In a random sample of
fivefive
microwave ovens, the mean repair cost was
$75.0075.00
and the standard deviation was
$14.0014.00.
Assume the population is normally distributed and use a t-distribution to construct a
9090%
confidence interval for the population mean
muμ.
What is the margin of error of
muμ?
Interpret the results.
In a random sample of fivefive microwave ovens, the mean repair cost was $75.0075.00 and the...
In a random sample of five microwave ovens, the mean repair cost was $65.00 and the standard deviation was $14.00. Assume the population is normally distributed and use a t-distribution to construct a 95% confidence interval for the population mean μ. What is the margin of error of μ? Interpret the results.
in a random sample of four microwave ovens, the mean repair cost was 65.00 and the standard deviation was 12.50 assume the population 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? interpret the results In a Round to two decimal places as needed.)
In a random sample of six microwave? ovens, the mean repair cost was ?$90.00 and the standard deviation was ?$13.00 Assume the variable is normally distributed and use a? t-distribution to construct a 90?% confidence interval for the population mean ?. What is the margin of error of ???
In a random sample of six microwave ovens, the mean repair cost was $65.00 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 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.) The margin of error is nothing. (Round to...
6.2.19-T Question Help In a random sample of four microwave ovens, the mean repair cost was $85.00 and the standard deviation was $13.00. 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. The 99% confidence interval for the population mean μ is (DD (Round to two decimal places as needed.) 6.2.21-T Question Help In a random sample...
Question Help In a random sample of live microwave ovens, the mean repair cost was $60.00 and the standard deviation was $12.00. Assume the population is normally distributed and use at distribution to construct a 90% confidence interval for the population mean . What is the margin of error of ? Interpret the results The 90% confidence interval for the population mean (Round to two decimal places as needed.) is C D Enter your answer in the edit fields and...
In a random sample of six mobile devices, the mean repair cost was $80.00 and the standard deviation was $12.00. 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 mu is
to estimate the population mean repair cost of microwave ovens in macomb, a random sample of 9 broken microwave ovens was selected and it is found that the sample mean repair cost was $80 with a sample standard deviation of $15. 1) what would be the “critical value” if you want to find a 95% confidence interval for the population mean? 2) what would be the “margin of error” if you want to find a 95% confidence interval for the...
6.2.20-T Question Help In a random sample of four mobile devices, the mean repair cost was $90.00 and the standard deviation was $13.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results.
In a random sample of four mobile devices, the mean repair cost was $60.00 and the standard deviation was $14.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results. The 90% confidence interval for the population mean is (DO (Round to two decimal places as needed.) The margin of error is $ (Round to two decimal places as needed.) Interpret...