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In a random sample of microwave​ ovens, the mean repair cost was ​$ and the standard deviation was ​$. 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 ​% confidence interval for the population mean μ. A 98​% confidence interval using the​ t-distribution was (687,913). Compare the results.

The margin of error of μ is:_______

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Complete Question

In a random sample of 13 microwave ovens, the mean repair cost was $90.00 and the standard deviation was $15.20. 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 98% confidence interval for the population mean u. A 98% confidence interval using the t-distribution was (78.7,101.3). Compare the results. The margin of error of u is

Answer:

The value is
E = 11.65

Explanation:

From the question we are told that

The sample is
n = 13

The sample mean is
\= x = \$ 90

The standard deviation is
\sigma = \$ 15.20

The lower limit of the 98% confidence interval is
a = 78.7

The upper limit of the 98% confidence interval is
b = 101.3

Generally the margin of error is mathematically represented as


E = (b-a)/(2)

=>
E = ( 101.3 - 78 )/(2)

=>
E = 11.65

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