The question is in the description. It is a statistics question that involves normal distribution

Question

You are given the following random sample from a population that you believe to be approximately normally distributed. 8.60 8.70 9.00 8.97 8.72 8.74 8.96 8.78

a. What is a 95% confidence interval for the population mean value?

b. What is a 95% lower confidence bound for the population variance?

Answer 1

a) To calculate the 95% confidence interval for the population mean and the 95% lower confidence bound for the population variance, follow these steps:

1. Calculate the sample mean (x̄) and sample variance (s^2).

Sample data: 8.60, 8.70, 9.00, 8.97, 8.72, 8.74, 8.96, 8.78

a. Sample mean (x̄): x̄ = (8.60 + 8.70 + 9.00 + 8.97 + 8.72 + 8.74 + 8.96 + 8.78) / 8 = 70.47 / 8 = 8.80875

b. Sample variance (s^2): First, calculate the sum of squared deviations from the mean: Σ(x_i - x̄)^2 = (8.60 - 8.80875)^2 + (8.70 - 8.80875)^2 + ... + (8.78 - 8.80875)^2 Σ(x_i - x̄)^2 = 0.043515625 + 0.011765625 + ... + 0.000828125 = 0.132171875

s^2 = Σ(x_i - x̄)^2 / (n - 1) = 0.132171875 / (8 - 1) = 0.132171875 / 7 ≈ 0.018881696

1. Calculate the 95% confidence interval for the population mean value.

To calculate the 95% confidence interval for the population mean, we need the sample mean (x̄), the sample standard deviation (s), the sample size (n), and the t-value associated with a 95% confidence level and n-1 degrees of freedom.

a. Sample standard deviation (s): √s^2 ≈ √0.018881696 ≈ 0.137406

b. Sample size (n): 8

c. Degrees of freedom (df): n - 1 = 8 - 1 = 7

d. Using a t-table or calculator, find the t-value for a 95% confidence level and 7 degrees of freedom. The t-value is approximately 2.365.

Now, calculate the margin of error (ME) using the t-value, sample standard deviation, and sample size: ME = t * (s / √n) ≈ 2.365 * (0.137406 / √8) ≈ 2.365 * 0.048529 ≈ 0.11474

Now, we can calculate the 95% confidence interval for the population mean: Lower limit: x̄ - ME = 8.80875 - 0.11474 ≈ 8.69401 Upper limit: x̄ + ME = 8.80875 + 0.11474 ≈ 8.92349

95% Confidence interval: (8.69401, 8.92349)

1. Calculate the 95% lower confidence bound for the population variance.

To calculate the 95% lower confidence bound for the population variance, we need the sample variance (s^2), the sample size (n), and the chi-square (χ^2) value associated with a 95% confidence level and n-1 degrees of freedom.

a. Sample variance (s^2): 0.018881696

b. Sample size (n): 8

c. Degrees of freedom (df): n - 1 = 8 - 1 = 7

d. Using a chi-square table or calculator, find the χ^2 value for a 95% lower confidence level and 7 degrees of freedom. The χ^2 value is approximately 2.167.

Now, calculate the 95% lower confidence bound for the population variance using the χ^2 value, sample variance, and degrees of freedom: Lower bound = (df * s^2) / χ^2 = (7 * 0.018881696) / 2.167 ≈ 0.06132

So, the 95% lower confidence bound for the population variance is approximately 0.06132.