Question

Decide whether the normal sampling distribution can be used. If it can be​ used, test the claim about the difference between two population proportions p 1p1 and p 2p2 at the given level of significance alphaα using the given sample statistics. Assume the sample statistics are from independent random samples. ​Claim: p 1p1equals=p 2p2​, alphaαequals=0.100.10 Sample​ statistics: x 1x1equals=5959​, n 1n1equals=171171 and x 2x2equals=3636​, n 2n2equals=203203 Can a normal sampling distribution be​ used? No Yes
Identify the null and alternative hypotheses.

Answer 1

Answer:

Null Hypothesis, $H_0$ :  $p_1=p_2$  or $p_1-p_2=0$   

Alternate Hypothesis, $H_a$ : $p_1\neq p_2$  or  $p_1-p_2\neq 0$  

Yes, normal sampling distribution can be​ used.

Step-by-step explanation:

We are given that the sample statistics are from independent random samples. ​Claim: $p_1$ = $p_2$ ​, alpha(α) = 0.10 Sample​ statistics: $x_1$ =59​, $n_1$ =171 and $x_2$ =36​, $n_2$ = 203

We have to test the claim about the difference between two population proportions $p_1$ and $p_2$ at the given level of significance alpha(α) using the given sample statistics.

Let $p_1$ = population proportion of first group

     $p_2$ = population proportion of second group

SO, Null Hypothesis, $H_0$ :  $p_1=p_2$  or $p_1-p_2=0$   

Alternate Hypothesis, $H_a$ : $p_1\neq p_2$  or  $p_1-p_2\neq 0$  

The test statistics that will be used here is Two-sample z proportion statistics;

              T.S. = $\frac{(\hat p_1 - \hat p_2)-(p_1-p_2)}{ \sqrt{\frac{\hat p_1(1-\hat p_1) }{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2}} }$  ~ N(0,1)

where, $\hat p_1 = \frac{x_1}{n_1}$ = $\frac{59}{171}$ = sample proportion of first group

            $\hat p_2 = \frac{x_2}{n_2}$ = $\frac{36}{203}$ = sample proportion of second group

             $n_1$ = sample size of first group = 171

             $n_2$ = sample size of second group = 203

So, Yes here normal sampling distribution can be used because the proportion test is approximately followed by a normal z distribution.