Respuesta :
Answer:
99% confidence interval for the average start up cost is [76.575 , 123.425].
Step-by-step explanation:
We are given that of the 14 stores Victoria investigated, the average start up cost is 100 thousand dollars with a standard deviation of 29.1 thousand dollars.
So, the pivotal quantity for 99% confidence interval for the population average start up cost is given by;
P.Q. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average start up cost = 100 thousand dollars
[tex]\sigma[/tex] = sample standard deviation = 29.1 thousand dollars
n = sample of stores = 14
[tex]\mu[/tex] = population average start up cost
So, 99% confidence interval for the average start up cost, [tex]\mu[/tex] is ;
P(-3.012 < [tex]t_1_3[/tex] < 3.012) = 0.99
P(-3.012 < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 3.012) = 0.99
P( [tex]-3.012 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]3.012 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.99
P( [tex]\bar X-3.012 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X +3.012 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-3.012 \times {\frac{s}{\sqrt{n} }[/tex] , [tex]\bar X +3.012 \times {\frac{s}{\sqrt{n} }[/tex] ]
= [ [tex]100-3.012 \times {\frac{29.1}{\sqrt{14} }[/tex] , [tex]100+3.012 \times {\frac{29.1}{\sqrt{14} }[/tex] ]
= [76.575 , 123.425]
Therefore, 99% confidence interval for the population average start up cost for a candy store is [76.575 , 123.425].
