Question

Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .10 significance level. Let p_M and p_F be the proportion of Men and Women who own cats respectively.

Based on a sample of 80 men, 40% owned cats
Based on a sample of 80 women, 51% owned cats
What is the test statistic and the critical value? Reject or Fail to Reject Null hypothesis?

Answer 1

The test statistics is
`t = -1.40`

The critical value is
`Z_(\alpha ) = 2.33`

The null hypothesis is rejected

Explanation:

From the question we are told that

The sample size for men is
`n_1 = 80`

The sample proportion of men that own a cat is
`\r p _M = 0.40`

The sample size for women is
`n_2 = 80`

The sample proportion of women that own a cat is
`\r p_F = 0.51`

The level of significance is
`\alpha = 0.10`

The null hypothesis is
`H_o : \r p _M = \ r P_F`

The alternative hypothesis is
`H_a : \r p _M < \r p_F`

Generally the test statistic is mathematically represented as

`t = \frac{(\r p_M - \r p_F)}{\sqrt{((p_M*(1-p_M))/(n_1 ) } + (p_F*(1-pF))/(n_2 ) }`

=>
`t = \frac{(0.40 - 0.51)}{\sqrt{((0.40 *(1-0.41))/(80) } + (0.51*(1-0.51))/(80 ) }`

=>
`t = -1.40`

The critical value of
`\alpha` from the normal distribution table is

`Z_(\alpha ) = 2.33`

The p-value is obtained from the z-table ,the value is

`p-value = P( Z < -1.40) = 0.080757`

=>
`p-value = 0.080757`

Given that the
`p-value < \alpha` then we reject the null hypothesis