Respuesta :
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;
[tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
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, [tex]\mu[/tex] is given by;
P(-2.032 < [tex]t_3_4[/tex] < 2.032) = 0.95
P(-2.032 < [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.032) = 0.95
P(-2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]{Xbar - \mu}[/tex] < 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95
P(X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] , X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ]
= [ 1080 - 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] , 1080 + 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] ]
= [ 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.
