Question

The average total dollar purchase at a convenience store is less than that at a supermarket. Despite smaller-ticket purchases, convenience stores can still be profitable because of the size of the operation, the volume of business, and the markup. A researcher is interested in estimating the average purchase amount for convenience stores in suburban Long Island. To do so, she randomly sampled 24 purchases from several convenience stores in suburban Long Island and tabulated the amounts to the nearest dollar. Use the following data to construct a 90% confidence interval for the population average amount of purchases. Assume that the population standard deviation is 3.23 and the population is normally distributed.
$2 $11 $8 $7 $9 $3
5 4 2 1 10 8
14 7 6 3 7 2
4 1 3 6 8 4

Answer 1

Answer:

90% confidence interval for the population average amount of purchases is [$4.54 , $6.71].

Step-by-step explanation:

We are given the randomly sampled 24 purchases from several convenience stores in suburban Long Island and tabulated the amounts to the nearest dollar;

2, 11, 8, 7, 9, 3,  5, 4, 2, 1, 10, 8,  14, 7, 6, 3, 7, 2,  4, 1, 3, 6, 8, 4

Assume that the population standard deviation is 3.23 and the population is normally distributed.

Firstly, the Pivotal quantity for 90% confidence interval for the population average is given by;

                             P.Q. =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\bar X$ = sample average amount of purchases = $\frac{\sum X}{n}$  = $5.625

            $\sigma$ = population standard deviation = $3.23

            n = sample of purchases = 24

            $\mu$ = population average amount of purchases

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 90% confidence interval for the population average, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                              of significance are -1.645 & 1.645}  

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

                                      = [ $5.625-1.645 \times {\frac{3.23}{\sqrt{24} } }$ , $5.625+1.645 \times {\frac{3.23}{\sqrt{24} } }$ ]

                                      = [4.54 , 6.71]

Therefore, 90% confidence interval for the population average amount of purchases is [$4.54 , $6.71].