In: Statistics and Probability
Refer to the accompanying data set of mean drive-through service times at dinner in seconds at two fast food restaurants. Construct a 99%
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.
Restaurant
X Y
79 103
118 129
119 152
150 113
262 175
176 131
125 109
149 127
167 131
210 128
329 133
308 137
178 225
110 217
155 291
146 128
98 90
237 140
244 244
189 143
149 148
202 205
170 145
117 141
60 140
205 147
182 161
119 130
139 172
168 132
186 236
199 232
232 249
193 237
354 229
302 168
212 85
192 107
182 52
192 169
105 80
152 140
175 140
156 94
176 129
160 144
168 128
122 184
138 148
312 125
99% confidence interval for restaurant x at dinner:
Sample size = n = 50
Sample mean = = 179.36
Standard deviation = s = 63.2161
We have to construct 99% confidence interval.
Formula is
Here E is a margin of error.
Degrees of freedom = n - 1 = 50 - 1 = 49
Level of significance = 0.01
tc = 2.680 ( Using t table)
So confidence interval is ( 179.36 - 23.9591 , 179.36 + 23.9591) = > ( 155.4009 , 203.3191)
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99% confidence interval for restaurant x at dinner:
Sample size = n = 50
Sample mean = = 152.86
Standard deviation = s = 49.6572
We have to construct 99% confidence interval.
Formula is
Here E is a margin of error.
Degrees of freedom = n - 1 = 50 - 1 = 49
Level of significance = 0.01
tc = 2.680 ( Using t table)
So confidence interval is ( 152.86 - 18.8202 , 152.86 + 18.8202) = > ( 134.0398 , 171.6802)
Restaurant X has large confidence interval as compare to restaurant Y.