Question

(1 point) There has been an alcohol war between Bob and Len for some time now. The following results were observed. Does the data indicate that the average purchases at LenÄcAAs exceed on average, those at BobAcAAs by more than 8.50 dollars? The data is given in the file below [ the link to 'Minitab File' is broken ] And is summarized here Bobs Booże Lens Liquor Emporium 46 average cost 52.52 ariance of cost6.99 39 61.9 2.08 (a) Form the correct hypothesis (b) () Using technology available to you, test to see if the variances are equal or not? A. They appear to be equal B. There appears to be more variation in Lens sales than in the Bobs. C. There appears to be more variation in Bobs sales than in the Lens. (b) (ii)Report the p-value of the test you ran in (b)( use three decimals in your answer P-value 0.033 (c) Test the statistical hypotheses in (a) by carrying out the appropriate statistical test. Find the value of the test statistic for this test, use two decimals in your answen Absolute value of the test statistic 32.53 (d) Determine the P-value for the difference of means test, and report it to three decimal places P 1.000 (e) Determine the appropriate degrees of freedom for this test. As an integer. DF- 82 () Based on the above calculations, we should reject the null hypothesis. Use a-0.045

Answer 1

Solution:-

a)

(C) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_Len - u_Bob< 8.5
Alternative hypothesis: u_Len - u_Bob > 8.5

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.045. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.45309
DF = 83

t = [ (x₁ - x₂) - d ] / SE

t = 1.94

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of 1.94.

Therefore, the P-value in this analysis is 0.028.

Interpret results. Since the P-value (0.028) is less than the significance level (0.045), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that averages purchases at Len exceed on average, those at Bob more than 8.50 dollars.