A supervisor records the repair cost for 17 randomly selected stereos. A sample mean of $66.34 and standard deviation of $15.22
1 answer:
Answer:
Critical value: T = 1.337
The 80% confidence interval for the mean repair cost for the stereos is between $45.991 and $86.689
Step-by-step explanation:
We have the standard deviation for the sample. So we use the t-distribution to solve this question.
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 = 17 - 1 = 16
80% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 16 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.337, which is the critical value
The margin of error is:
M = T*s = 1.337*15.22 = 20.349
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 66.34 - 20.349 = $45.991
The upper end of the interval is the sample mean added to M. So it is 66.34 + 20.349 = $86.689
The 80% confidence interval for the mean repair cost for the stereos is between $45.991 and $86.689
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Given
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