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 Lenâ€TMs exceed, on average, those at Bobâ€TMS by more than 8.50 dollars? The data is given in the file below. Use a = 0.09 for all calculations. Download .csv file And is summarized here Bobs Booze Lens Liquor Emporium n 46 39 average cost 49.97 58.81 variance of cost 12.53 3.08 (a) Form the correct hypothesis A. Ho : M Len – H Bob = 8.5 HA : M Len – u Bob > 8.5 B. H. : H Len = M Bob HA : H Len < M Bob C. Ho : Len – Bob = 8.5 HA : H Len – Bob # 8.5 D. H : H Len = M Bob HA : H Len > M Bob E. H. : H Len – H Bob = 8.5 HA : H Len – H Bob < 8.5 F. Ho : H Len = Bob HA : M Len + U Bob (b) (i) 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 Bobs sales than in the Lens. C. There appears to be more variation in Lens sales than in the Bobs. (b) (ii) Report the p-value of the test you ran in (b)(i) use at least three decimals in your answer. P-value = (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 at least two decimals in your answer. Value of the test statistic = (d) Determine the P-value for the difference of means test, and report it to at least three decimal places. P= !!!

Answer 1

Note: The null hypothesis is always equal to a value and the alternative hypothesis expresses what we are trying to test.

So the correct hypothesis is:

a) H0: U Len - U Bob = 8.5 vs HA: U Len - U Bob>8.5

b) i) from the data the variances look unequal. There appears to be more variation in Bob's than lens

The test to see if the variances are equal is the F test.

Test statistic: (S1^2)/(S2^2) = (12.53)/(3.08) = 4.068

Df = n1-1, n2-1 = 45,38

ii)P-value of this two sided F test: 2P(F45,38> 4.068) = 0.000024

Since the p-value <significance level of 0.09 we can conclude that the variance are unequal.

c) Since variances are unequal we carry out the t test as follows:

Test statistic: 58.81- 49.97/√(3.08^2/39 + 12.53^2/46) = 4.623

d) df for the t test = n1+n2-2 = 83

P-value of this one sided t test is given as: P(t83>4.623) <0.000 (found by extrapolation from the t-tables for df = 83 at alpha = 0.001 and 0.005)