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)
2.36, 1.8331
To find the calculated value of the test statistic, we need to perform a t-test since we don't have the population standard deviation.
The formula for the t-test statistic is:
t = (sample mean - population mean) / (sample standard deviation / √n)
Where:
- sample mean is the mean of the sample = $22.10
- population mean is the mean of the population = $17.10
- sample standard deviation is the square root of the sample variance = √45 = 6.71
- n is the sample size = 10
Calculating the t-test statistic:
t = ($22.10 - $17.10) / (6.71 / √10)
= $5 / (6.71 / 3.16)
≈ 2.36
Now, to find the critical value for a 5% significance level, we need to refer to the t-distribution table with (n-1) degrees of freedom. In this case, we have 10 - 1 = 9 degrees of freedom.
For a one-tailed test at a 5% significance level (since we want to test if the mean is in excess of $17.10), the critical value is approximately 1.833.
Therefore, the calculated value of the test statistic and the critical value respectively are:
(2.36, 1.8331)