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 is the degree of freedom

Answer 1

To find the 90% confidence interval for the mean, we will first calculate the sample mean and the standard error.

Sample mean (x̄):
x̄ = (54 + 56 + 59 + 59 + 65 + 66 + 67 + 68 + 69 + 76 + 84 + 86 + 89 + 94 + 101) / 15
= 1061 / 15
= 70.733

Standard deviation (s):
First, we need to find the sample variance, which can be done using the formula:

s^2 = ∑(xi - x̄)^2 / (n - 1)

Where xi is each data point, x̄ is the sample mean, and n is the sample size.

s^2 = [(54 - 70.733)^2 + (56 - 70.733)^2 + (59 - 70.733)^2 + (59 - 70.733)^2 + (65 - 70.733)^2 + (66 - 70.733)^2 + (67 - 70.733)^2 + (68 - 70.733)^2 + (69 - 70.733)^2 + (76 - 70.733)^2 + (84 - 70.733)^2 + (86 - 70.733)^2 + (89 - 70.733)^2 + (94 - 70.733)^2 + (101 - 70.733)^2] / (15 - 1)
= [110.449 + 106.329 + 123.529 + 123.529 + 24.649 + 20.169 + 14.989 + 10.489 + 8.009 + 34.969 + 183.169 + 206.249 + 243.769 + 408.809 + 724.249] / 14
= 3126.018 / 14
= 223.286

Standard deviation (s):
s = √s^2
= √223.286
= 14.944

Standard error (SE):
SE = s / √n
= 14.944 / √15
= 3.860

Now we can calculate the 90% confidence interval using the following formula:

CI = x̄ ± (Z * SE)

Where Z is the Z-score corresponding to the desired level of confidence. For a 90% confidence level, Z = 1.645.

CI = 70.733 ± (1.645 * 3.860)
= 70.733 ± 6.346

Therefore, the 90% confidence interval for the mean is (64.387, 77.079) to three decimals.

The degree of freedom for this sample is given by (n - 1), where n is the sample size. In this case, the degree of freedom is 15 - 1 = 14.