Question

For Questions 7-10 use the data on the "HS AP Data" tab. For Questions 7-9, create a Yes/No variable in the "Pass Rate > 50 (Y/N)" column equal to 1 if the school's"Pass_Rate_All" value is greater than 50 to calculate the following sample statistics.

Sample Proportion P-Hat of schools with Pass Rate > 50 = .35

Sample size n = 51

Variance of Sample Proportion: P-Hat*(1 - P-Hat)/n = 0.0123

9. Using the sample statistics above, provide a test of null hypothesis H0: P = 0.3 versus the alternative hypothesis HA: P < 0.3
with α = 0.2.

a. Z-Critical =

b. P-Hat-Critical =

c. Conclusion (Reject H0/Fail to Reject H0)

Answer 1

Question 9

Solution:

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

H0: p = 0.3 versus Ha: p < 0.3

This is a lower tailed test.

Part a

We are given

Level of significance = α = 0.2

So, critical value by using z-table or excel is given as below:

Z-Critical = -0.8416

Part b

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

n = sample size = 51

p̂ = x/n = 0.35

p = 0.3

q = 1 - p = 0.7

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.35 – 0.30)/sqrt(0.30*0.70/51)

Z = 0.8250

Test statistic = P-Hat-Critical = 0.8250

Part c

Here, test statistic > Z-Critical = -0.8416

So, we do not reject the null hypothesis

Fail to reject H0

There is not sufficient evidence to conclude that the population proportion of the schools with pass rate more than 50 is less than 0.3.