STAT 500 Midterm 2 Answers (Penn State University) Latest Verified Review 2023 Practice Questions and Answers for Exam Preparation, 100% Correct with Explanations, Highly Recommended, Download to Score A+

STAT 500 Midterm 2 Answers (Penn State University) Latest Verified Review 2023 Practice Questions and Answers for Exam Preparation, 100% Correct with Explanations, Highly Recommended, Download to Score A+ 1. In 2000, 130 were found to be overweight out of 750 men who are 20-34 years old. In 2012, 160 were found to be overweight out of 700 men who are 20-34 years old. Complete parts a. and b. by hand calculations. a. (7 points) Estimate with 95% confidence the change in the population proportion from the survey of year 2000 to the survey of year 2012 that were overweight and interpret the results. The number of successes and failures in both populations are ≥ 5.So, normal distribution is okay to use. (p_1 ) ̂= 130/(750 ) = 0.173; (p_2 ) ̂= 160/(700 ) = 0.229. 95% confidence interval for p_1 - p_2: (p_1 ) ̂ - (p_2 ) ̂ ± z_(.025 ) √(((p_1 ) ̂ (1-(p_1 ) ̂) )/n_1 + ((p_2 ) ̂ (1-(p_2 ) ̂) )/n_2 ) i.e, (0.173- 0.229) ± 1.96 √(( (0.173)(1-0.173) )/750+( (0.229)(1-0.229) )/700 ) i.e, -0.056 ± 0.0414 , i.e, (-0.0974, -0.0146) We can say, with 95% confidence, that the change in the population proportion from the survey of year 2000 to the survey of year 2012 that were overweight falls in between -0.0974 and -0.00146. b. (8 points) Conduct a hypothesis test for a difference between the two proportions at a 5% level of significance. Include hypotheses, p-value, decision and conclusion. H_0 : p_1 - p_2=0 vs H_0 : p_1 - p_2≠0 p ̂ = (130+160)/(750+700) = 290/1450 = 0.2 Test Statistic: z^* = ((p_1 ) ̂ - (p_2 ) ̂)/√((p ) ̂(1-p ̂ )((1 )/n_1 + (1 )/n_2 )) = (0.173 – 0.229)/√((0.2)(0.8)((1 )/750+ (1 )/700)) = -2.67 p-value = 2 . P(Z < -2.67) = 2 (.0039) = .0078; α= .05. p-value < α. So, we reject H_0 at α= .05. At 5% level of significance, we believe that there is a difference between the two proportions.