The standard deviation of a simple random sample of 15 customer waiting times is found to be 4.8 minutes. Find the test statistic to test the claim that the standard deviation of all customer waiting times is greater than 3.5 minutes. Use a 0.01 significance level.
To test the claim that the standard deviation of all customer waiting times is greater than 3.5 minutes, we can use the chi-square test statistic.
The test statistic for testing the claim about the standard deviation is calculated using the formula:
χ^2 = (n-1) * s^2 / σ^2,
where n = sample size, s = sample standard deviation, and σ = hypothesized population standard deviation.
In this case,
n = 15 (sample size),
s = 4.8 (sample standard deviation), and
σ = 3.5 (hypothesized population standard deviation).
Substituting these values into the formula, we get:
χ^2 = (15-1) * (4.8^2) / (3.5^2)
= 14 * 23.04 / 12.25
= 26.88 / 12.25
= 2.19 (rounded to two decimal places).
Therefore, the test statistic to test the claim that the standard deviation of all customer waiting times is greater than 3.5 minutes is 2.19.