Question

A computer manufacturing company has sent a mail survey to 2,800 of its randomly selected customers that have purchased a new laptop. The survey asked the customers whether or not they were satisfied with the computer. 800 customers responded to the survey. 640 customers indicated that they were satisfied, while 160 customers indicated they were not satisfied with their new computer. Construct a 96 percent confidence interval estimate of the true proportion of customers satisfied with their new computer.

Answer 1

Answer:

The 96% confidence interval for the population proportion of customers satisfied with their new computer is (0.77, 0.83).

Step-by-step explanation:

We have to calculate a 96% confidence interval for the proportion.

We consider the sample size to be the customers that responded the survey (n=800), as we can not assume the answer for the ones that did not answer.

The sample proportion is p=0.8.

$p=X/n=640/800=0.8$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.8*0.2}{800}}\\\\\\ \sigma_p=\sqrt{0.0002}=0.014$

The critical z-value for a 96% confidence interval is z=2.054.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=2.054 \cdot 0.014=0.03$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.8-0.03=0.77\\\\UL=p+z \cdot \sigma_p = 0.8+0.03=0.83$

The 96% confidence interval for the population proportion is (0.77, 0.83).