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 And why?
To find a confidence interval (CI) for the mean, we need to determine whether to use the t-distribution or the standard normal distribution (Z-distribution). This decision is based on sample size and whether the population standard deviation is known or unknown.
Since the population standard deviation is unknown, we will use the t-distribution.
To calculate the confidence interval, we first need to find the sample mean and sample standard deviation.
Sample mean (x̄) = (54 + 56 + 59 + 59 + 65 + 66 + 67 + 68 + 69 + 76 + 84 + 86 + 89 + 94 + 101) / 15 = 73.733
Sample standard deviation (s) = sqrt(((54-73.733)^2 + (56-73.733)^2 + (59-73.733)^2 + (59-73.733)^2 + (65-73.733)^2 + (66-73.733)^2 + (67-73.733)^2 + (68-73.733)^2 + (69-73.733)^2 + (76-73.733)^2 + (84-73.733)^2 + (86-73.733)^2 + (89-73.733)^2 + (94-73.733)^2 + (101-73.733)^2) / (15-1))
After evaluating the above expression, we find that s ≈ 14.641.
The degrees of freedom for the t-distribution is given by n - 1, where n is the sample size. Here, the sample size is 15, so the degrees of freedom (df) = 15 - 1 = 14.
Now that we have all the required values, we can calculate the confidence interval using the following formula:
CI = x̄ ± (t * (s / sqrt(n)))
Here, t is the critical value from the t-distribution, based on the desired confidence level and the degrees of freedom.
For a 90% confidence interval, the critical value for a two-tailed test with df = 14 is approximately 1.761.
Plugging all the values into the formula, we get:
CI = 73.733 ± (1.761 * (14.641 / sqrt(15)))
Evaluating the above expression, we find:
CI ≈ 73.733 ± 8.706
Therefore, the 90% confidence interval for the mean is (65.027, 82.439) when rounded to three decimals.
In conclusion, we use the t-distribution for this problem because the population standard deviation is unknown.