Question

In a sample of 1000 randomly, selected consumers who had opportunities to send in a rebate claim form after purchasing a product, 250 of these people said they never did so. Reasons cited for their behavior included too many steps in the process, amount too small, missed deadline, fear of being placed on a mailing list, lost receipt, and doubts about receiving the money. Calculate a 95% confidence Interval for the true proportion of such consumers who never apply for a rebate.

Answer 1

Answer: $=(0.2232,0.2768)$

Step-by-step explanation:

The confidence interval for population proportion is given by :-

$p\ \pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

Given : Sample size : $n=1000$

The proportion of people selected consumers who had opportunities to send in a rebate claim form after purchasing a product said they never did so =$\dfrac{250}{1000}=0.25$

Significance level : $1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Now, the 95% confidence Interval for the true proportion of such consumers who never apply for a rebate will be :-

$0.25\ \pm (1.96)\sqrt{\dfrac{0.25(1-0.25)}{1000}}\\\\\approx0.25\pm0.0268\\\\=(0.2232,\ 0.2768)$

Hence, a confidence Interval for the true proportion of such consumers who never apply for a rebate .