Question

A survey of undergraduates revealed the follwoing information: WOMEN MENsample mean weight 124.7 183.3sample standard deviation of weight 23.32 25.41sample proportion Roman Catholic 0.40 0.32Sample mean GPA 3.34 3.24Sample standard deviation of GPA 0.35 0.44Sample size 20 25Assume the populations are normally distributed. Suppose you want to determine whether the proportion of SCU women who are Roman Catholic is greater than the proportion of SCU men that are Roman Catholic.a. What are the null and alternative hypothesis to run this test?b. What is the calculated value of the test statistic?c. What is the p-value of the calculated test statistic?d. What is the conclusion of the hypothesis test, at 5% the significance level?

Answer 1

Answer:

the answers are below:

Step-by-step explanation:

a. null hypothesis:

H0: Pw - Pm = 0 (so Pw = Pm)

alternate hypothesis:

H1: Pw - Pm > 0 (so Pw > Pm)

where Pw is the proportion of women

Pm is the proportion of men

b.) proportion of women = o.40

proportion of men =  0.32

sample size of women = 20

sample size of men = 25

$z = 0.4 - 0.32/ \sqrt{((0.4 *0.6)/20) * (0.32 * 0.68)/25)}$

$z = 0.56$

c.) p value =

p(z>0.56)

= 0.7123

= 1 - 0.7123

= o.2877 which can be approximated to be 0.288

d. alpha value was set at 0.05

the p value is greater than alpha.

therefore it is not statistically significant.

we conclude that the proportion of roman catholic women is not greater than men.