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

Answer:

 The  test statistics is $t = -1.40$

  The critical value is  $Z_{\alpha } = 2.33$

   The null hypothesis is rejected

Step-by-step 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

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

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

=>   $t = -1.40$

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