Question

Hello Everyone!

For this week’s discussion, I have chosen option 2. I would like to investigate the proportion of women, aged 22-30 and whom are in the U.S. Coast Guard who have families. A study conducted in 2012 reported that 39% of women aged 22-30 whom are active duty in the U.S. Coast Guard have families.

Take a random sample of 150 and determined that 42 out of 150 women had families.

H* _{0}* = .39

H* _{1}*> .39

Test this hypothesis at a significance level of 0.03.

**PLEASE USE STAT CRUNCH AND SHOW AS MUCH WORK AS
POSSIBLE. ALSO PLEASE DO NOT HANDWRITE THE ANSWER AS I NEED TO COPY
AND PASTE INTO A DISCUSSION BOARD.**

Answer #1

The steps in Statcrunch would be:

Stat --> Proportion Stats --> One Sample --> Summary

Then you need to put inputs as below:

The results you will get is:

From above output:

p - Value = 0.9971

Since this p - value is greater than 0.03 significance level, we do not reject the null hypothesis. Hence,

Conclusion: There is not sufficient evidence to conclude that p > 0.39

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