b) The 95% one sided confidence interval given below includes the value 0.75. Therefore, we are not be able to conclude that the standard deviation is greater than 0.75.

16(.09) 2 2 26.3 0.07   9.4.4 a) In order to use the 2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true standard deviation of titanium percentage, . However, the solution can be found by performing a hypothesis test on 2.

2) H0 : 2 = (0.25)2 3) H1 : 2  (0.25)2 4)  20 =

( n − 1)s2 2

5) Reject H0 if 20  12−  / 2,n −1 where  = 0.05 and  20.995,50 = 32.36 or 20  2 ,2,n −1 where  = 0.05 and  20.005,50 = 71.42 for n = 51

6) n = 51, s = 0.37  20 =

( n − 1)s2 2

=

50(0.37) 2 (0.25) 2

= 109.52

9-18

Applied Statistics and Probability for Engineers, 7h edition 7) Because 109.52 > 71.42 reject H0. The standard deviation of titanium percentage is significantly different from 0.25 at  = 0.01. P-value/2 < 0.005, then P-value < 0.01 b) 95% confidence interval for : First find the confidence interval for 2 : For  = 0.05 and n = 51, 2 / 2 , n−1 =  20.025,50 = 71.42 and 12−  / 2,n −1 =  20.975,50 = 32.36

50(0.37) 2 50(0.37) 2 2  71.42 32.36

0.096  2  0.2115 Taking the square root of the endpoints of this interval we obtain, 0.31 <  < 0.46 Because 0.25 falls below the lower confidence bound we conclude that the population standard deviation is not equal to 0.25.

9.4.5 a) In order to use the 2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the standard deviation of tire life, . However, the answer can be found by performing a hypothesis test on 2.

2) H0 : 2 = 40002 3) H1 : 2 < 40002 4)  20 =

(n − 1) s 2

2

5) Reject H0 if  0  1− ,n−1 where  = 0.05 and  0.95,15 = 7.26 for n = 16 2

2

2

6) n = 16, s2 = (3645.94)2  20 =

(n − 1) s 2

2

=

15(3645.94) 2 = 12.46 40002

7) Because 12.46 > 7.26 fail to reject H0. There is not sufficient evidence to conclude the true standard deviation of tire life is less than 4000 km at  = 0.05.

P-value = P(2 <12.46) for 15 degrees of freedom. Thus, 0.5 < 1-P-value < 0.9 and 0.1 < P-value < 0.5 b) The 95% one sided confidence interval below includes the value 4000. Therefore, we are not able to conclude that the variance is less than 40002.

15(3645.94) 2 = 27464625 7.26   5240

2 

Section 9-5 9.5.1

a) A two-sided test because the alternative hypothesis is p not = 0.4

X 98 = = 0.3564 N 275 x − np0 98 − 275(0.4) z0 = = = −1.4771 np0 (1 − p0 ) 275(0.4)(0.6)

b) sample p =

9-19

Applied Statistics and Probability for Engineers, 7h edition P-value = 2(1 – (1.4771)) = 2(1 – 0.9302) = 0.1396 c) The normal approximation is appropriate because np > 5 and n(1 - p) > 5.

9.5.2

a) 1) The parameter of interest is the true fraction of satisfied customers.

2) H0 : p = 0.9 3) H1 : p  0.9 4) z 0 =

x − np0

z0 =

or

np0 (1 − p0 )

pˆ − p 0

p 0 (1 − p 0 ) n

;

Either approach will yield the same conclusion 5) Reject H0 if z0 < − z/2 where  = 0.05 and −z/2 = −z0.025 = −1.96 or z0 > z/2 where  = 0.05 and z/2 = z0.025 = 1.96

850 = 0.85 1000 x − np0 850 − 1000(0.9) z0 = = = −5.27 np0 (1 − p0 ) 1000(0.9)(0.1)

6) x = 850 n = 1000

pˆ =

7) Because -5.27 < -1.96 reject the null hypothesis and conclude the true fraction of satisfied customers differs from 0.9 at  = 0.05.

The P-value: 2(1-(5.27)) ≤ 2(1-1) ≈ 0 b) The 95% confidence interval for the fraction of surveyed customers is:

pˆ − z / 2 .85 − 1.96

pˆ (1 − pˆ )  p  pˆ + z / 2 n

pˆ (1 − pˆ ) n

0.85(0.15) 0.85(0.15)  p  .85 + 1.96 1000 1000 0.827  p  0.87

Because 0.9 is not included in the confidence interval, reject the null hypothesis at  = 0.05.

9.5.3

a) 1) The parameter of interest is the true success rate 2) H0 : p = 0.78 3) H1 : p > 0.78 4) z 0 =

x − np0

np0 (1 − p0 )

or

z0 =

pˆ − p 0

p 0 (1 − p 0 ) n

; Either approach will yield the same conclusion

5) Reject H0 if z0 > z . Since the value for  is not given. We assume  = 0.05 and z = z0.05 = 1.65

289  0.83 350 x − np0 289 − 350(0.78) = = 2.06 np0 (1 − p0 ) 350(0.78)(0.22)

ˆ= 6) x = 289 n = 350 p

z0 =

9-20

Applied Statistics and Probability for Engineers, 7h edition 7) Because 2.06 > 1.65 reject the null hypothesis and conclude the true success rate is greater than 0.78, at  = 0.05.

P-value = 1 - 0.9803 = 0.0197 b) The 95% lower confidence interval:

pˆ (1 − pˆ ) p n

pˆ − z .83 − 1.65

0.83(0.17) p 350 0.7969  p

Because the hypothesized value is not in the confidence interval (0.78 < 0.7969), reject the null hypothesis.

9.5.4

a) 1) The parameter of interest is the true proportion of engineering students planning graduate studies 2) H0 : p = 0.50 3) H1 : p  0.50

x − np0

4) z 0 =

np0 (1 − p0 )

or

z0 =

pˆ − p 0

p 0 (1 − p 0 ) n

; Either approach will yield the same conclusion

5) Reject H0 if z0 < − z/2 where  = 0.05 and −z/2 = −z0.025 = −1.96 or z0 > z/2 where  = 0.05 and z/2 = z0.025 = 1.96 6) x = 117 n = 484 117 𝑝̂ = = 0.2417 484

z0 =

x − np0

np0 (1 − p0 )

=

117 − 484(0.5) 484(0.5)(0.5)

= −11.36

7) Because −11.36 > −1.65 reject the null hypothesis and conclude that the true proportion of engineering students planning graduate studies differs from 0.5, at  = 0.05.

P-value = 2[1 − (11.36)]  0 b) 𝑝̂ =

117 484

= 0.2417  0.242

pˆ − z / 2 0.242 − 1.96

pˆ (1 − pˆ )  p  pˆ + z / 2 n

pˆ (1 − pˆ ) n

0.242(0.758) 0.242(0.758)  p  0.242 − 1.96 484 484 0.204  p  0.280

Because the 95% confidence interval does not contain the value 0.5 we conclude that the true proportion of engineering students planning graduate studies differs from 0.5.

9-21

Applied Statistics and Probability for Engineers, 7h edition 9.5.5

The problem statement implies that H0: p = 0.6, H1: p > 0.6 and defines an acceptance region as 315 p  = 0.63 and rejection region as p  0.63 500

a) The probability of a type 1 error is     .

0.63 − 0.6    = P( pˆ  0.63 | p = 0.6 ) = P Z  = P(Z  1.37 ) = 1 − P ( Z  1.37) = 0.08535 0 . 6 ( 0 . 4)    500  

b)  = P( P  0.63 | p = 0.75) = P(Z  −6.196) = 0.

9.5.6

1) The parameter of interest is the true percentage of customers received fully charged batteries, p.

2) H0 : p = 0.85 3) H1 : p > 0.85 4)

x − np0 z0 = np0 (1 − p0 )

z0 =

pˆ − p0 p0 (1 − p0 ) n

or Either approach will yield the same conclusion 5) Reject H0 if z0 > z where  = 0.05 and

z0.05 = 1.65

6) x = 96 n = 100

96 = 0.96 100 pˆ − p0 z0 = = p0 (1 − p0 ) n pˆ =

0.96 − 0.85 = 3.08 0.85(1 − 0.85) 100

7) Because 3.08 > 1.65 reject the null hypothesis. There is enough evidence to conclude that the proportion of customers received fully charged batteries is at least as high as the previous model.

9-22

Applied Statistics and Probability for Engineers, 7h edition Section 9-7

9.7.1 Under the null hypothesis there are 50 observations from a binomial distribution with n = 6 and p = 0.25.

Use the binomial distribution to obtain the expected frequencies from the 50 observations.

Value Observed Expected

0 4 8.8989

1 21 17.7979

2 10 14.8315

3 13 6.5918

4 or more 2 1.8799

The expected frequency for cell “4 or more” is less than 3. Combine this cell with its neighboring cell to obtain the following table.

Value

0

1

2

Observed Expected

4 8.8989

21 17.7979

10 14.8315

3 or more 15 8.4717

The degrees of freedom are k − p − 1 = 4 − 0 − 1 = 3 a) 1) The variable of interest is the form of the distribution for the random variable X. 2) H0: The form of the distribution is binomial with n = 6 and p = 0.25 3) H1: The form of the distribution is not binomial with n = 6 and p = 0.25 4) The test statistic is k

( O i − E i )2

i =1

Ei

20 = 

5) Reject H0 if 2o  20.05,3 = 7.81 for  = 0.05 6)

 02 =

(4 − 8.8989)2 +  + (15 − 8.2397)2 = 9.8776

8.8989 8.2397 7) Because 9.8776 > 7.81 reject H0. We conclude that the distribution is not binomial with n = 6 and p = 0.25 at  = 0.05.

0.02<p-value<0.05 b) P-value = 0.0197 (from computer software)

9.7.2

The expected frequency is found from the Poisson distribution P( X = x) =

e − x with  = 1.2 x!

Value

0

1

2

3

Observed Frequency Expected Frequency

24

30

31

11

4 or more 4

30.12

36.14

21.69

8.67

3.37

Because the expected frequencies of all categories are greater than 3, there is no need to combine categories.

9-23

Applied Statistics and Probability for Engineers, 7h edition

The degrees of freedom are k − p − 1 = 5 − 0 − 1 = 4 a) 1) Interest is on the form of the distribution for X. 2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) The test statistic is k

(Oi − Ei )2

i =1

Ei

 02 =  5) Reject H0 if  o2   02.05, 4 = 9.49 for  = 0.05 6)

 02 =

(24 − 30.12)2 + (30 − 36.14)2 + (31 − 21.69)2 + (11 − 8.67)2 + (4 − 3.37)2 = 7.0267 30.12

36.14

21.69

8.67

3.37

7) Because 7.0267 < 9.49 fail to reject H0. b) The P-value is between 0.1 and 0.2 using Table IV. From computer software, the P-value = 0.1345

9.7.3 The estimated mean = 49.6741. Based on a Poisson distribution with  = 49.674 the expected frequencies are shown in the following table. All expected frequencies are greater than 3.

The degrees of freedom are k − p − 1 = 26 − 1 − 1 = 24

Vechicles per minute 40 or less 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61

Frequency 14 24 57 111 194 256 296 378 250 185 171 150 110 102 96 90 81 73 64 61 59 50

9-24

Expected Frequency 277.6847033 82.66977895 97.77492539 112.9507307 127.5164976 140.7614945 152.0043599 160.6527611 166.2558608 168.5430665 167.4445028 163.091274 155.7963895 146.0197251 134.3221931 121.3151646 107.6111003 93.78043085 80.31825 67.62265733 55.98491071 45.5901648

Applied Statistics and Probability for Engineers, 7h edition 62 63 64 65 or more

42 29 18 15

36.52661944 28.80042773 22.35367698 62.60833394

a) 1) Interest is the form of the distribution for the number of cars passing through the intersection.

2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) The test statistic is k

20 = 

( O i − E i )2 Ei

i =1 2 2 5) Reject H0 if  o   0.05,24 = 36.42 for  = 0.05

6) Estimated mean = 49.6741

 02 = 1012 .8044 7) Because 1012.804351 >> 36.42, reject H0. We can conclude that the distribution is not a Poisson distribution at  = 0.05. b) P-value ≈ 0 (from computer software) 9.7.4

Estimated mean = 10.131

Value Rel. Freq Observe d (Days) Expecte d (Days)

5 or less 0.067

6

7

8

9

10

11

12

13

14

15 or more

0.067

0

0.100

0.133

0.200

0.133

0.133

0.067

0.033

0.067

2

2

0

3

4

6

4

4

2

1

2

1.8687

1.794 2

2.596 7

3.288 4

3.701 6

3.750 1

3.453 8

2.915 9

2.272 4

1.644 4

2.7138

Because there are several cells with expected frequencies less than 3, a revised table follows. We note that there are other reasonable alternatives to combine cells. For example, cell 7 could also be combined with cell 8.

Value Observed (Days) Expected (Days)

7 or less 4

8

9

10

11

3

4

6

4

12 or more 9

6.2596

3.2884

3.7016

3.7501

3.4538

9.5465

The degrees of freedom are k − p − 1 = 6 − 1 − 1 = 4 a) 1) Interest is on the form of the distribution for the number of calls arriving to a switchboard from noon to 1pm during business days.

2) H0: The form of the distribution is Poisson

9-25

Applied Statistics and Probability for Engineers, 7h edition 3) H1: The form of the distribution is not Poisson 4) The test statistic is k

( O i − E i )2

i =1

Ei

20 = 

5) Reject H0 if    2 o

2 0.05, 4

= 9.49 for  = 0.05

6) 𝜒0 2 =

(4−6.2596)2 6.2596

+ ⋯+

(9−9.5465)2 9.5465

= 2.4040

7) Because 2.4040 < 9.49, fail to reject H0. b) The P-value is between 0.5 and 0.9 using Table IV. P-value = 0.685 (from computer software)

Section 9-8 9.8.1 1) Interest is on the species distribution.

2) H0: Species distribution is independent of year.

3) H1: Species distribution is not independent of year.

4) The test statistic is: r

c

 =  2 0

(O − E )

2

ij

ij

Eij

i =1 j =1

5) The critical value is  0.05 , 4 = 9.488 for  = 0.05.

2

6) The calculated test statistic is  0 = 146 .3648 2

7) Because  0   0.05, 4 , reject H0. The evidence is sufficient to claim that species distribution is not independent of 2

2

year at  = 0.05. P-value = P (  0  146 .3648 ) ≈ 0 2

9.8.2 1) Interest is on the survival distrbution.

2) H0: Survival is independent of ticket class.

3) H1: Survival is not independent of ticket class.

4) The test statistic is: r

c

 =  2 0

i =1 j =1

(O − E )

2

ij

ij

Eij

5) The critical value is  0.05 , 3 = 7.815 for  = 0.05.

2

6) The calculated test statistic is  0 = 187 .7932 2

7) Because  0   0.05, 3 , reject H0. The evidence is sufficient to claim that survival is not independent of ticket class 2

2

at  = 0.05. P-value = P (  0  187 .7932 ) ≈ 0 2

9.8.3

1) Interest is on the distribution of breakdowns among shift.

9-26

Applied Statistics and Probability for Engineers, 7h edition 2) H0: Breakdowns are independent of shift.

3) H1: Breakdowns are not independent of shift.

4) The test statistic is: r

c

 =  2 0

(O − E )

2

ij

ij

Eij

i =1 j =1

5) The critical value is  .05 , 6 = 12 .592 for  = 0.05 2

6) The calculated test statistic is  0 = 11 .65 2

7) Because  20   20.05,6 fail to reject H0. The evidence is not sufficient to claim that machine breakdown and shift are dependent at  = 0.05. P-value = P (  0  11 .65) = 0.070 (from computer software) 2

9.8.4

1) Interest is on the distribution of calls by surgical-medical patients.

2) H0: Calls by surgical-medical patients are independent of Medicare status.

3) H1: Calls by surgical-medical patients are not independent of Medicare status.

4) The test statistic is: r

c

 =  2 0

(O − E )

2

ij

i =1 j =1

ij

Eij

5) The critical value is  .01,1 = 6.637 for  = 0.01 2

6) The calculated test statistic is  0 = 0.033 2

7) Because  0    0.01,1 fail to reject H0. The evidence is not sufficient to claim that surgical-medical patients and 2

2

Medicare status are dependent. P-value = P (  0  0.033 ) = 0.85 2

9.8.5

1) Interest is on the distribution of statistics and OR grades.

2) H0: Statistics grades are independent of OR grades.

3) H1: Statistics and OR grades are not independent.

4) The test statistic is: r

c

 =  2 0

(O − E )

2

ij

ij

Eij

i =1 j =1

5) The critical value is  .01, 9 = 21 .665 for  = 0.01 2

6) The calculated test statistic is  0 = 25 .55 2

7)  0   0.01,9 Therefore, reject H0 and conclude that the grades are not independent at  = 0.01.

2

2

P-value = P (  0  25 .55 ) = 0.002 (from computer software) 2

9.8.6

1) Interest is on the distribution of deflections.

2) H0: Deflection and range are independent.

3) H1: Deflection and range are not independent.

4) The test statistic is: r

c

 =  2 0

i =1 j =1

(O − E )

2

ij

ij

Eij

5) The critical value is  0.05 , 4 = 9.488 for  = 0.05 2

9-27

Applied Statistics and Probability for Engineers, 7h edition 6) The calculated test statistic is  0 = 2.46 2

7) Because  0    0.05, 4 fail to reject H0. The evidence is not sufficient to claim that the data are dependent at  = 2

2

0.05. The P-value = P (  0  2.46 ) = 0.652 (from computer software).

2

9.8.7

1) Interest is on the distribution of failures of an electronic component.

2) H0: Type of failure is independent of mounting position.

3) H1: Type of failure is not independent of mounting position.

4) The test statistic is: r

c

 =  2 0

(O − E )

2

ij

ij

Eij

i =1 j =1

5) The critical value is  0.01, 3 = 11 .344 for  = 0.01 2

6) The calculated test statistic is  0 = 10 .71 2

7) Because  0    0.01,3 fail to reject H0. The evidence is not sufficient to claim that the type of failure is dependent 2

2

on the mounting position at  = 0.01. P-value = P (  0  10 .71) = 0.013 (from computer software).

2

9.8.8

1) Interest is on the distribution of opinion on core curriculum change.

2) H0: Opinion of the change is independent of the class standing.

3) H1: Opinion of the change is not independent of the class standing.

4) The test statistic is: r

c

 =  2 0

(O − E )

2

ij

ij

Eij

i =1 j =1

5) The critical value is  0.05 , 3 = 7.815 for  = 0.05 2

6) The calculated test statistic is  .0 = 26 .97 .

2

7)  0   0.05, 3 , reject H0 and conclude that opinion on the change and class standing are not independent. P2

2

value = P (  0  26 .97 )  0 2

9.8.9

a) 1) Interest if on the distribution of successes.

2) H0: successes are independent of size of stone.

3) H1: successes are not independent of size of stone.

4) The test statistic is: r

c

 =  2 0

(O − E )

i =1 j =1

2

ij

ij

Eij

5) The critical value is  .05 ,1 = 3.84 for  = 0.05 2

6) The calculated test statistic  0 = 13.766 with details below.

2

7)  0   0.05,1 , reject H0 and conclude that the number of successes and the stone size are not independent.

2

1 2 All

2

1 55 66.06 234 222.94 289 289.00

2 25 13.94 36 47.06 61 61.00

All 80 80.00 270 270.00 350 350.00

9-28

Applied Statistics and Probability for Engineers, 7h edition Cell Contents:

Count Expected count Pearson Chi-Square = 13.766, DF = 1, P-Value = 0.000 b) P-value = P(  0  13 .766 ) < 0.005 2

Section 9-9

9.9.1

a) 1) The parameter of interest is median titanium content.

2) H 0 : ~ = 8.5 3) H : ~  8.5 1

4) The test statistic is the observed number of plus differences or r+ = 7 for  = 0.05.

5) Reject H0 if the P-value corresponding to r+ = 7 is less than or equal to  = 0.05.

6) Using the binomial distribution with n = 20 and p = 0.5, P-value = 2P( R*  7 | p = 0.5) = 0.1315 7) We fail to reject H0. There is not enough evidence to conclude that the median of the titanium content differs from 8.5. b) 1) Parameter of interest is the median titanium content

2) H 0 : ~ = 8.5 3) H : ~  8.5 1

4) Test statistic is z0 =

r + − 0.5n 0.5 n

5) Reject H0 if the |Z0| > Z0.025 = 1.96 for =0.05 6) Computation: z = 7 − 0.5(20) = −1.34 0

0.5 20

7) We fail to reject H0. There is not enough evidence to conclude that the median titanium content differs from 8.5. The P-value = 2P( Z < -1.34) = 0.1802.

9.9.2

a) 1) The parameter of interest is the median of pH.

2) H 0 : ~ = 7.0 3) H : ~  7.0 1

4) The test statistic is the observed number of plus differences or r+ = 8 for  = 0.05.

5) Reject H0 if the P-value corresponding to r+ = 8 is less than or equal to  = 0.05.

6) Using the binomial distribution with n = 10 and p = 0.5, the P-value = 2P(R+  8 | p = 0.5) = 0.1 7) We fail to reject H0. There is not enough evidence to conclude that the median of the pH differs from7.0. b) 1) The parameter of interest is median of pH.

2) H 0 : ~ = 7.0 3) H : ~  7.0 1

4) The test statistic is z0 =

r * − 0.5n 0.5 n

5) Reject H0 if |Z0| > 1.96 for =0.05.

r * − 0.5n 8 − 0.5(10) = = 1.90 6) r*=8 and z0 = 0.5 n 0.5 10 7) Fail to reject H0. There is not enough evidence to conclude that the median of the pH differs from 7.0.

P-value = 2[1 - P(|Z0| < 1.90)] = 2(0.0287) = 0.0574

9-29

Applied Statistics and Probability for Engineers, 7h edition

9.9.3

a) 1) The parameter of interest is the mean ball diameter 2) H0:  0 = 0.265

3) H0:  0  0.265 4) w = min(w+, w-) 5) Reject H0 if w  w0.05,n =9 = 5 for  = 0.05 *

6) Usually zeros are dropped from the ranking and the sample size is reduced. The sum of the positive ranks is w+ = (1+4.5+4.5+4.5+4.5+8.5+8.5) = 36. The sum of the negative ranks is w - = (4.5+4.5) = 9. Therefore, w = min(36, 9) = 9. observation 1 6 9 3 2 4 5 7 8 12 10 11

Difference xi - 0.265 0 0 0 0.001 -0.002 0.002 0.002 0.002 0.002 -0.002 0.003 0.003

Signed Rank 1 -4.5 4.5 4.5 4.5 4.5 -4.5 8.5 8.5

7) Conclusion: because w- = 9 is not less than or equal to the critical value w0.05,n =9 = 5 , we fail to reject the null *

hypothesis that the mean ball diameter is 0.265 at the 0.05 level of significance. b)

Z0 =

W + − n(n + 1) / 4 36 − 9(10) / 4 = = 1.5993 n(n + 1)(2n + 1) / 24 9(10)(19) / 24

and Z0.025 = 1.96. Because Z0 = 1.5993< Z0.025 = 1.96 we fail to reject the null hypothesis that the mean ball diameter is 0.265 at the 0.05 level of significance. Also, the P-value = 2[1 – P(Z0 < 1.5993)] = 0.1098.

9.9.4

a) 1) Parameter of interest is the median impurity level.

2) H 0 : ~ = 2.5 3) H : ~  2.5 1

4) The test statistic is the observed number of plus differences or r+ = 2 for  = 0.05.

5) Reject H0 if the P-value corresponding to r+ = 2 is less than or equal to  = 0.05.

6) Using the binomial distribution with n = 22 and p = 0.5, the P-value = P(R+  2 | p = 0.5) = 0.0002 7) Conclusion, reject H0. The data supports the claim that the median is impurity level is less than 2.5. b) 1) Parameter of interest is the median impurity level

2) H 0 : ~ = 2.5 3) H : ~  2.5 1

4) Test statistic is z0 =

r + − 0.5n 0.5 n

5) Reject H0 if the Z0 < Z0.05 = -1.65 for =0.05

9-30

Applied Statistics and Probability for Engineers, 7h edition 6) Computation: z = 2 − 0.5(22) = −3.84 0

0.5 22

7) Reject H0 and conclude that the median impurity level is less than 2.5.

The P-value = P(Z < -3.84) = 0.000062

9.9.5

1) The parameter of interest is the mean dying time of the primer 2) H0:  0 = 1.5 3) H0:  0  1.5 4) w5) Reject H0 if w

−

 w0*.05,n =17 = 41 for  = 0.05

6) The sum of the positive rank is w+ = (4+4+4+4+4+9.5+9.5+13.5+13.5+13.5+13.5+16.5+16.5) = 126. The sum of the negative rank is w- = (4+4+9.5+9.5) = 27.

Observation 1.5 1.5 1.5 1.6 1.6 1.6 1.4 1.6 1.4 1.6 1.3 1.7 1.7 1.3 1.8 1.8 1.8 1.8 1.9 1.9

Difference xi - 1.5 0 0 0 0.1 0.1 0.1 -0.1 0.1 -0.1 0.1 -0.2 0.2 0.2 -0.2 0.3 0.3 0.3 0.3 0.4 0.4

Sign Rank 4 4 4 -4 4 -4 4 -9.5 9.5 9.5 -9.5 13.5 13.5 13.5 13.5 16.5 16.5

7) Conclusion: Because w- = 27 is less than the critical value w0.05,n =17 = 41 , we reject the null hypothesis that the *

mean dying time of the primer exceeds 1.5.

9.9.6

1) The parameter of interest is mean hardness 2) H0:  0 = 60 3) H0:  0  60 4) w5) Reject H0 if w

−

 w0*.05,n =7 = 3 for  = 0.05

6) The sum of the positive rank is w+ = (3.5+5.5) = 9. The sum of the negative rank is w- = (1+2+3.5+5.5+7) = 19. observation

Difference xi - 60

9-31

Sign Rank

Applied Statistics and Probability for Engineers, 7h edition 4 8 3 1 6 2 5 7

0 -1 -2 3 -3 5 -5 -7

-1 -2 3.5 -3.5 5.5 -5.5 -7

7) Conclusion: Because w- = 19 is not less than or equal to the critical value w0.05,n =7 = 3 , we fail to reject the null *

hypothesis that the mean hardness reading is greater than 60.

Section 9-10

9.10.1 a) 1) The parameter of interest is the mean molecular weight of a raw material from a new supplier, .

2, 3) The null and alternative hypotheses that must be tested are as follows (δ=50): H0 :  = 3550 and H0 :  = 3450 H1 :  ≤ 3550 H1 :  ≥ 3450 b) Test the first hypothesis (H0 :  = 3550 vs. H1 :  ≤ 3550): 4) The test statistic is:

t0 =

x − 0 s/ n

5) Reject H0 if t0 < -t,n-1 where t0.05,9 = 1.833 for n = 10 and  = 0.05.

6) x = 3550 , s = 25 , n = 10

t0 =

3550 − 3550 =0 25 / 10

7) Because 0 > -1.833, fail to reject H0.

Test the second hypothesis (H0 :  = 3450 vs. H1 :  ≥ 3450): 4) The test statistic is:

t0 =

x − 0 s/ n

5) Reject H0 if t0 > t,n-1 where t0.05,9 = 1.833 for n = 10 and  = 0.05.

6) x = 3550 , s = 25 , n = 10

t0 =

3550 − 3450 = 12.65 25 / 10

7) Because 12.65 > 1.833, reject H0.

9-32

Applied Statistics and Probability for Engineers, 7h edition

There is enough evidence to conclude that the mean molecular weight is greater than 3450. However, there is not enough evidence to conclude that it is less than 3550. As a result, we cannot conclude that the new supplier provides a molecular weight that is equivalent to the current one at  = 0.05.

9.10.2 a) 1) The parameter of interest is the mean absorption rate of the new product, .

2, 3) The null and alternative hypotheses that must be tested are as follows (δ = 0.50): H0 :  = 18.50 and H0 :  = 17.50 H1 :  ≤ 18.50 H1 :  ≥ 17.50 b) Test the first hypothesis (H0 :  = 18.50 vs. H1 :  ≤ 18.50): 4) The test statistic is:

t0 =

x − 0 s/ n

5) Reject H0 if t0 < –t,n-1 where t0.05,19 = 1.729 for n = 20 and  = 0.05.

6) x = 18 .22 , s = 0.92 , n = 20

t0 =

18.22 − 18.50 = −1.361 0.92 / 20

7) Because -1.361 > -1.729, fail to reject H0.

Test the second hypothesis (H0 :  = 17.50 vs. H1 :  ≥ 17.50): 4) The test statistic is:

t0 =

x − 0 s/ n

5) Reject H0 if t0 > t,n-1 where t0.05,19 = 1.729 for n = 20 and  = 0.05.

6) x = 18 .22 , s = 0.92 , n = 20

t0 =

18.22 − 17.50 = 3.450 0.92 / 20

7) Because 3.450 > 1.729, reject H0.

There is enough evidence to conclude that the mean absorption rate is greater than 17.50. However, there is not enough evidence to conclude that it is less than 18.50. As a result, we cannot conclude that the new product has an absorption rate that is equivalent to the absorption rate of the current one at  = 0.05.

9.10.3 a) 1) The parameter of interest is the mean bond strength , .

2) H0 :  = 9750 3) H1 :  ≥ 9750

9-33

Applied Statistics and Probability for Engineers, 7h edition b) 4) The test statistic is:

t0 =

x − 0 s/ n

5) Reject H0 if t0 > t,n-1 where t0.05,5 = 2.015 for n = 6 and  = 0.05.

6) x = 9360 , s = 42 .6 , n = 6

t0 =

9360 − 9750 = −22.425 42.6 / 6

7) Because -22.425 < 2.015, fail to reject H0.

There is not enough evidence to conclude that the mean bond strength of cement product is at least 9750 psi and that the process by which the cement product is manufactured is equivalent to the standard.

9.10.4 a) 1) The parameter of interest is the mean breaking strength , .

2) H0 :  = 9.5 3) H1 :  ≥ 9.5 b) 4) The test statistic is:

t0 =

x − 0 s/ n

5) Reject H0 if t0 > t,n-1 where t0.05,49 = 1.677 for n = 50 and  = 0.05.

6) x = 9.31 , s = 0.22 , n = 50

t0 =

9.31 − 9.50 = −6.107 0.22 / 50

7) Because -6.107 < 1.677, fail to reject H0.

There is not enough evidence to conclude that the mean breaking strength of the insulators is at least 9.5 psi and that the process by which the insulators are manufactured is equivalent to the standard.

Section 9-11 9.11.1

 02 = −2[ln(0.15) + ln(0.83) + ... + ln(0.13)] = 37.40 with 2m = 2(8) = 16 degrees of freedom.

The P-value for this statistic is less than 0.01 (≈ 0.0018). In conclusion, we reject the shared null hypothesis.

9.11.2

 02 = −2[ln(0.12) + ln(0.08) + ... + ln(0.06)] = 46.22 with 2m = 2(10 ) = 20 degrees of freedom.

The P-value for this statistic is less than 0.01 (≈ 0.0007). In conclusion, we reject the shared null hypothesis.

9.11.3

9-34

Applied Statistics and Probability for Engineers, 7h edition H0 : μ = 22 H1 : μ ≠ 22

 02 = −2[ln(0.065 ) + ln(0.0924 ) + ... + ln(0.021)] = 30 .57 with 2m = 2(5) = 10 degrees of freedom.

The P-value for this statistic is less than 0.01 (≈ 0.0006). As a result, we reject the shared null hypothesis. There is sufficient evidence to conclude that the mean package weight is not equal to 22 oz.

9.11.4 H0 : σ = 0.2 H1 : σ < 0.2

 02 = −2[ln(0.15) + ln(0.091) + ... + ln(0.06)] = 33.66 with 2m = 2(6) = 12 degrees of freedom.

The P-value for this statistic is less than 0.01 (≈ 0.0007). As a result, we reject the shared null hypothesis. There is sufficient evidence to conclude that the standard deviation of fill volume is less than 0.2 oz.

Supplemental Exercises 9.S5 a) Degrees of freedom = n – 1 = 16 – 1 = 15.

S 4.61 = = 1.1525 N 16 98.33 − 100 t0 = = −1.4490 4.61 / 16 t 0 = −1.4490 with df = 15, so 2(0.05) < P-value < 2(0.1). That is, 0.1 < P-value < 0.2.

b) SE Mean =

95% CI of the mean is x − t 0.025,15 S    x + t 0.025,15 S n n

4.61 4.61    98.33 + (2.131) 16 16 95 .874    100 .786 98.33 − (2.131)

c) Because the P-value >  = 0.05 we fail to reject the null hypothesis at the 0.05 level of significance. d) t 0.05,15 = 1.753 . Because t0 = -1.4490 < t 0.05,15 = 1.753 we fail to reject the null hypothesis at the 0.05 level of significance.

9.S6 a) SE Mean =



=

1.5 = 0.401 , so n = 14 N

N 26.541 − 26 z0 = = 1.3495 1.5 / 14 P-value = 1 −  ( Z 0 ) = 1 −  (1.3495 ) = 1 − 0.9114 = 0.0886 b) A one-sided test because the alternative hypothesis is  > 26.

9-35

Applied Statistics and Probability for Engineers, 7h edition c) Because z0 < 1.65 and the P-value = 0.0886 >  = 0.05, we fail to reject the null hypothesis at the 0.05 level of significance. d) 95% CI of the mean is x − z 0.025     x + z 0.025  n n

1.5 1.5    26.541 + (1.96) 14 14 25 .7553    27 .3268

26.541 − (1.96)

9.S7 a) Degree of freedom = n – 1 = 25 – 1 = 24. b) SE Mean =

t0 =

s

N 84.331 − 85

=

s 25

= 0.631 , so s = 3.155

= −1.06 3.155 / 25 t 0 = −1.06 with df = 24, so 0.1 < P-value < 0.25

c) Because the P-value >  = 0.05 we fail to reject the null hypothesis at the 0.05 level of significance. d) 95% upper CI of the mean is   x + t 0.05, 24 S n

  84.331 + (1.711)

  85 .4106

3.155 25

e) If the null hypothesis is changed to mu = 100 versus mu > 100,

t0 =

84.331 − 100

= −24.832 3.155 / 25 t 0 = −24 .832 and t 0.05, 24 = 1.711 with df = 24.

Because t 0  t 0.05, 24 we fail to reject the null hypothesis at the 0.05 level of significance.

9.S8 a) The null hypothesis is  = 300 versus  < 300

x = 275 .333 , s = 42.6645, and n = 6 t0 =

275.333 − 300

= −1.4162 42.665 / 6 and t 0.05,5 = 2.015 . Because t 0  −t 0.05,5 = −2.015 we fail to reject the null hypothesis at the 0.05 level of significance. b) Yes, because the sample size is very small the central limit theorem’s conclusion that the distribution of the sample mean is approximately normally distributed is a concern.

9-36

Applied Statistics and Probability for Engineers, 7h edition c) 95% two-sided CI of the mean is x − t 0.025,5 S    x + t 0.025,5 S n n

275.333 − (2.571)

42.665 42.665    275.333 + (2.571) 6 6 230.5490 < 𝜇 < 320.1109

9.S9 n 50

a)

Test statistic

b )

100

c)

500

1000 d )

0.095 − 0.10

z0 = 𝑍0 =

= −0.12 (

0.10(1 − 0.10) / 50 0.095 − 0.10 √0.10(1 − 0.10)/100

z0 = z0 =

= −0.166

0.095 − 0.10 0.10(1 − 0.10) / 500 0.095 − 0.10 0.10(1 − 0.10) / 1000

= −0.37 = −0.53

P-value 0.4522 2=0.9 044 0.4364 12=0.

87282 0.3556 92=0.

71138 0.2981 2=0.5 962

conclusion Fail to reject H0

Fail to reject H0

Fail to reject H0

Fail to reject H0

e) The P-value decreases as the sample size increases.

9.S10

p (1 − p ) Sample Mean = p Sample Variance = n

a)

Sample Size, n 50

Sampling Distribution Normal

Sample Mean p

b)

80

Normal

p

c)

100

Normal

p

Sample Variance p(1 − p) 50 p(1 − p) 80 p(1 − p) 100

d) As the sample size increases, the variance of the sampling distribution decreases.

9.S11

a)  = 0.05

 0.5 − 0.6  =  (1.65 − 2.0) =  (−0.35) = 0.3632  0.5(0.5) / 100   Power = 1 −  = 1 − 0.3632 = 0.6368

n = 100  =  z 0.05 +

0.5−0.6 n = 150 𝛽 = 𝜙 (𝑍0.05 + √0.5(0.5)/150 ) = 𝜙(1.65 − 2.45) = 𝜙(−0.8) = 0.2119

Power = 1 −  = 1 − 0.2119 = 0.7881

9-37

Applied Statistics and Probability for Engineers, 7h edition  0.5 − 0.6  =  (1.65 − 3.46 ) =  (−1.81) = 0.03515   0 .

5 ( 0 .

5 ) / 300   Power = 1 −  = 1 − 0.03515 = 0.96485

n = 300  =  z 0.05 +

As the sample size is increased, the power of the test is observed to increase. b)  = 0.01

 0.5 − 0.6  =  (2.33 − 2.0) =  (0.33) = 0.6293  0.5(0.5) / 100   Power = 1 −  = 1 − 0.6293 = 0.3707

n = 100  =  z 0.01 +

 0.5 − 0.6  =  (2.33 − 2.45) =  (−0.12 ) = 0.4522   0 .

5 ( 0 .

5 ) / 100   Power = 1 −  = 1 − 0.4522 = 0.5478

n = 150  =  z 0.01 +

 0.5 − 0.6  =  (2.33 − 3.46 ) =  (−1.13) = 0.1292  0.5(0.5) / 300   Power = 1 −  = 1 − 0.1292 = 0.8702

n = 300  =  z 0.01 +

Decreasing the value of  decreases the power of the test for the different sample sizes. c)  = 0.05

 0.5 − 0.8  =  (1.65 − 6.0) =  (−4.35)  0.0  0.5(0.5) / 100   Power = 1 −  = 1 − 0  1

n = 100  =  z 0.05 +

The true value of p has a large effect on the power. The greater is the difference of p from p0, the larger is the power of the test. d)

 z / 2 p 0 (1 − p 0 ) − z  p(1 − p)   n=   p − p0  

2

2

 2.58 0.5(1 − 0.50) − 1.65 0.6(1 − 0.6)   = (4.82) 2 = 23.2  24 =   0 .

6 − 0 .

5  

 z / 2 p 0 (1 − p 0 ) − z  p(1 − p)   n=   p − p0  

2

2

 2.58 0.5(1 − 0.50) − 1.65 0.8(1 − 0.8)   = (2.1) 2 = 4.41  5 =   0.8 − 0.5   The true value of p has a large effect on the sample size. The greater is the distance of p from p0, the smaller is the sample size that is required.

9-38

Applied Statistics and Probability for Engineers, 7h edition

9.S12

a) Rejecting a null hypothesis provides a stronger conclusion than failing to reject a null hypothesis.

Therefore, place what we are trying to demonstrate in the alternative hypothesis. For alternative H1<150, there might be a chance that Ho is not rejected which is not a strong conclusion, so we try to demonstrate the chances of Ho getting rejected.

Assume that the data follow a normal distribution. b) 1) the parameter of interest is the mean weld strength, .

2) H0 :  = 150 3) H1 :   150 4) The test statistic is:

x − 0

t0 =

s/ n

5) Because no value of  is given, we calculate the P-value 6) x = 153.7 , s= 11.3, n=20

t0 =

153.7 − 150 11.3 / 20

= 1.46

P-value = P(t > 1.46) = 0.05 < P-value < 0.10 7) There is some modest evidence to support the claim that the weld strength exceeds 150 psi. If we used  = 0.01 or 0.05, we would fail to reject the null hypothesis, thus the claim would not be supported. If we used  = 0.10, we would reject the null in favor of the alternative and conclude the weld strength exceeds 150 psi.

9.S13

Assume the data follow a normal distribution.

1) The parameter of interest is the standard deviation of the concentration, .

2) H0 : 2 = 42 3) H1 : 2 < 42 4) The test statistic is:

20 =

( n − 1)s2 2

5) Because no value of alpha is specified we calculate the P-value 6) s = 0.004 and n = 10

 02 =

(

)

9(0.004) 2 = 0.000009 (4) 2

P − value  0 P-value = P   0.00009 7) Conclusion: The P-value is approximately 0. Therefore we reject the null hypothesis and conclude that the standard deviation of the concentration is less than 4 grams per liter.

9.S14

2

Assume the data follow a normal distribution. a) 1) The parameter of interest is the standard deviation, .

2) H0 : 2 = (0.00002)2 3) H1 : 2 < (0.00002)2 4) The test statistic is:

20 =

( n − 1)s2 2

9-39

Applied Statistics and Probability for Engineers, 7h edition . reject H0 if  20  1.24 for  = 0.01 5) 20.99,7 = 124

6) s = 0.00001 and  = 0.01

7(0.00001) 2  = = 1.75 (0.00002) 2 2 0

7) Because 1.75 > 1.24 we fail to reject the null hypothesis. There is insufficient evidence to conclude the standard deviation is at most 0.00002 mm. b) Although the sample standard deviation is less than the hypothesized value of 0.00002, it is not significantly less (when  = 0.01) than 0.00002. The value of 0.00001 could have occurred as a result of sampling variation.

9.S15 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true mean concentration of suspended solids, .

2) H0 :  = 50 3) H1 :  < 50 4) Because n >> 30 we can use the normal distribution z0 =

x−

s/ n 5) Reject H0 if z0 < -1.65 for  = 0.05 6) x = 59.87 s = 12.50 n = 60 z0 =

59.87 − 50

12.50 / 60

= 6.12

7) Because 6.12 > –1.65 fail to reject the null hypothesis. There is insufficient evidence to conclude that the true mean concentration of suspended solids is less than 50 ppm at  = 0.05. b) P-value =  (6.12 )  1 c) We can divide the real line under a standard normal distribution into eight intervals with equal probability. These intervals are [0, 0.32), [0.32, 0.675), [0.675, 1.15), [1.15, ) and their negative counterparts. The probability for each interval is p = 1/8 = 0.125 so that the expected cell frequencies are E = np = (60) (0.125) = 7.5.

The sample mean and standard deviation are 59.86667 and 12.49778, respectively, and these are used to determine the boundaries of the intervals. That is, the first interval is from negative infinity to 59.86667 + 12.49778(-1.15) = 45.50. The second interval is from this value to 59.86667 + 12.49778(-0.675) = 51.43.

The table of ranges and their corresponding frequencies is completed as follows.

Interval Obs. Frequency. Exp. Frequency. x  45.50 9 7.5 45.50< x  51.43 5 7.5 51.43< x  55.87 7 7.5 55.87< x  59.87 11 7.5 59.87< x  63.87 4 7.5 63.87< x  68.31 9 7.5 68.31< x  74.24 8 7.5 x  74.24 6 7.5 The test statistic is:

9-40

Applied Statistics and Probability for Engineers, 7h edition

 2o =

(9 − 7.5) 2 (5 − 7.5) 2 (8 − 7.5) 2 (6 − 7.5) 2 + ++ + = 5.06 7.5 7.5 7.5 7.5

and we reject if this value exceeds

 20.05,5 = 11.07 . Because it does not, we fail to reject the

hypothesis that the data are normally distributed.

9.S16

Divide the real line under a standard normal distribution into eight intervals with equal probability. These intervals are [0, 0.318), [0.318, 0.675), [0.675, 1.150), [1.15, ) and their negative counterparts. The probability for each interval is p = 1/8 = 0.125 so the expected cell frequencies are E = np = (100) (0.125) = 12.5.

The sample mean and standard deviation are 5421.77 and 33.43616, respectively, and these are used to determine the boundaries of the intervals. That is, the first interval is from negative infinity to 5421.77 + 33.43616(-1.15) = 5383.30. The second interval is from this value to 5421.77 + 33.43616(-0.67449) = 5399.218. The table of ranges and their corresponding frequencies is completed as follows.

Interval Obs. Frequency. Exp. Frequency. x  5383.307 12 12.5 5383.307 < x  5399.218 14 12.5 5399.218 < x  5411.116 12 12.5 5411.116 < x  5421.770 12 12.5 5421.770 < x  5432.424 14 12.5 5432.424 < x  5444.322 9 12.5 5444.322 < x  5460.233 14 12.5 x  5460.233 13 12.5 The test statistic is: (12 - 12.5)2 (14 − 12.5) 2 (14 - 12.5)2 (13 − 12.5) 2  02 = + ++ + = 1.6 12.5 12.5 12.5 12.5 We reject the null hypothesis if this value exceeds  20.05,5 = 11.07 . Because  o   0.05,5 , fail to reject 2

2

the null hypothesis that the data are normally distributed. The P-value = P(  2  1.6) = 0.90 .

9.S17 a) 1) The parameter of interest is the true mean sugar concentration, .

2) H0 :  = 11.5 3) H1 :   11.5 4) t0 =

x− s/ n

5) Reject H0 if |t0| > t/2,n-1 where t/2,n-1 = 2.093 for  = 0.05 6) x = 11 .47 , s = 0.022 n=20

t0 =

11.47 − 11.5 0.022 / 20

= −6.10

7) Because 6.10 > 2.093 reject the null hypothesis. There is sufficient evidence that the true mean sugar concentration is different from 11.5 at  = 0.05.

9-41

Applied Statistics and Probability for Engineers, 7h edition

From Table V the t0 value in absolute value is greater than the value corresponding to 0.0005 with 19 degrees of freedom. Therefore 2(0.0005) = 0.001 > P-value b) d =

 |  −  0 | | 11.4 − 11.5 | = = = 4.54   0.022

Using the OC curve, Chart VII e) for  = 0.05, d = 4.54, and n =20 we find   0 and Power  1. c) d =

 |  − 0 | | 11.45 − 11.5 | = = = 2.27   0.022

Using the OC curve, Chart VII e) for  = 0.05, d = 2.27, and 1 -  > 0.9 ( < 0.1), we find that n should be at least 5. d) 95% two sided confidence interval

 s   s  x − t0.025,19      x + t0.025,19    n  n  0.022   0.022  11.47 − 2.093     11.47 + 2.093   20   20  11.46    11.48 We conclude that the mean sugar concentration content is not equal to 11.5 because that value is not inside the confidence interval. e) The normality plot below indicates that the normality assumption is reasonable.

Probability Plot of Sugar Concentration Normal

99

95 90

Percent

80 70 60 50 40 30 20 10 5

1

11.40

9.S18 a) z0 =

11.42

11.44 11.46 11.48 Sugar Concentration

11.50

11.52

x − np0 53 − 225(0.25) = = −0.5004 np0 (1 − p0 ) 225(0.25)(0.75)

The P-value = (–0.5004) = 0.3084 b) Because the P-value = 0.3084 >  = 0.05 we fail to reject the null hypothesis at the 0.05 level of significance. c) The normal approximation is appropriate because np > 5 and n(p-1) > 5.

9-42

Applied Statistics and Probability for Engineers, 7h edition

ˆ= d) p

53 = 0.2356 225

The 95% upper confidence interval is:

p  pˆ + z

pˆ (1 − pˆ ) n

p  0.2356 + 1.65

0.2356(0.7644) 225

p  0.2823 e) P-value = 2(1 - (0.5004)) = 2(1 - 0.69146) = 0.61708.

9.S19 a) In order to use the 2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true variance of the ratio between the numbers of symmetrical and total synapses, 2. 2) H0 : 2 = 0.02 3) H1 : 2  0.02 4)  20 =

( n − 1)s2 2

5) Reject H0 if 20  12−  / 2,n −1 where  = 0.05 and  0.975,30 = 16 .79 or  0    / 2,n−1 where  = 0.05 and 2

2

2

 02.025,30 = 46.98 for n = 31 6) n = 31, s = 0.198  20 =

(n − 1) s 2

2

=

30(0.198) 2 = 58.81 0.02

7) Because 58.81 > 46.98 reject H0. The true variance of the ratio between the numbers of symmetrical and total synapses is different from 0.02 at  = 0.05. b) P-value/2 < 0.005 so that P-value < 0.01

9.S20 a) In order to use the 2 statistic in hypothesis testing and confidence interval construction, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true variance of tissue assay, 2.

2) H0 : 2 = 0.6 3) H1 : 2  0.6 4)  20 =

( n − 1)s2 2

5) Reject H0 if 20  12−  / 2,n −1 where  = 0.01 and  0.995,11 = 2.60 or  0    / 2,n−1 where  = 0.01 and 2

 02.005,11 = 26.76 for n = 12 6) n = 12, s = 0.75858  20 =

(𝑛−1)𝑠 2 𝜎2

=

11(0.758)2 0.6

=10.55

9-43

2

2