In a random sample of six cell phones, the mean full retail price was $538.00 and the standard deviation was $184.00. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 90% confidence interval for the population mean mu. Interpret the results. Identify the margin of error. Construct a 90% confidence interval for the population mean. Interpret the results. Select the correct choice below and fill in the answer box to complete your choice.

The margin of error is 370.8.

The 90% confidence interval for the population mean is between $167.2 and $908.8

The correct interpretation is that we are 90% sure that the true mean price for all cellphones in within the interval end-points, so option B.

Step-by-step explanation:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 6 - 1 = 5

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.0150

The margin of error is:

M = T*s = 2.0150*184 = 370.8.

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 538 - 370.8 = $167.2

The upper end of the interval is the sample mean added to M. So it is 538 + 370.8 = $908.8

The 90% confidence interval for the population mean is between $167.2 and $908.8

The correct interpretation is that we are 90% sure that the true mean price for all cellphones in within the interval end-points, so option B.

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step-by-step explanation: