A clothing store analyzed the value of purchases made on credit card by a sample of 30 credit card customers. The sample mean was found to be R 165.45 with a sample standard deviation of R 38.60. The lower limit of a 95% confidence interval for the population variance of credit card purchases are:

Answer 1

To find the lower limit of the 95% confidence interval for the population variance, we can use the chi-squared distribution.

The chi-squared distribution is used when estimating the population variance. The formula for the confidence interval is:

(𝑛−1)𝑠²/𝜒²ₕ,𝑛₋₁ ≤ 𝜎² ≤ (𝑛−1)𝑠²/𝜒²ₗ,𝑛₋₁

Where:
n = sample size
s = sample standard deviation
𝜒²ₕ,𝑛₋₁ = chi-squared critical value at the upper tail (for a 95% confidence level and n-1 degrees of freedom)
𝜒²ₗ,𝑛₋₁ = chi-squared critical value at the lower tail (for a 95% confidence level and n-1 degrees of freedom)

Using the given values:
n = 30
s = R 38.60

Since we want the lower limit of the confidence interval, the lower tail critical value will be used.

To find the chi-squared critical value at the lower tail, we need to determine the degrees of freedom, which is n-1.

Degrees of freedom = 30 - 1 = 29

Using a chi-squared table or software, the critical value at a 95% confidence level with 29 degrees of freedom is approximately 14.68 (rounded to two decimal places).

Now we can substitute the values into the formula:

(30-1)(38.60)²/14.68 ≤ 𝜎²

Simplifying the equation:

29(38.60)²/14.68 ≤ 𝜎²

Calculating:

22616.84 ≤ 𝜎²

Therefore, the lower limit of the 95% confidence interval for the population variance of credit card purchases is R 22616.84.