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.
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?
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