Statistics
posted by John on .
A telephone company's records indicate that private customers pay on average $17.10 per month for longdistance telephone calls. A random sample of 10 customers' bills during a given month produced a sample mean of $22.10 expended for longdistance 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)

Remember that standard deviation is the square root of the variance.
Formula:
t = (sample mean  population mean)/(standard deviation divided by the square root of the sample size)
t = (22.10  17.10)/(6.7082/√10)
I'll let you finish the calculation.
The test is onetailed at .05 level of significance. Look at the appropriate table and determine the critical value.
Hint: degrees of freedom = n  1 = 9