Question

Victoria is interested in opening her own candy store. She has researched and found that of the 14 stores she investigated, the average start up cost is 100 thousand dollars with a standard deviation of 29.1 thousand dollars. Construct a 99% confidence interval for the average start up cost for a candy store.

Answer 1

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. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average start up cost = 100 thousand dollars

             $\sigma$ = sample standard deviation = 29.1 thousand dollars

             n = sample of stores = 14

             $\mu$ = population average start up cost

So, 99% confidence interval for the average start up cost, $\mu$ is ;

P(-3.012 < $t_1_3$ < 3.012) = 0.99

P(-3.012 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 3.012) = 0.99

P( $-3.012 \times {\frac{s}{\sqrt{n} }$ < ${\bar X - \mu}$ < $3.012 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

P( $\bar X-3.012 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +3.012 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-3.012 \times {\frac{s}{\sqrt{n} }$ , $\bar X +3.012 \times {\frac{s}{\sqrt{n} }$ ]

                                                 = [ $100-3.012 \times {\frac{29.1}{\sqrt{14} }$ , $100+3.012 \times {\frac{29.1}{\sqrt{14} }$ ]

                                                 = [76.575 , 123.425]

Therefore, 99% confidence interval for the population average start up cost for a candy store is [76.575 , 123.425].