Question

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 36 feel more enriched as a result of her class. Now, what do you think? Conduct a hypothesis test at the 5% level. Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom.)

What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round your answers to three decimal places.)

What is the p-value? (Round your answer to four decimal places.)

Construct a 95% confidence interval for the true proportion. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to four decimal places.)

Answer 1

Polution- Given X=36, 0 : 644 P: Gov. :06 Hypothesis :- Hoi p=0.60 He: P +0.60 Sample proportion (fs): 1 = 36 = 0.5625 L) Here we cese - distribution Test Stotistic (7): Ô-P - 0.5625 - 0.6 0.601-0.6) 64 la : -0.61 p-tralue = P (2 >121) :&P(2> 1-0.611) - 2P (22 061) = 0[1- P(2 <0·6)] = 2(1-0.1891) :2 (0.2709) - 0.5418 p-value > d=0.05 so we fail to roject Ho 967. Of is Ộ = 24 Pico d=1-0.95 -0.05 20.095 : 1.96 = 0.5625 +1.96 10.5625 (1-0.5625) 64 =0.5695+ 0.0680 0.5098 0.6245),