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.
1 answer:
Answer:
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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