Solution manual for statistical reasoning for everyday life 5th edition jeff bennett

Solution Manual for Statistical

Reasoning for Everyday Life, 5th Edition, Jeff Bennett

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Section 8.1

Statistical Literacy and Critical Thinking

1 A distribution of sample means is the distribution that results when we find the means of all possible samples of a given size, and a distribution of sample proportions is the distribution that results when we find the proportions of all such samples.

2 The sampling error is the error introduced because a random sample is used to estimate a population parameter. It does not include other sources of error, such as biased sampling, bad survey questions, or recording mistakes. If the sample size were increased, then the sampling error would decrease because more data would be collected about the population.

3 A sample mean is the mean of a particular sample drawn from a population. A sample proportion is a fraction (or percentage) with which some variable occurs in a sample.

Notation for SampleNotation for Population

4 The larger the sample size, the more closely the sampling distribution will approximate a normal distribution. A common guideline is to assume that the distribution of sample means or proportions is close to normal if the sample size is greater than 30. For a distribution of sample means: the mean is the population mean, μ, and the standard deviation is n  . For a distribution of sample proportions: the mean is the population proportion, p, and the standard deviation is (1)/ ppn .

5 This statement does not make sense. A sampling distribution is a distribution of all possible samples of a particular size, and there are far more than three possible samples of size n = 10 from a population of 1500.

6 This statement does not make sense. The Nielsen survey is for a sample, so it measures only a sample proportion, not a population proportion.

7 This statement makes sense. As discussed in this section, the distribution of sample proportions should be nearly normal for a sample size of a few thousand, and its mean will be the actual population proportion. Therefore, most individual samples will have proportions clustered around the actual population proportion.

8 This statement makes sense. The notation refers to a sample mean, and for a sample of weights, a mean in pounds makes sense. The notation refers to a sample proportion, so it makes sense that the proportion of babies with some condition might be 0.45, or 45%.

9 The sample size, n = 575. The sample mean, x = 12.2 volts. The sample standard deviation, s = 1.4 volts.

10 The population size, N = 3427. The sample size, n = 50. The sample proportion, ˆ p = 0.10.

11 The population size, N = 50,000. The sample size, n = 200. The sample proportion, ˆ p = 0.04.

12 The sample size, N = 50,000. The sample size, n = 200. The sample mean, x = 328 mg. The sample standard deviation, s = 12 mg.

Concepts and Applications

13 No, the results from the study do not apply to females. It is possible (and it is now known) that the behavior of females is different from the behavior of males.

14 The mean from the larger sample is likely to be closer to the population mean, but it is possible that the mean from the smaller sample is closer to the population mean.

15 a) The sample mean = 12.19 is (12.19 – 12.00)/0.02 = 9.5 standard deviations from the mean of the distribution of sample means. That is, is 9.5 standard deviations above the mean of the distribution.

b) From Table A-1, the probability of a sample mean greater than 12.19 ounces is very small, 0.01%. So, the probability of a random sample with a sample mean of at least 12.19 ounces is almost 100%.

c) Since the probability of selecting another sample with a mean greater than 12.19 ounces is so small, we assume that the consumers are not being cheated. It appears that the cans are being filled with an amount that is greater than the actual mean of 12.00 ounces. 16

The sample mean is 2.96 standard deviations above the population mean.

b) From Table A-1, z = -2.96 corresponds to a probability of 0.0015 or 0.15%.

17 a) The population proportion is p = 1,014/25,344= 0.040.

b) The sample proportion is ˆ p = 18/250 = 0.072.

c) Random samples collected with sound sampling methods vary, and the sample proportions differ from the population proportion because of sampling error.

18 a) The population proportion is p = 269/2444 = 0.110.

b) The sample proportion is ˆ p = 8/50 = 0.160.

c) It does not necessarily mean that the sampling method is flawed. The discrepancy could be attributable to sampling error, and it could occur with a good sampling method.

b) The mean of the sample means in part (a) is

c) The mean of the population is (1 + 2 + 5)/3 = 8/3 = 2.7 interceptions, which is the same as the mean of the sample means in part (a). Yes, these means will always be equal.

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CHAPTER 8, INFERENCES FROM SAMPLES TO POPULATIONS

b) The mean of the sample means is (56.0 + 52.5 + ... + 46.0)/16 = 836/16 = 52.25 years.

c) The mean of the population is (56 + 49 + 58 + 46)/4 = 297/5 = 52.25 years, which is equal to the mean of the sample means in part (b). These means will always be equal.

According to Table A-1, the probability of getting another sample with a birth weight of 2966 grams or less is 0.1563 or 15.63% (technology: 15.73%). No, this probability is not unusual.

According to Table A-1, the probability of getting another sample with a birth weight of 3272 grams or less is 0.9773. The probability of getting another sample with a birth weight of 3272 grams or more is 1 - 0.9773 = 0.0227 or 2.27% (technology: 2.29%). Yes, this probability is unusual.

According to Table A-1, the probability of getting another sample with a birth weight of 3390 grams or less is 0.9999. The probability of getting another sample with a birth weight of 3390 grams or more is 1 - 0.9999 = 0.0001 or 0.01% (technology: 0.003%). Yes, this probability is very unusual.

According to Table A-1, the probability of

getting another sample with a birth weight of 2919 grams or less is 0.0668 or 6.68% (technology: 6.63%). No, this probability is not unusual.

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

25 The sample proportion of left-handed adults is 45/500 = 0.09. Using the population proportion of 0.10, the standard deviation of the distribution of sample proportions is (1)0.10(10.10) 500 0.0134 pp

= . . . = −0.75. From Table A-1, the probability of getting a sample with 45 or fewer left-handed adults is 0.2266 or 22.66% (technology: 0.2280). No, this probability is not unusual.

26 The sample proportion of left handed adults is 15/250 = 0.06. Using the population proportion of 0.10, the standard deviation of the distribution of sample proportions is (1)0.10(10.10) 250 0.0190 pp n 

= . . . =−2.11. From Table A-1, the probability of getting a sample with 15 or fewer left-handed adults is 0.0174 or 1.74% (technology: 0.0175). Yes, this probability is unusual.

27 The sample proportion of left-handed adults is 92/820 = 0.11. Using the population proportion of 0.10, the standard deviation of the distribution of sample proportions is (1)0.10(10.10) 820 0.0105 pp n 

= = . . . =0.95. From Table A-1, the probability of getting a sample with 92 or fewer left-handed adults is 0.8289. The probability of getting a sample with 92 or more left-handed adults is 1 –0.8289 = 0.1711 or 17.11%. No, this probability is not unusual.

28 The sample proportion of left-handed adults is 84/650 = 0.13. Using the population proportion of 0.10, the standard deviation of the distribution of sample proportions is (1)0.10(10.10) 650 0.0118 pp n 

= = . . . =2.54. From Table A-1, the probability of getting a sample with 84 or fewer left-handed adults is 0.9945. The probability of getting a sample with 84 or more left-handed adults is 1 –0.9945 = 0.0055 or 0.55%. Yes, this probability is unusual.

Section 8.2

Statistical Literacy and Critical Thinking

1 Because you have only the one sample, you use the sample mean as your best estimate of the population mean.

2 The margin of error for 95% confidence is found by evaluating: 1.96 s E n  . We find the confidence interval by adding and subtracting this margin of error from the sample mean. The confidence interval ranges from ( x - margin of error) to ( x + margin of error).

3 We have 95% confidence that the confidence interval limits actually contain the true value of the population mean. This means that if we were to repeat the process of constructing such confidence intervals, 95% of them will have limits that contain the true population mean.

4 Once you have identified a particular margin of error, E, that you are seeking, you can calculate the minimum sample size with the formula:

 . Often σ is estimated by the sample standard deviation s.

5 The statement does not make sense. The center of a confidence interval is not necessarily the population mean.

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6 The statement does not make sense. A 95% confidence interval does not give the percentage of the population that is between the confidence interval limits.

7 The statement does not make sense. The population mean,  , is a fixed constant and is not a random variable. Either the population mean falls within the confidence interval or it does not, and there is no probability associated with this.

8 The statement makes sense. The larger the sample size, the smaller the margin of error.

Concepts and Applications

9 We have 95% confidence that the limits of 98.1oF and 98.3oF actually do contain the true population mean body temperature. We expect that 95% of such samples will result in confidence interval limits that contain the population mean.

10 The margin of error is ( x - margin of error) to ( x + margin of error) = (2965 – 245) to (2965 + 245) = 2720 grams to 3210 grams, or 2720 grams < μ < 3210 grams.

11 The media often omit reference to the confidence level, which is typically 95%. The word “mean” should be used instead of the word “average.”

12 No, it is not valid to criticize the sample for being just 0.01% of the population. A sample can provide a good estimate of a population mean if the sample size is large enough based on the standard deviation and desired margin of error, but the size of the population is generally not a factor.

volts, or 111.1 volts < μ < 113.3 volts.

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25 The population mean µ is estimated by the sample mean x = 5.639 g. The margin of error is = .

. The

lb or 1.64 lb < µ < 2.18 lb.

29 Use the software of your choice to find the mean and standard deviation of the sample of n = 36 weights. You should find x = 183.6 lb. and s = 112.9 lb. The margin of error is = . √ = . × . √ =36.9 lb. and the 95% confidence interval is ± =183.6 ±36.9=146.7 to 220.5 lb. or 146.7 lb. < µ < 220.5 lb.

30 Use the software of your choice to find the mean and standard deviation of the sample of n = 40 weights. You should find x = 172.5 ng/mL and s = 119.5 ng/mL. The margin of error is = . √ = . × . √ =37.0 and the 95% confidence interval is ± =172.5 ±37.0=136 to 210 ng/mL or 136 ng/mL < µ < 210 ng/mL.

31 Use the software of your choice to find the mean and standard deviation of the sample of n = 20 Chips Ahoy cookies.

a) x = 23.4 chocolate chips

b) s = 2.3 chocolate chips

c) The best estimate for the mean number of chocolate chips for the population of all Chips Ahoy cookies is the sample mean, 23.4 chocolate chips.

d) The margin of error is = . √ = . × . √ =1.0 and the 95% confidence interval is ± =23.4 ±1.0=22.4 to 24.4 chocolate chips or 22.4 chocolate chips < µ < 24.4 chocolate chips.

e) We can be 95% confident that the interval from 22.4 to 24.4 contains the population mean µ. That is, we expect 95% of confidence intervals computed from such samples to contain the population mean. However, since the sample is very small, the estimate might not be too reliable.

32 Use the software of your choice to find the mean and standard deviation of the sample of n = 31 household TV sets.

a) x = 2.0 sets

b) s = 1.3 sets

c) The best estimate for the mean number of household TV sets for the population of all American families is the sample mean, 2.0 sets.

d) The margin of error is = . √ = . × . √ =0.5 and the 95% confidence interval is ± =2.0 ±0.5=1.5 to 2.5 sets or 1.5 sets < µ < 2.5 sets.

e) We can be 95% confident that the interval from 1.5 to 2.5 contains the population mean µ. That is, we expect 95% of confidence intervals computed from such samples to contain the population mean number of household TV sets.

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Section 8.3

Statistical Literacy and Critical Thinking

1 Because you have only the one sample, you use the sample proportion as your best estimate of the population proportion.

2 For 95% confidence, you use this formula to calculate the margin of error: ˆˆ(1) 1.96 pp E n  . Your 95% confidence interval then ranges from your sample proportion minus the margin of error to your sample proportion plus the margin of error.

3 You expect that 95% of all possible samples will have a 95% confidence interval that contains the population proportion.

4 Once you have identified a particular margin of error, E, that you are seeking, you can calculate the minimum required sample size with the formula 2 1 n E  .

5 This statement does not make sense. With 95% confidence, the true population proportion is between 56% – 4% and 56% + 4%, or between 52% and 60%.

6 This statement makes sense. We expect 95% of all possible samples to contain the true population proportion, but 5% will not, and this case was apparently one of those.

7 This statement makes sense. The margin of error was added to and subtracted from the sample proportion to construct the confidence interval.

8 This statement does not make sense. You can have a small margin of error with relatively small samples compared to the full population.

Concepts and Applications

9 We have 95% confidence that the limits of 0.803 and 0.831 actually do contain the true population proportion. We expect that 95% of such samples will result in confidence interval limits that do contain the population proportion.

10 The 95% confidence interval is

or 0.671 < p < 0.723.

11 The media often omit reference to the confidence level, which is typically 95%.

12 n is the sample size, and n = 3011.

p is the sample proportion, and

p = 0.73. p is the population proportion, and its value is unknown.

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The margin of error is =1.96 ( ) = 2 . ( . ) =0.0297. The 95% confidence interval is ± =0.091 ±0.0297=0.0613 to 0.1207 or 0.061 < p < 0.121.

25 The sample proportion is = =0.836. The margin of error is =1.96 ( ) = 1.96 . ( . ) =0.0589. The 95% confidence interval is ± =0.836 ±0.0589=0.775 to 0.895; or 0.777 < p < 0.895.

26 a) The best estimate is the sample proportion = =0.550 .

b) The margin of error is =1.96 ( ) = 1.96 . ( . ) =0.0236

The 95% confidence interval is ± =0.550 ±0.0236=0.526 to 0.574; or 0.526 < p < 0.574.

c) Yes, there is strong evidence since the confidence interval lies entirely above the 50% point, it appears that the majority is in favor of immediate government action.

27 a) The sample proportion is ˆ p = 0.98.

b) The margin of error is =1.96 ( ) = 1.96 . ( . ) =0.014

The 95% confidence interval is ± =0.98 ±0.014=0.966 to 0.994; or 0.966 < p < 0.994.

c) No, this does not represent a random sample. To be a random sample, all films must have an equal chance of being chosen.

28 The sample proportion is = =0.0258. The margin of error is

=1.96 ( ) = .196 . ( . ) =0.0037. The 95% confidence interval is ± =0.0258 ±0.0040=0.0218 to 0.0298 or 0.0218 < p < 0.0298.

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CHAPTER 8, INFERENCES FROM SAMPLES TO POPULATIONS

29 For the first poll, = =0.520. The margin of error is =1.96 ( ) =

1.96 . ( . ) =0.025. The 95% confidence interval is ± =0.520 ±0.025= 0.495 to 0.545; or 0.495 < p < 0.545.

For the second poll, = =0.514. The margin of error is =1.96 ( ) =

1.96 . ( . ) =0.020. The 95% confidence interval is ± =0.514 ±0.020= 0.494 to 0.534; or 0.494 < p < 0.534.

For the third poll, = =0.515. The margin of error is =1.96 ( ) =

1.96 . ( . ) =0.017. The 95% confidence interval is ± =0.515 ±0.017= 0.498 to 0.532; or 0.498 < p < 0.532. Because all of these confidence intervals include values below 0.5, Martinez cannot be confident of winning a majority.

30 a) =1.96 ( ) = 1.96 . ( . ) =0.00145. This level of precision allows the Bureau to detect very small changes in the unemployment rate and is reasonable for that purpose.

b) E would be reduced by 1/2 to 0.000725 since n appears in the denominator of the square root.

c) E would be increased by a factor of 2 to 0.0029.

31 a) The 95% confidence interval for the true population proportion approving the President’s performance is 0.540.04 0.50 to 0.58  .

b) Since =1.96 ( ) = 1.96 . ( . ) =0.04, we conclude that . ( . ) =0.02 or ( . )( . ) =0.0004. This implies that =

. )( . ) . =621. The sample size may have been determined as 1/E2 = 1/0.042 = 625. The difference is small.

32 a) The actual margin of error is =1.96 ( ) = 1.96 . ( . ) =0.037

This is consistent with the stated margin of error of 4 percentage points.

b) n = 1/E2 = 1/0.022 = 2500.

Chapter 8 Review Exercises

1 a) The best estimate for the population proportion of yellow peas is the sample proportion of yellow peas, = =0.262

b) The margin of error is =1.96 ( ) = 1.96 . ( . ) =0.036. The 95% confidence interval is ± =0.262 ±0.036=0.226 0.298; or 0.226 < p < 0.298.

c) The margin of error is =0.036.

d) We have 95% confidence that the limits of 0.226 and 0.298 contain the true value of the population proportion. If such samples of size 580 were randomly selected many times, the resulting confidence interval limits would contain the true population proportion in 95% of those samples.

e) The distribution of the sample proportions should be approximately normal or bell-shaped.

f) n = 1/E2 = 1/0.042 = 625.

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2 a) = . = . × =96.04, we will use n = 97.

b) If the sample size is larger than necessary, the estimate will be better, i.e., the confidence interval will be shorter than planned. If the sample size is smaller than necessary, the estimate will be worse, i.e., the confidence interval will be longer than planned.

c) Because college students are a more homogeneous group than the population as a whole, the standard deviation for college students is likely to be smaller than 15. If we use an actual value that is smaller than 15, the sample size will be smaller than 100.

3 a) The margin of error is = . √ = . × . √ =20.21. The 95% confidence interval is ± = 279.5 ±20.21=259.3 to 299.7

b) The margin of error, from part (a), is E = 20.21.

c) We can be 95% confident that the interval from 259.3 to 299.7 contains the population mean white blood cell count. If such samples of size 40 were randomly selected many times, the resulting confidence interval limits would contain the true population mean in 95% of those samples.

d) = 1.96s E

2 = 1.96 ×65.2 10

2 =163.3, we will use n = 164.

4 a) n = 1/E2 = 1/0.022 = 2500.

b) No. Your sample size will be large enough, but it has a high potential for being biased. The people who refuse to answer could well constitute a segment of the population with a different perspective, and that perspective will be incorrectly excluded from the sample.

Chapter 8 Quiz

1 The distribution of the sample means is approximately a normal distribution or bell-shaped.

2 The distribution of the sample proportions is approximately a normal distribution or bell-shaped.

3 x is the sample mean and ˆ p is the sample proportion. These are the values used to estimate the population mean and proportion when a single sample is drawn.

4 0.30 < p < 0.40

5 The best estimate of the population mean is the mean of the sample, (b).

6 The sample proportion is ˆ p = 235/475 = 0.495. The margin of error is =1.96 ( ) = 1.96 ( . ) =0.0450. The 95% confidence interval is ± =0.495 ±0.0450=0.450 0.540; or 0.450 < p < 0.504.

7 The margin of error is one-half of the length of the confidence interval. Since the length of the interval is 98.6 – 98.0 = 0.6, the margin of error is E = 0.3. The sample mean x is 98.0 + 0.3 = 98.6 – 0.3 = 98.3.

8 The 95% confidence interval is ± =0.656 ±0.120=0.536 0.776 or 0.536 < p < 0.776.

9 The 95% confidence interval is ± = 68.2 ±5.8=62.4 74.0 cm or 62.4 cm < μ < 74.0 cm.

10 Any confidence interval estimate of a population proportion should have limits that are themselves proportions (usually decimal values), not values with units such as dollars.

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