Question

Assume that the following data were generated using a valid and appropriate methodology. XYZ Data Analytics, Inc. Received properly completed surveys from 523 members of the target population for a new product. One hundred seventy nine of those respondents indicated they would buy the new product at the proposed price

Answer 1

Using the z-distribution, as we are working with a proportion, the 95% confidence interval for the proportion of consumers who would buy the product at it's proposed price is (0.3016, 0.3830).

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

In this problem, we have a 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so the critical value is z = 1.96.

179 out of 523 members indicated they would buy the new product at the proposed price, hence:

$\pi = \frac{179}{523} = 0.3423$

Then the bounds of the interval are found as follows:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3423 - 1.96\sqrt{\frac{0.3423(0.6577)}{523}} = 0.3016$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3423 + 1.96\sqrt{\frac{0.3423(0.6577)}{523}} = 0.3830$

More can be learned about the z-distribution at brainly.com/question/25890103