Question

A ski company in Vail owns two ski shops, one on the west side and one on the east side of Vail. Ski hat sales data (in dollars) for a random sample of 5 Saturdays during the 2004 season showed the following results. Is there a significant difference in sales dollars of hats between the west side and east side stores at the 10 percent level of significance?

Saturday Sales Data ($) for Ski Hats

Saturday East Side Shop West Side Shop
1 548 523
2 493 721
3 609 695
4 567 510
5 432 532

Required:
a. State the decision rule for 5 percent level of significance.
b. Find the test statistic tcalc.

Answer 1

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,
`\overline x_1` = 529.8

The standard deviation for the East Side Shop, s₁ = 68.751

The mean for the West Side Shop,
`\overline x_2` = 596.2

The standard deviation for the West Side Shop, s₂ = 102.7701

`t_0=\frac{(\bar{x}_(1)-\bar{x}_(2))}{\sqrt{(s_1^(2) )/(n_(1))+\frac{s{2}^(2)}{n_(2)}}}`

Therefore, we have;

The null hypothesis is H₀; μ₂ - μ₁ = 0

The alternative hypothesis is Hₐ; μ₂ - μ₁ ≠ 0

`t_0=\frac{(529.8-596.2)}{\sqrt{(68.751^(2) )/(5)+(102.7701^(2))/(5)}} \approx -1.2008`

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;

`t_0=\frac{(\bar{x}_(1)-\bar{x}_(2))}{\sqrt{(s_1^(2) )/(n_(1))+\frac{s{2}^(2)}{n_(2)}}}`

`t_0=\frac{(529.8-596.2)}{\sqrt{(68.751^(2) )/(5)+(102.7701^(2))/(5)}} \approx -1.2008`