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 value.

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:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁ = P₂
Alternative hypothesis: P₁ > P₂

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = 0.2519

SE = 0.07946

A)

z = 2.78

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 2.78.

B)

P-value = P(z > 2.78)

Use z-calculator or z-table to determine the p-values.

P-value = 0.0027

Interpret results. Since the P-value (0.0027) is less than the significance level (0.10), we have to reject he null hypothesis.

(C) Reject the null hypothesis.

(D) There is enough evidence to support the claim that the percentage of women who were attacked by relatives is greater than the percentage of men who were attacked by relatives.