Answer:
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 ]
<em>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.</em>
Therefore, the store manager believe is not correct.