## Question 683:

1

No answer provided yet.So we're asked to generate a 90% confidence interval around the sample mean of 178 minutes. The sample size is reasonably large at 40, so it would likely be OK if we used the normal distribution to build the interval. Since we don't know the population standard deviation, the more conservative approach would be to use the t-distribution. Recall that to use the normal distribution we need to know the population standard deviation or the sample size should be relatively large. I'll provide the computations using the t-distribution, then show the results using the normal or z-distribution.

A confidence interval is just the mean plus/minus the margin of error. The margin of error is equal to the standard error of the mean (SE) times a critical value from the t or normal distribution (depending on which method is used).

The critical value can be found using the Inverse t-Distribution Calculator, using .10 as the proportion of area and 39 degrees of freedom (1 less than the sample size) and we get the value 1.6849.

The SE is made up of the standard deviation divided by the square root of the sample size = 12/SQRT(40) = 1.8974. The margin of error is then 1.8974*1.6849 = 3.1968. Adding and subtracting that on the mean gets us a 90% confidence interval between 174.80 and 181.20 minutes.

If we were to use the z-value, we'd just change the critical value from 1.6849 to 1.6448 and we'd get an interval between 174.88 and 181.12. We can see that there is very little difference between the two.

The validity of this method is fine using the t-distribution and potentially not valid as explained above when using the normal distribution. However, we see, the results are very close.