Question

For the given​ data, (a) find the test​ statistic, (b) find the standardized test​ statistic, (c) decide whether the standardized test statistic is in the rejection​ region, and​ (d) decide whether you should reject or fail to reject the null hypothesis. The samples are random and independent. ​
Claim: μ1 < μ2​, α = 0.01. Sample​ statistics: x^-1 = 1240, n1 = 40, x^-2 = 1200 and n2 = 80. Population​ statistics: σ1 = 65 and σ2 = 110.
(a) The test statistic for μ1 - μ2 is _________.
(b) The standardized test statistic for μ1 - μ2 is __________. (Round to two decimal places as needed.)
(c) Is the standardized test statistic in the rejection region?
O Yes Ο Nο
(d) Should you reject or fail to reject the null hypothesis?

Answer 1

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

  The test statistic for μ1 - μ2 is   $\= x_1 - \= x_2 = 40$

b

  The standardized test statistic for μ1 - μ2 is   $z = 2.5$

c

  No

d

    Fail to reject null hypothesis  $H_O : \mu_1 \ge \mu_2 ; H_a : \mu_1 < \mu_2$ At the 1% significance level , there is insufficient evidence to support the claim

Step-by-step explanation:

From the question we are told that

The given data is  

         $\= x_1 = 1240$

          $\= x_2 = 1200$

           $n_1 = 40$

           $\alpha = 0.01.$

          $n_2 = 80.$

           $\sigma 1 = 65 \ and\ \sigma2 = 110.$

Now the test statistic is mathematically evaluated as

           $\= x_1 - \= x_2 = 1240 -1200$

          $\= x_1 - \= x_2 = 40$

 The standardized test​ statistic is mathematically represented as

        $z = \frac{\= x_1 - \= x_2}{\sqrt{\frac{\sigma_1^2}{n_1} } + \frac{\sigma_2^2}{n_2} }$

substituting values

     $z = \frac{\= 1240 - \= 1200}{\sqrt{\frac{65^2}{40} } + \frac{110^2}{80} }$

     $z = 2.5$

Now  the standardized test​ statistic is not in the rejection region because the z value of  $\alpha$  is  2.33 and  the standardized test​ statistic  is greater than that hence it is not in the rejection region

This implies that the test statisties failed to reject the null hypothesis at significance level of 0.01 , there insufficient evidence to support the claim