Question

Refer to the accompanying data set of mean​ drive-through service times at dinner in seconds at two fast food restaurants. Construct a 90​% confidence interval estimate of the mean​ drive-through service time for Restaurant X at​ dinner; then do the same for Restaurant Y. Compare the results.

Construct a 90​% confidence interval of the mean​ drive-through service times at dinner for Restaurant X. nothing secless than muless than nothing sec ​(Round to one decimal place as​ needed.)

Restaurant X	Restaurant Y
88	102
121	121
115	157
150	113
261	174
184	139
127	111
150	128
161	126
209	127
330	134
304	134
172	230
113	210
154	289
149	131
97	99
234	136
238	238
183	146
154	143
196	200
165	149
122	145
63	134
202	144
177	156
112	132
145	168
171	134
186	246
198	231
236	253
192	236
349	235
301	167
211	84
194	108
180	51
184	171
110	75
148	146
175	139
154	103
166	130
154	147
165	135
123	185
138	149
309	132

Answer 1

A 90% confidence interval for Restaurant X at dinner.

Sample size = n = 50

Sample mean = = 178.4

Standard deviation = s = 61.8916

We have to construct 90% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 50 - 1 = 49

Level of significance = 0.10

tc = 1.677    ( Using t table)

So confidence interval is ( 178.4 - 14.6745 , 178.4 + 14.6745) = > ( 163.7255 , 193.0745)

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A 90% confidence interval for Restaurant Y at dinner.

Sample size = n = 50

Sample mean = = 153.46

Standard deviation = s = 49.5165

We have to construct 90% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 50 - 1 = 49

Level of significance = 0.10

tc = 1.677    ( Using t table)

So confidence interval is ( 153.46 - 11.7404 , 153.46 + 11.7404) = > ( 141.7196 , 165.2004)