Question

Assuming CLT applies for the sample data given by 54, 56, 59, 59, 65, 66, 67, 68, 69, 76, 84, 86, 89, 94, 101 Find a 90% CI to three decimals for the mean
What distribution do you use T or Z

Answer 1

To determine which distribution to use, we need to know the population standard deviation. Since this information is not provided, we should use the t-distribution.

To find the 90% confidence interval (CI), we first need to calculate the sample mean and the sample standard deviation.

Sample mean (x̄) = (54 + 56 + 59 + 59 + 65 + 66 + 67 + 68 + 69 + 76 + 84 + 86 + 89 + 94 + 101) / 15 = 73.267

Sample standard deviation (s) = sqrt(((54-73.267)^2 + (56-73.267)^2 + (59-73.267)^2 + (59-73.267)^2 + (65-73.267)^2 + (66-73.267)^2 + (67-73.267)^2 + (68-73.267)^2 + (69-73.267)^2 + (76-73.267)^2 + (84-73.267)^2 + (86-73.267)^2 + (89-73.267)^2 + (94-73.267)^2 + (101-73.267)^2) / (15-1)) = 15.137

Next, we need to determine the critical value for a 90% confidence level with df = n-1 = 15-1 = 14. Using a t-table or statistical software, we find the critical value to be approximately 1.761.

Now we can calculate the margin of error (ME) using the formula ME = critical value * (s / sqrt(n)):
ME = 1.761 * (15.137 / sqrt(15)) ≈ 8.129

Finally, we can construct the confidence interval:
CI = (x̄ - ME, x̄ + ME)
CI = (73.267 - 8.129, 73.267 + 8.129)
CI ≈ (65.138, 81.396)

Therefore, the 90% confidence interval for the mean is approximately (65.138, 81.396), rounded to three decimals.