Question

Decide whether the normal sampling distribution can be used. If it can be​ used, test the claim about the population proportion p at the given level of significance using the given sample statistics.
Claim: p>0.12; α=0.05; Sample statistics: Modifying above p with caret equals 0.08, n is equal to 250

Answer 1

Answer:

There is no sufficient evidence to support the claim

Step-by-step explanation:

From the question we are told that

     The level of significance is  $\alpha = 0.05$

     The  sample proportion is  $\r p = 0.08$

     The  sample size is  $n = 250$

     

Generally for normal sampling distribution can  be used

     $n * p > 5$

So  

     $n* p = 250 * 0.12 = 30$

Since  

     $n * p > 5$ then  normal sampling distribution can  be used

The null hypothesis is  $H_o : p = 0.12$

  The alternative hypothesis is  $H_a : p > 0.12$

The  test statistic is evaluated as

            $t = \frac{\r p - p }{ \sqrt{ \frac{p(1- p)}{n} } }$

substituting values

            $t = \frac{0.08 - 0.12 }{ \sqrt{ \frac{0.12 (1- 0.12)}{250 } } }$

            $t = -1.946$

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

        $p-value = P(t > -1.9462) =0.97512$

Since the $p-value > \alpha$

    Then we fail to reject the null hypothesis

Hence it means there is no sufficient evidence to support the claim