A telephone company's records indicate that private customers pay on average $17.10 per month for long-distance telephone calls. A random sample of 10 customers' bills during a given month produced a sample mean of $22.10 expended for long-distance calls and a sample variance of 45. A 5% significance test is to be performed to determine if the mean level of billing for long distance calls per month is in excess of $17.10. The calculated value of the test statistic and the critical value respectively are:

(2.36, 1.8331)

(1.17, 2.2622)

(2.36, 2.2622)

(1.17, 1.8331)

(0.025, 1.8125)

To determine whether the mean level of billing for long-distance calls per month is in excess of $17.10, you need to conduct a hypothesis test using the sample mean and given information.

Null hypothesis (H0): The mean level of billing for long-distance calls per month is $17.10.
Alternative hypothesis (Ha): The mean level of billing for long-distance calls per month is greater than $17.10.

To perform the hypothesis test, you will need to calculate the test statistic and compare it to the critical value. The test statistic formula for a one-sample t-test is given by:

t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))

Using the given data:
- Sample mean (x̄) = $22.10
- Population mean (μ) = $17.10
- Sample variance (s²) = 45
- Sample size (n) = 10

First, calculate the sample standard deviation (s) using the sample variance (s²):
s = sqrt(45) ≈ 6.71

Now, calculate the test statistic (t):
t = (22.10 - 17.10) / (6.71 / sqrt(10)) ≈ 2.36

Next, you need to find the critical value for a one-tailed test at a 5% significance level with 9 degrees of freedom (n-1). You can use a t-table or a t-distribution calculator to find the critical value.

The critical value for a one-tailed 5% significance level with 9 degrees of freedom is approximately 1.8331.

Therefore, the calculated value of the test statistic and the critical value respectively are (2.36, 1.8331).

The correct answer is: (2.36, 1.8331)