Question

) Find 95% confidence limits for the mean of a normality distributed population from which the following sample was taken and taken 15, 17,10,18,16,9,7,11,13,14.

Answer 1

[AI-generated answer — Bot GPT 3.5]

To find the 95% confidence limits for the mean of a normally distributed population based on the given sample, we can use the t-distribution since the population standard deviation is not given.

First, let's calculate the sample mean and sample standard deviation:

Sample mean (x̄) = (15 + 17 + 10 + 18 + 16 + 9 + 7 + 11 + 13 + 14) / 10 = 13

Sample standard deviation (s) = √[Σ(xi - x̄)² / (n - 1)]
= √[((15-13)² + (17-13)² + (10-13)² + (18-13)² + (16-13)² + (9-13)² + (7-13)² + (11-13)² + (13-13)² + (14-13)²] / 9
= √[(4² + 4² + (-3)² + 5² + 3² + (-4)² + (-6)² + (-2)² + 0² + 1²) / 9]
= √[(16 + 16 + 9 + 25 + 9 + 16 + 36 + 4 + 0 + 1) / 9]
= √(116 / 9)
≈ √12.8889
≈ 3.59

Next, we need to find the t-value for a 95% confidence level with 9 degrees of freedom (n - 1 = 10 - 1 = 9). From the t-distribution table, for a 95% confidence level and 9 degrees of freedom, the t-value is approximately 2.262.

Now, we can calculate the margin of error:
Margin of error = t-value * (s / √n) = 2.262 * (3.59 / √10) ≈ 2.163

Finally, we can calculate the confidence interval:
Confidence interval = (x̄ - margin of error, x̄ + margin of error) = (13 - 2.163, 13 + 2.163) ≈ (10.837, 15.163)

Therefore, the 95% confidence limits for the mean of the normally distributed population based on the given sample are approximately 10.837 and 15.163.