Answer to Question #99115 in Statistics and Probability for John

Question

The marketing manager of a firm that produces laundry products decides to test market a new laundry product in each of the firm's two sales regions. He wants to determine whether there will be a difference in mean sales per market per month between the two regions. A random sample of 18 supermarkets from Region 1 had mean sales of 87.1 with a standard deviation of 6.5. A random sample of 12 supermarkets from Region 2 had a mean sales of 80.8 with a standard deviation of 7.4. Does the test marketing reveal a difference in potential mean sales per market in Region 2? Let μ1 be the mean sales per market in Region 1 and μ2 be the mean sales per market in Region 2. Use a significance level of α=0.01 for the test. Assume that the population variances are not equal and that the two populations are normally distributed.

Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.

Answer 1

For the first sample,, we have

$N1=18$

$\overline{X1} =87.1$

$S1=6.5$

For the second sample, we have

$N2=12$

$\overline{X2} =80.8$

$S2=7.4$

Null hypothesis, $H0:μ1=μ2$

Alternate hypothesis: $Ha: μ1≠μ2$

Test statistic, $T=(\overline{X1}-\overline{X2})/\sqrt{\smash[b]{(S1^2/N1)+(S2^2/N2)}}$

Putting all the value, we can get the value of test statistic as , $T=2.395$

Now, we have to find the degree of freedom,

For unequal variance degree of freedom is given by,

$υ=(S1^2/N1+S2^2/N2)/{(S1^2/N1)^2/(N1−1)+(S2^2/N2)^2/(N2−1)}$

Putting all the values, the value of v will be $v\approx23$

So, for $\alpha=0..01$ , the critical region will be $t<-2.807 ,t>2.807$

It can be seen that the test statistics does not fall in the critical region, so the null hypothesis cannot be rejected.

To reject the null hypothesis the test statistic should lie in the critical region.