n an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected from 64 randomly selected customers. The sample mean was calculated as $27.75 with a sample standard deviation of $4. We wish to develop a 91% confidence interval for the mean amount spent per customer for dinner at this restaurant. Assume that the standard deviation of the amounts spent by ALL customers is $4.30. The resulting 91% confidence interval is given by 27.75±0.91127.75±0.911 Based upon the above confidence interval, can it be concluded – with at least 91% confidence – that the average customer spends more than $25 at the restaurant? A) yes B) No Why? A) because the entire interval is above $25 B) because $27.75 > $25 C) because some numbers in the interval are less than $25 D) Because there isnt enough data
In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected from 64 randomly selected customers. The sample mean was calculated as $27.75 with a sample standard deviation of $4. We wish to develop a 91% confidence interval for the mean amount spent per customer for dinner at this restaurant. Assume that the standard deviation of the amounts spent by ALL customers is $4.30.
The resulting 91% confidence interval is given by 27.75±0.91127.75±0.911
Based upon the above confidence interval, can it be concluded – with at least 91% confidence – that the average customer spends more than $25 at the restaurant?
A) yes
B) No
Why?
A) because the entire interval is above $25
B) because $27.75 > $25
C) because some numbers in the interval are less than $25
D) Because there isnt enough data
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