Victoria is interested in opening her own candy

Ответ:

mai1261

21.01.2022

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].

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