Question

A random survey of 85 women who were victims of violence found that 28 were attacked by relatives. A random survey of 46 men found that 5 were attacked by relatives. At =α0.10, can it be shown that the percentage of women who were attacked by relatives is greater than the percentage of men who were attacked by relatives? Use p1 for the proportion of women who were attacked by relatives. Use the P-value method with tables.

A) Compute the test.

B) Find the P value.

C) Reject or do not reject the null hypothesis.

D) There is or is not enough evidence to support the claim?

Answer 1

Solution :

For sample 1, we have that the sample size is N1​=85, the number of favorable cases is X1​=28, so then the sample proportion is

For sample 2, we have that the sample size is N2​=46, the number of favorable cases is X2​=5, so then the sample proportion is

The value of the pooled proportion is computed as

Also, the given significance level is α=0.10.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: p1​=p2​

Ha: p1​>p2​

This corresponds to a right-tailed test, for which a z-test for two population proportions needs to be conducted.

(2) Rejection Region

Based on the information provided, the significance level is α=0.10, and the critical value for a right-tailed test is zc​=1.28.

The rejection region for this right-tailed test is R={z:z>1.28}

(3) Test Statistics

The z-statistic is computed as follows:

The p-value is p=P(Z>2.778)=0.0027

(4) The decision about the null hypothesis

Since it is observed that z=2.778>zc​=1.28, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0027, and since p=0.0027<0.10, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho isrejected. Therefore, there is enough evidence to claim that population proportion p1​ is greater than p2​, at the 0.10 significance level

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