Question

A clothing company produces denim jeans. The jeans are made and sold with either a regular cut or a boot-cut. To estimate the proportion of all customers in Tacoma, WA, who prefer boot-cut jeans, a marketing researcher examined sales receipts for a random sample of 178 customers who purchased jeans from the firm’s Tacoma store. 56 of the customers in the sample purchased boot-cut jeans. Construct the 99% confidence interval to estimate the proportion of all customers in Tacoma, Washington, who prefer boot-cut jeans and interpret the confidence interval (please write the interval boundaries to THREE decimal places

Answer 1

Answer:

99% confidence interval for the proportion of all customers in Tacoma, Washington, who prefer boot-cut jeans is [0.225 , 0.405].

Step-by-step explanation:

We are given that a marketing researcher examined sales receipts for a random sample of 178 customers who purchased jeans from the firm’s Tacoma store. 56 of the customers in the sample purchased boot-cut jeans.

Firstly, the Pivotal quantity for 99% confidence interval for the population proportion is given by;

                           P.Q. =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of customers who purchased boot-cut jeans = $\frac{56}{178}$ = 0.315

           n = sample of customers = 178

           p = population proportion of customers who prefer boot-cut jeans

Here for constructing 99% confidence interval we have used One-sample z test for proportions.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99  {As the critical value of z at 0.5% level

                                                 of significance are -2.58 & 2.58}  

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$, $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$]

    = [ $0.315-2.58 \times {\sqrt{\frac{0.315(1-0.315)}{178} } }$ , $0.315+2.58 \times {\sqrt{\frac{0.315(1-0.315)}{178} } }$ ]

    = [0.225 , 0.405]

Therefore, 99% confidence interval for the proportion of all customers in Tacoma, Washington, who prefer boot-cut jeans is [0.225 , 0.405].

The interpretation of the above confidence interval is that we are 99% confident that the proportion of all customers in Tacoma, Washington, who prefer boot-cut jeans will lie between 0.225 and 0.405.