Hint: For this problem pay attention to how you determine when a question is ask

Question

Hint: For this problem pay attention to how you determine when a question is asking about the difference between proportions and when it is asking about the difference between means.

To answer this question, complete the following items.

a. (0.25 point) Find the 90% confidence interval for the difference in financial aid rates between men and women and interpret this interval in the context of the problem.

b. (0.25 point) Write the two hypotheses for the problem.

c. (0.25 point) Compute the appropriate statistical hypothesis test and report the test statistic and p-value.

d. (0.25 point) Write the conclusion for the problem regarding the null hypothesis using statistical language.

2.) Given below are samples of enrollment from medical schools that specialize in research and from those that specialize in primary care. You need to decide if there is a difference in the average enrollment for the two specializations.

Research 474     577     605     663     783     467     670     414     813     443     565     696     692     694     277     419     884

Primary Care 783     605     427     728     546     474     371     107     442     587     293     277     662     555     527     320

a. (0.25 point) Find the 95% confidence interval for the difference between the means and interpret the interval in the context of the problem.

b. (0.25 point) Write the two hypotheses for this problem.

c. (0.25 point) Find the test-statistic (i.e. t) and p-value. Sketch the distribution of test-statistics and indicate t and the shaded region corresponding to the p-value.

d. (0.25 point) In the context of the problem, make a conclusion and substantiate with statistical arguments.

Explanation / Answer

Question 1

Part a

Formula for confidence interval for difference between two proportions is given as below:

Confidence interval = (p1 – p2) -/+ z*sqrt[(p1q1/n1) + (p2q2/n2)]

Here,

n1 = 250

n2 = 300

p1 = 200/250

p1 = 0.8

q1 = 1 – p1 = 1 – 0.8 = 0.2

p2 = 225/300

p2 = 0.75

q2 = 1 – p2 = 1 – 0.75 = 0.25

z = 1.28

lower limit = (0.8 – 0.75) – 1.28* sqrt [(0.8*0.2/250)+(0.75*0.25/300)]

lower limit = 0.05 – 1.28*0.03557

lower limit = 0.05 - 0.0455296

lower limit = 0.0044704

upper limit = (0.8 – 0.75) + 1.28* sqrt [(0.8*0.2/250)+(0.75*0.25/300)]

upper limit = 0.05 + 1.28*0.03557

upper limit = 0.05 + 0.0455296

upper limit = 0.0955296

Confidence interval = (0.0044704, 0.0955296)

Part b

The null and alternative hypothesis is given as below:

Null hypothesis: H0: There is no significant difference in financial aid rates between men and women.

Alternative hypothesis: Ha: There is a significant difference in financial aid rates between men and women.

Part c

Here, we have to use the two sample z test for the population proportion.

The test statistic formula is given as below:

Z = (p1 – p2) / sqrt[(p1q1/n1) + (p2q2/n2)]

Z = (0.8 – 0.75) / sqrt [(0.8*0.2/250)+(0.75*0.25/300)]

Z = 0.05 / 0.03557

Z = 1.4057

Test statistic value = 1.4057

P-value = 0.079909741 by using the z-table or excel

Part d

Conclusion

Here, we get the p-value as 0.07991 which is greater than the given level of significance or alpha value 0.01, so we do not reject the null hypothesis that there is no significant difference in financial aid rates between men and women.

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