Question

1) At a large university, the dean of students wonders if the percentage of “first generation” students has changed since he arrived 10 years earlier. When he arrived, the proportion was 56%. He randomly samples 94 students, finding that 60 are first-generation.

a) State the null and alternative hypotheses using proper notation.

b) Use software to calculate the test statistic (z) and P-value.

c) At = 0.05, do you reject the null hypothesis? YES / NO

d) Write a sentence summarizing your finding in the context of the problem.

2) Using the data from problem 1, use software to calculate a 95% confidence interval. Write your answer in a form similar to (15.44%, 18.36%).

3) A March 2016 Gallup poll surveyed 1017 American adults and asked the question: “Do you approve of President Obama’s job in office?” The corresponding 95% confidence interval for the proportion who said “yes” was (47%, 55%). Circle all correct statements based on these results:

a) 95% of the people in the survey report their approval of the job the President is doing to be somewhere between 47% to 55%.

b) We are 95% confident that between 47% and 55% of the people in the study conducted by Gallup in September 2014 approve of the job that President Obama is doing.

c) It is plausible that the population proportion is equal to 49%.

d) The researchers used a method that fails to give a correct answer 5% of the time. e) The population is the 1017 American adults surveyed.

Answer 1

#1.

a)

H0: p = 0.56

Ha: p not equals to 0.56

b)

pcap = 60/94 = 0.6383

SE = sqrt(0.56*0.44/94) = 0.0512

Test statistic, z = (0.6383 - 0.56)/0.0512 = 1.5293

p-value = 0.1262

c)

As p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis.

d)

The percentage of "fist generation" students has not changed since he arrived.

#2.

n = 94

p = 0.6383

z-value of 95% CI = 1.9600

SE = sqrt(p*(1-p)/n)

= sqrt(0.6383*(1-0.6383)/94)

= 0.04956

ME = z*SE

= 1.96*0.04956

= 0.09713

Lower Limit = p - ME = 0.6383 - 0.09713 = 0.54116

Upper Limit = p + ME = 0.6383 + 0.09713 = 0.73543

CI = (54.12%, 73.54%)

95% CI (0.5412 , 0.7354 )