The owner of a pizza restaurant in France knows that the time customers spend in the restaurant on Saturday evening has mean 9090 minutes and standard deviation 1515 minutes. He has read that pleasant odors can influence customers, so he spreads a lavender odor throughout the restaurant. The times (in minutes) for customers on the next Saturday evening are given in the table. 9292 126126 114114 106106 8989 137137 9393 7676 9898 108108 124124 105105 129129 103103 107107 109109 9494 105105 102102 108108 9595 121121 109109 104104 116116 8888 109109 9797 101101 106106 Click to download the data in your preferred format. CSV Excel JMP Mac-Text Minitab14-18 Minitab18+ PC-Text R SPSS TI CrunchIt! A stemplot of the data is given. The distribution is roughly symmetric and single‑peaked, so the distribution of ?⎯⎯⎯�¯ should be close to Normal. Stem Leaf 77 66 88 88 88 99 99 22 33 44 99 55 77 88 1010 11 22 33 44 1010 55 55 66 66 77 88 88 99 99 99 1111 44 1111 66 1212 11 44 1212 66 99 1313 1313 77 Nicolas Guéguen and Christine Petr, "Odors and consumer behavior in a restaurant," International Journal of Hospitality Management, 25 (2006), pp. 335–339. SOLVE: What is the value of the test statistic ??�? Suppose that the standard deviation ?=15�=15 minutes is not changed by the odor. Also assume that there is an SRS. Give your answer to two decimal places. ?=
The owner of a pizza restaurant in France knows that the time customers spend in the restaurant on Saturday evening has
9292 | 126126 | 114114 | 106106 | 8989 | 137137 | 9393 | 7676 | 9898 | 108108 |
124124 | 105105 | 129129 | 103103 | 107107 | 109109 | 9494 | 105105 | 102102 | 108108 |
9595 | 121121 | 109109 | 104104 | 116116 | 8888 | 109109 | 9797 | 101101 | 106106 |
Click to download the data in your preferred format.
CSV Excel JMP Mac-Text Minitab14-18 Minitab18+ PC-Text R SPSS TI CrunchIt!
A stemplot of the data is given. The distribution is roughly symmetric and single‑peaked, so the distribution of ?⎯⎯⎯�¯ should be close to Normal.
Stem | Leaf | |||||||||
---|---|---|---|---|---|---|---|---|---|---|
77 | 66 | |||||||||
88 | ||||||||||
88 | 88 | 99 | ||||||||
99 | 22 | 33 | 44 | |||||||
99 | 55 | 77 | 88 | |||||||
1010 | 11 | 22 | 33 | 44 | ||||||
1010 | 55 | 55 | 66 | 66 | 77 | 88 | 88 | 99 | 99 | 99 |
1111 | 44 | |||||||||
1111 | 66 | |||||||||
1212 | 11 | 44 | ||||||||
1212 | 66 | 99 | ||||||||
1313 | ||||||||||
1313 | 77 |
SOLVE: What is the value of the test statistic ??�? Suppose that the standard deviation ?=15�=15 minutes is not changed by the odor. Also assume that there is an SRS. Give your answer to two decimal places.
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