Question

A sample of 129 potential customers was asked to use your new product and the product of the leading competitor. After one week, they were asked to indicate which product they preferred. Let X = the number of customers who said that they preferred your product. In the sample, X = 44. What is the lower endpoint of a 95% confidence interval for p, the proportion of potential customers who prefer your product to that of your competitor? (Apply the large-sample confidence interval procedure. You will need to calculate z* in Excel, and do not round in your intermediate calculations.) Express your answer in decimal form to two decimal places of accuracy.

Answer 1

Answer:

Given,

To determine the lower bound

n = 129

x = 44

sample proportion = x/n

= 44/129

= 0.3411

p = 0.3411

Now here for the 95% confidence interval , z value is 1.96

So consider,

Margin of error E = z*sqrt(p(1-p)/n)

substitute the values

= 1.96*sqrt(0.3411(1-0.3411)/129)

= 0.0818

Lower bound = p - E

= 0.3411 - 0.0818

= 0.2593

= 0.26

Lower bound = 0.26

Upper bound = 0.3411 + 0.0818

= 0.4229

= 0.42

Upper bound = 0.42