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:

Question

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.

Answer 2

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

Step 1: Determine the degrees of freedom (df).
The degrees of freedom for estimating the population variance is given by (n - 1), where n is the sample size. In this case, the sample size is 30, so (n - 1) = 30 - 1 = 29.

Step 2: Find the critical value from the chi-squared distribution table.
For a 95% confidence interval, we need to find the chi-squared value with an upper tail probability of 0.05. Looking at the chi-squared distribution table with 29 degrees of freedom, we find the critical value to be approximately 45.722.

Step 3: Calculate the lower limit.
The lower limit of the confidence interval is given by:

Lower limit = ((n - 1) * s^2) / chi-squared value

where s is the sample standard deviation.

Plugging in the values:
Lower limit = (29 * (38.60)^2) / 45.722
= 1160.46

Therefore, the lower limit of the 95% confidence interval for the population variance of credit card purchases is R 1160.46.