Question

The owner of a factory wants to estimate the mean cost of manufacturing a product. She takes a random sample of 36 products and finds the sample mean cost to be $90. Suppose the population standard deviation is known to be σ = $10.

A) Create a 90% confidence interval for the population mean.

B) Interpret your confidence interval from part (a).

C) Without calculating another confidence interval, if the sample of 36 products instead had a mean of $200, would a 90% confidence interval be narrower, wider or the same width as your confidence interval from part a)? Briefly explain.

Answer 1

A) Create a 90% confidence interval for the population mean.

Confidence interval for Population mean is given as below:

Confidence interval = x̄ ± Z*σ/sqrt(n)

From given data, we have

x̄ = 90

σ = 10

n = 36

Confidence level = 90%

Critical Z value = 1.6449

(by using z-table)

Confidence interval = x̄ ± Z*σ/sqrt(n)

Confidence interval = 90 ± 1.6449*10/sqrt(36)

Confidence interval = 90 ± 1.6449*1.6667

Confidence interval = 90 ± 2.7414

Lower limit = 90 - 2.7414 = 87.26

Upper limit = 90 + 2.7414 = 92.74

Confidence interval = (87.26, 92.74)

B) Interpret your confidence interval from part (a).

We are 90% confident that the average population cost will be lies between $87.26 and 92.74.

C) Without calculating another confidence interval, if the sample of 36 products instead had a mean of $200, would a 90% confidence interval be narrower, wider or the same width as your confidence interval from part a)? Briefly explain.

Without calculating another confidence interval, if the sample of 36 products instead had a mean of $200, a 90% confidence interval for population mean will be the same width as of the confidence interval in part a, because we know that there will no change in the width of the confidence interval only we change sample mean by keeping other terms same. There would be change if we change sample size, confidence level, or standard deviation.