Question

In a random sample of 13 microwave​ ovens, the mean repair cost was ​$85.00 and the standard deviation was ​$15.30. Using the standard normal distribution with the appropriate calculations for a standard deviation that is​ known, assume the population is normally​ distributed, find the margin of error and construct a 95​% confidence interval for the population mean. A 95​% confidence interval using the​ t-distribution was (75.8, 94.2 ). Compare the results.

Answer 1

Answer: Margin of error = 8.32, and Confidence interval using normal distribution is narrower than confidence interval using t-distribution.

Step-by-step explanation:

Since we have given that

n = 13

Mean repair cost = $85.00

Standard deviation = $15.30

At 95% confidence interval,

z= 1.96

Since it is normally distributed.

Margin of error is given by

$z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{15.30}{\sqrt{13}}\\\\=8.32$

95% confidence interval would be

$\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=85\pm 1.96\times \dfrac{15.30}{\sqrt{13}}\\\\=85\pm 8.32\\\\=(85-8.32,85+8.32)\\\\=(76.68,93.32)$

A 95​% confidence interval using the​ t-distribution was (75.8, 94.2 ).

Confidence interval using normal distribution is narrower than confidence interval using t-distribution.