Question

A random sample of 2300 adult women indicated that 1341 of them asked for medical assistance last time they felt sick. The same sized sample of men indicated that 1293 of them asked for medical assistance Use the z-values rounded to three decimal places to obtain the answers a) Is it reasonable to conclude that women ask for medical assistance more often than men? Use ? = 0.05. Find the P-value for this test. Round your answer to three decimal places (e.g. 98.765) P-value b) Suppose that pi0.6 and p2 -0.55. Determine the sample size needed to detect this difference with a probability of at least 0.95 nE

Answer 1

Let's used minitab:

Step 1: Click on Stat >>> Basic Statistics >>>2 Proportions...

Step 2: Select Summarized data

Fill the given information

Look the following picture ...

Then click on Option:

Look the following image:

Then click on OK again click on Ok

So we get the following output

From the above output he P - value = 0.081

z = 0.49 , p-value = 0.622

c. Determine the rejection rule:

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.622 > 0.05 so we used 2nd rule.

That is we fail to reject null hypothesis

Conclusion: At 5% level of significance there are not sufficient evidence to say that the sample data indicates there are no evidence that the proportions of homeowners and renters differ.