Question

In: Statistics and Probability

In a random sample of five microwave​ ovens, the mean repair cost was​$75.00 and the standard...

In a random sample of five microwave​ ovens, the mean repair cost was​$75.00 and the standard deviation was ​$12.50

Assume the population is normally distributed and use a​ t-distribution to construct a 90​% confidence interval for the population mean u

What is the margin of error of μ​?

Interpret the results.

Solutions

Expert Solution

Given:

Sample size = n = 5 Sample mean = Sample standard deviation = s = 12.50   

Assuming that the population is normally distributed, 90% confidence interval for the population mean can be constructed using the formula:

can be obtained from t tables for = 0.10 / 2 = 0.05 and n - 1 = 5 - 1 = 4 df

Substituting in the formula,

90% confidence interval for the population mean

=

90% confidence interval for the population mean is obtained as

The margin of error of =

It implies that we are 90% confidennt that the true population mean would lie in this interval. If 100 additional samples are drawn, then the true population mean will fall in the interval 90 out of 100 times.


Related Solutions

In a random sample of 5 microwave ovens, the mean repair cost was $75.00 and the...
In a random sample of 5 microwave ovens, the mean repair cost was $75.00 and the standard deviation was $12.50. Construct a 95% C.I. for the population mean. Assume Normality. (Please show your work)
In a random sample of six microwave​ ovens, the mean repair cost was ​$75.00 and the...
In a random sample of six microwave​ ovens, the mean repair cost was ​$75.00 and the standard deviation was ​$11.00. Assume the population is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean μ. What is the margin of error of μ​? Interpret the results.
In a random sample of five microwave​ ovens, the mean repair cost was ​$65.00 and the...
In a random sample of five microwave​ ovens, the mean repair cost was ​$65.00 and the standard deviation was ​$13.50. Assume the population is normally distributed and use a​ t-distribution to construct a 90​% confidence interval for the population mean mu. What is the margin of error of mu​? Interpret the results. The 90​% confidence interval for the population mean mu is ​( nothing​, nothing​). ​(Round to two decimal places as​ needed.)
In a random sample of five microwave​ ovens, the mean repair cost was ​$60.00 and the...
In a random sample of five microwave​ ovens, the mean repair cost was ​$60.00 and the standard deviation was ​$14.00. Assume the variable is normally distributed and use a​ t-distribution to construct a 95​% confidence interval for the population mean u. What is the margin of error of u​? The 95​% confidence interval for the population mean u is l(_, _ ) ​(Round to two decimal places as​ needed.
In a random sample of five microwave​ ovens, the mean repair cost was $60.00 and the...
In a random sample of five microwave​ ovens, the mean repair cost was $60.00 and the standard deviation was $11.50 Assume the population is normally distributed and use a​ t-distribution to construct a 90​% confidence interval for the population mean μ. What is the margin of error of μ​? Interpret the results. (a)The confidence interval for the population mean is _, _   ​(Round to one decimal place as​ needed.) (b)The margin of error is _ ​(Round to two decimal places...
In a random sample of 1313 microwave​ ovens, the mean repair cost was ​$80.0080.00 and the...
In a random sample of 1313 microwave​ ovens, the mean repair cost was ​$80.0080.00 and the standard deviation was ​$15.5015.50. 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 9090​% confidence interval for the population mean muμ. A 9090​% confidence interval using the​ t-distribution was left parenthesis 72.3 comma 87.7 right parenthesis(72.3,87.7). Compare the results. The margin of error...
In a random sample of six microwave​ ovens, the mean repair cost was ​$70.00 and the...
In a random sample of six microwave​ ovens, the mean repair cost was ​$70.00 and the standard deviation was ​$12.00. Assume the variable is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean u. What is the margin of error of u​? The 99​% confidence interval for the population mean u is (_, _ ) ​(Round to two decimal places as​ needed.) The margin of error is ___. ​(Round to two decimal places...
In a random sample of four microwave​ ovens, the mean repair cost was $85.00 and the...
In a random sample of four microwave​ ovens, the mean repair cost was $85.00 and the standard deviation was $12.00. Assume the population is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean mu. What is the margin of error of mu? Interpret the results. The 99​% confidence interval for the population mean mu is?
In a random sample of five mobile​ devices, the mean repair cost was ​$90.00 and the...
In a random sample of five mobile​ devices, the mean repair cost was ​$90.00 and the standard deviation was ​$11.00. Assume the population is normally distributed and use a​ t-distribution to find the margin of error and construct a 95​% confidence interval for the population mean. Interpret the results. The 95​% confidence interval for the population mean ​(76.34,103.66). ​(Round to two decimal places as​ needed.) The margin of error is ​$____?? I NEED THIS ANSWER, OR EXPLAINED HOW TO FIGURE...
In a random sample of 60 computers, the mean repair cost was $120 with a standard...
In a random sample of 60 computers, the mean repair cost was $120 with a standard deviation of $30. Construct the 95% confidence interval for the population mean repair cost.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT