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).
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], 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:
[tex]\pi = \frac{179}{523} = 0.3423[/tex]
Then the bounds of the interval are found as follows:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3423 - 1.96\sqrt{\frac{0.3423(0.6577)}{523}} = 0.3016[/tex]
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3423 + 1.96\sqrt{\frac{0.3423(0.6577)}{523}} = 0.3830[/tex]
More can be learned about the z-distribution at https://brainly.com/question/25890103