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.
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
1 answer
1 answer