Question

A survey of 50 retail stores revealed that the average price of a microwave was $375 with a sample standard deviation of $20. Assuming the population is normally distributed, what is the 99% confidence interval to estimate the true cost of the microwave?

a. $367.42 to $382.58
b. $315.00 to $415.00
c. $323.40 to $426.60
d. $335.82 to $414.28

Answer 1

True cost of the microwave is in 99% confidence interval:
`c. $323.40 to $426.60`

Step-by-step explanation:

Relevant data:

`n=50\\\mu=375\\\sigma=20\\\alpha=0,001`

As we want to know the 99% confidence interval, the significance level is:

`(1-\alpha).100\%=99\%\\1-\alpha=0.99\\\alpha=0.01`

We need to estimate a confidence interval by a two tailed normal bell. Then we have:

`Z_(\alpha/2)=Z_(0.005)`

The z-value for a probability of 0.005 in a normal standard distribution is 2.576

Confidence interval is given by;:

`\=x\±Z_(\alpha/2)\sigma\\375\±Z_(\0.005)(20)\\375\±(2.58)(20)\\375\±51.60`

`375+51.60=426.60\\375-51.60=323.40`

True cost of the microwave is in 99% confidence interval:
`c. $323.40 to $426.60`