1. In order to get more female customers, a new clothing store offers free gourmet coffee and pastry to its customers. The average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260. Use this sample information to construct a 95% confidence interval for the average daily revenue. The store manager believes that the coffee and pastry strategy would lead to an average daily revenue of $1,200. Is the manager correct based on the 95% confidence interval?

No, the manager is not correct based on the 95% confidence interval.

Step-by-step explanation:

We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .

The Pivotal quantity for 95% confidence interval is given by;

                ~

where, X bar = sample mean = $1080

                s  = sample standard deviation = $260

                 n = sample size = 35 {five-week}

So, 95% confidence interval for average daily revenue, is given by;

P(-2.032 < < 2.032) = 0.95

P(-2.032 < < 2.032) = 0.95

P(-2.032 * < < 2.032 * ) = 0.95

P(X bar - 2.032 * < < X bar + 2.032 * ) = 0.95

95% confidence interval for = [ X bar - 2.032 * , X bar + 2.032 * ]

                                            = [ 1080 - 2.032 * , 1080 + 2.032 * ]

                                             = [ 990.70 , 1169.30 ]

No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.

Therefore, the store manager believe is not correct.

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step-by-step explanation:

answer:

a=9

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