Answer:
At 10 percent significance level, there is not enough statistical evidence to suggest that there is a difference between mean
a. The decision rule for 5% level of confidence is;
For test statistic is less than 2.306 we fail to reject the null hypothesis
For a test statistic larger than 2.306 we reject the null hypothesis
b. The test statistic is approximately -1.2008
Explanation:
The given data are;
No; 1, 2, 3, 4, 5
East Side Shop; 548, 493, 609, 567, 432
West Side Shop; 523, 721, 695, 510, 532
Therefore, we have;
The mean for the East Side Shop,
= 529.8
The standard deviation for the East Side Shop, s₁ = 68.751
The mean for the West Side Shop,
= 596.2
The standard deviation for the West Side Shop, s₂ = 102.7701
Therefore, we have;
The null hypothesis is H₀; μ₂ - μ₁ = 0
The alternative hypothesis is Hₐ; μ₂ - μ₁ ≠ 0
At degrees of freedom, df = n₁ + n₂ - 2 = 8, we have the critical-t = 1.895
Given that the critical-t is more than test statistic, we do not reject the null hypothesis, therefore, there is not enough statistical evidence to suggest that there is a difference between mean
Required;
a. At 5% level, we have critical-t = 2.306
Therefore, the decision rule for 5% level of confidence is that we fail to reject the null hypothesis when the magnitude of the test statistic is less than 2.306 and we reject the null hypothesis for a test statistic larger than 2.306
b. The test statistic is given as follows;