Question

The average selling price of a smartphone purchased by a random sample of 41customers was ​$296. Assume the population standard deviation was $34.

a. Construct a 90​% confidence interval to estimate the average selling price in the population with this sample.

b. What is the margin of error for this​ interval?

Answer 1

Given Information:

Sample Mean (X) = $ 296

Sample Size (n) = 41

Population Standard Deviation (σ) = $ 34

Confidence interval: 90%

A.

To find: the confidence interval

Following formula can be used to find the confidence interval

The z score at 90% confidence interval is 1.64

Where,

X is the sample mean

Z is the z score

σ is the population standard deviation

n is the sample size

By putting given values in the above formula, we get:

Confidence interval = 296 ± 8.7125

Lower Specification Limit – 296-8.715 = $ 287.285

Upper Specification Limit – 296 + 8.715 = $ 304.715

B. The margin of error: