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Eliza Alvarez, Meggy Punsalan, Sherica Torres

SA 100 – C

Feb 25, 2013

SA 100 GROUP ASSIGNMENT # 5 1. An educational researcher sought to estimate the average number of close friends that students on a particular campus made during their first year of school. Questioning a random sample of 50 students completing their freshman year, he finds a sample mean of 3 close friends and a sample standard deviation of 1. Construct a 95% confidence interval to estimate the mean number of close friends made by the population of college students during their first year on campus. X(mean)= 3 ĎƒX=1/√50 ĎƒX= 0.14 95% CI=X+/- (1.96)ĎƒX 95% CI=2.72 to 3.27 2. An administrator in charge of undergraduate education on a large campus wanted to estimate the average number of books required by instructors. Using bookstore data, she drew a random sample of 25 courses for which she obtained a sample mean of 2.8 books and a sample standard deviation of .4. Construct a 99% confidence interval to estimate the mean number of books assigned by instructors on campus. X(mean)= 2.8 ĎƒX=0.4/√25 ĎƒX= 0.08 99% CI=X+/- (2.58)ĎƒX 99% CI=2.59 to 3.01 3. A major research organization conducted a national survey to determine what percent of Filipinos feel that things are getting better for them economically. Asking 1,200 respondents called at random if their own economic situation was better today than last year, 45% reported that they were in fact better off than before. A) Find the standard error of the proportion. B) Find the 95% confidence interval for the population proportion. C) Find the 99% confidence interval for the population proportion. P=.45 A) Sp=  Sp=  

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Standard error of the proportion: Sp=0.014 B) 95% CI=P+/- (1.96)sp 95% CI=0.422 to 0.477


C) 99% CI=P+/- (2.58)sp 99% CI=0.414 to 0.486 4.

5.

Estimate the standard error of the mean with the following sample of 34 scores on a 10-item objective test of political name recognition: 10 1 4 8 10 7 5 5 5 6 6 10 7 6 3 8 5 7 4 7 4 6 5 5 6 5 6 4 7 3 5 4 8 5 With the sample mean in problem 4, find a) the 95% confidence interval and (b) the 99% confidence interval.

6. Suppose that the population standard deviation for a normally distributed standardized test of achievement is known to be 7.20. What would the standard error of the sample mean be if we were to draw a random sample of 16 test scores? ĎƒX=7.20/√16 Standard error of sample mean: ĎƒX= 1.8 7. Suppose that the random sample in problem 6 yielded these observed scores: 6 5 6 12 10 9

12 20

5 10 23 20

11 28

13 18

a) Find the 95% confidence interval for the mean. b) Find the 99% confidence interval for the mean. X(mean)= 13 (A) 95% CI=X+/- (1.96)ĎƒX 95% CI=9.47 to 16.53 (B) 99% CI=X+/- (2.58)ĎƒX 99% CI=8.36 to 17.64 8. To estimate the proportion of students on a particular campus who favor a campus wide ban on alcohol, a social researcher interviewed a random sample of 200 students from the college population. She found that 36% of the sample favored banning alcohol. With this information, (a) find the standard error of the proportion and (b) find the 95% confidence interval for the population proportion. P=.36 A) Sp=  Sp=  

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Staandard error of the proportion: Sp=0.0339


B) 95% CI=P+/- (1.96)sp 95% CI=0.294 to 0.426 9. Estimate the standard error of the mean with the following sample of 34 scores on a 10item objective test of political name recognition: 10 1 4 8 10 7 5 5 5 6 6 10 7 6 3 8 5 7 4 7 4 6 5 5 6 5 6 4 7 3 5 4 8 5 X (mean): 5.794 X2=33.57 standard deviation: s = √∑X2 / N – X2 standard deviation: s = √1277 / 34 – 33.57 standard deviation: s = 1.997 Sx= s/√N-1 Sx= 1.997/√34-1 Sx= 1.997/√33 Standard error of mean: Sx= 0.348 10. With the sample mean in problem 9, find (a) the 95% confidence interval and (b) the 99% confidence interval. df = N – 1 = 34 – 1 = 33 A) α = 1 – level of confidence = 1 – 0.95 = 0.05 t = 2.042 Margin of error = tsx = (2.042)(0.348) = 0.711 95% CI = X +/- tsx 95% CI = 5.79 +/- 0.711 95% CI = 5.079 to 6.501 B) α = 1 – level of confidence = 1 – 0.99 = 0.01 t = 2.750 Margin of error = tsx = (2.750)(0.348) = 0.957 99% CI = X +/- tsx 99% CI = 5.79 +/- 0.957 99% CI = 4.833 to 6.747

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