# Statistics

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)

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1. 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 one-tailed at .05 level of significance. Look at the appropriate table and determine the critical value.
Hint: degrees of freedom = n - 1 = 9

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