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nekit [7.7K]
3 years ago
9

A survey of 50 retail stores revealed that the average price of a microwave was $375 with a sample standard deviation of $20. As

suming 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
Business
1 answer:
ohaa [14]3 years ago
6 0

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

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7 0
3 years ago
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azamat

Answer:

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4 0
3 years ago
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