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?

Fail to reject null hypothesis [tex]H_O : \mu_1 \ge \mu_2 ; H_a : \mu_1 < \mu_2[/tex] 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

[tex]\= x_1 = 1240[/tex]

[tex]\= x_2 = 1200[/tex]

[tex]n_1 = 40[/tex]

[tex]\alpha = 0.01.[/tex]

[tex]n_2 = 80.[/tex]

[tex]\sigma 1 = 65 \ and\ \sigma2 = 110.[/tex]

Now the test statistic is mathematically evaluated as

[tex]\= x_1 - \= x_2 = 1240 -1200[/tex]

[tex]\= x_1 - \= x_2 = 40[/tex]

The standardized test statistic is mathematically represented as

[tex]z = \frac{\= x_1 - \= x_2}{\sqrt{\frac{\sigma_1^2}{n_1} } + \frac{\sigma_2^2}{n_2} }[/tex]

substituting values

[tex]z = \frac{\= 1240 - \= 1200}{\sqrt{\frac{65^2}{40} } + \frac{110^2}{80} }[/tex]

[tex]z = 2.5[/tex]

Now the standardized test statistic is not in the rejection region because the z value of [tex]\alpha[/tex] 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