Question

A market researcher for Eric’s Electronics wants to study TV viewing habits of residents in Chicago. The individuals in the survey are asked to keep track of their weekly TV viewing time. A random sample of 50 respondents is selected, and the average viewing time per week for the 50 individuals in the sample is 17.5 hours. The population standard deviation is known to be 5.0 hours. Assume that TV viewing is a normally distributed random variable. (7 points) Construct a 90% confidence interval estimate for the mean amount of television watched per week by individuals in Chicago. Interpret your result.

Answer 1

Explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Since the sample size is large and the population standard deviation is known, we would use the following formula and determine the z score from the normal distribution table.

Margin of error = z × σ/√n

Where

σ = population standard Deviation

n = number of samples

From the information given

1) x = 17.5

σ = 5

n = 50

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.9 = 0.1

α/2 = 0.1/2 = 0.05

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.05 = 0.95

The z score corresponding to the area on the z table is 1.645. Thus, confidence level of 90% is 1.645

Margin of error = 1.645 × 5/√50 = 1.16

Confidence interval = 17.5 ± 1.16