Question

Chegg construct the 95% confidence interval for the difference between the positive rates of men and women.

Answer 1

Complete Question

In a random sample of 700 men tested for the coronavirus, 63 were positive. Another independent random sample of 2950 women tested for the coronavirus resulted in 7 positive cases.Construct the 95% confidence interval for the difference between the positive rates of men and women

Answer:

The 95% confidence interval is  $0.833< p_1 - p_2 < 0.1059$

Step-by-step explanation:

From the question we are told that

   The sample size of men  is  $n_1 = 700$

   The number men that tested positive is   $x_1 = 63$

    The sample size of women is $n_2 = 2950$

    The number of women that tested positive is  $x_2 = 7$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the proportion of men that tested positive is mathematically represented as

        $\^ p_1 = \frac{ x_1 }{n_1}$

=>    $\^ p_1 = \frac{ 68 }{700}$

=>    $\^ p_1 = 0.097$

Generally the proportion of women that tested positive is mathematically represented as

        $\^ p_2 = \frac{ x_2 }{n_2}$

=>    $\^ p_2 = \frac{ 7 }{2950}$

=>    $\^ p_2 = 0.00237$

Generally the pooled population proportion is mathematically represented as

         $\^ p = \frac{x_1 + x_2 }{ n_1 + n_2}$

=>       $\^ p = \frac{ 63 + 7 }{ 700 + 2950}$

=>      $\^ p = 0.0192$

Generally the standard error is mathematically represented as

     $SE = \sqrt{\^ p (1- \^ p ) [ \frac{1}{n_1} + \frac{1}{n_2} ]}$

=>   $SE = \sqrt{ 0.0192(1- 0.0192 ) [ \frac{1}{700} + \frac{1}{2950} ]}$

=>   $SE = 0.00577$

Generally the margin of  error is mathematically represented as

    $E = Z_{\frac{\alpha }{2} } * SE$

=> $E = 1.96 * 0.00577$

=> $E = 0.0113$

Generally 95% confidence interval is mathematically represented as  

      $(\^ p_1 - \^ p_2 )-E < p_1 - p_2 < \^( p_1 - \^ p_2) + E$

=>   $(0.097 - 0.00237 )-0.0113< p_1 - p_2 < (0.097 - 0.00237 )+ 0.0113$

=>   $0.833< p_1 - p_2 < 0.1059$