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

Respuesta :

Answer:

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

Explanation:

Relevant data:

[tex]n=50\\\mu=375\\\sigma=20\\\alpha=0,001[/tex]

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

[tex](1-\alpha).100\%=99\%\\1-\alpha=0.99\\\alpha=0.01[/tex]

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

[tex]Z_{\alpha/2}=Z_{0.005}[/tex]

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

Confidence interval is given by;:

[tex]\=x\±Z_{\alpha/2}\sigma\\375\±Z_{\0.005}(20)\\375\±(2.58)(20)\\375\±51.60[/tex]

[tex]375+51.60=426.60\\375-51.60=323.40[/tex]

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