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(Solved): In a random sample of five mobile devices, the mean repair cost was \( \$ 70.00 \) and the standard ...




In a random sample of five mobile devices, the mean repair cost was \( \$ 70.00 \) and the standard deviation was \( \$ 14.00
In a random sample of five mobile devices, the mean repair cost was \( \$ 70.00 \) 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 \( 99 \% \) confidence interval for the population mean. Interpret the results. The \( 99 \% \) confidence interval for the population mean \( \mu \) is (Round to two decimal places as needed.) The margin of error is \( \$ \) (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. A. If a large sample of mobile devices are taken approximately \( 99 \% \) of them will have repair costs between the bounds of the confidence interval. B. With \( 99 \% 6 \) confidence, it can be said that the repair cost is between the bounds of the confidence interval. C. With \( 99 \% \) confidence, it can be said that the population mean repair cost is between the bounds of the confidence interval. D. It can be said that \( 99 \% \) of mobile devices have a repair cost between the bounds of the confidence interval.


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