A supervisor records the repair cost for 22 randomly selected VCRs. A sample mean of $75.50 and standard deviation of $18.07 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the VCRs. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

The t value for 99% CI for 21 df is 2.831.

The critical value that should be used in constructing the confidence interval is (64.593, 86.407).

Step-by-step explanation:

Now the sample size is less than 30 and also population standard deviation is not known.

Then we will use t distribution to find CI

t value for 99% CI for 21 df is TINV(0.01,21)=2.831

The margin of error is

Hence CI is

2 over 5

step-by-step explanation:

the 2: 3 ratio is the length of the ac segment to the cb segment total length. this implies that the ratio between segment ac length and segment ab length is 2: 5. point c is 2/5 from a to b, so the required fraction is 2/5.

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