Question

1. A survey on men and women’s shopping behavior collected data from a sample of 50 men and 50 women. The following data on online shopping spending every month.

Men	Women
148	272
211	176
256	251
309	235
190	145
205	179
203	30
208	135
231	200
125	270
149	174
205	123
195	199
178	195
196	192
198	102
110	110
199	184
181	228
168	316
218	170
222	234
206	163
168	245
239	174
130	126
246	227
149	86
262	96
142	185
174	288
181	154
198	217
147	184
143	154
185	217
200	222
166	175
171	265
133	196
295	172
242	113
299	240
209	235
189	269
173	243
109	131
291	134
208	56
227	164

Treat the men as population 1 and the women as population 2. Formulate the null and alternative hypotheses and test whether there is evidence of a difference in the variances of monthly online shopping spending between men and women. Please follow the hypothesis testing steps in arriving at a decision. You must show all the calculations in the spreadsheet. (15 points)

Answer 1

The provided sample variances are and and the sample sizes are given by and .

The following null and alternative hypotheses need to be tested:

This corresponds to a two-tailed test, for which a F-test for two population variances needs to be used.

The significance level is α=.05, and the the rejection region for this two-tailed test is R={F:F<0.567 or F>1.762}.

Test Statistics

The F-statistic is computed as follows:

As F_lower = 0.567 < F = 0.774 < F_U​pper = 1.762, it is then concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that there is difference in the variances of monthly online shopping spending between men and women.

Excel formulae: Data is in column DC and DD; DC3:DC52; DD3:DD52. F-statistics is ratio of sample variance.

Sample Variance (Men)	Sample Variance (Women)	F-Value	Lower Critical Value	Upper Critical Value
=STDEV.S(DC3:DC52)	=STDEV.S(DD3:DD52)	=DE3/DF3	=F.INV(0.025,49,49)	=F.INV(0.975,49,49)