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

Answer:

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

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: $c. 323.40 to 426.60$