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## Two-tailed t-test

Past data have shown that if the payphones at the airport are emptied every 14 days, the coin collectors will be 70% full on the average. The phone company tries to schedule the collection visits at 70% full because money is lost if the phones get full and unusable, but visiting the phones too frequently is also an expense. The company keeps data on fill amounts during collection, in case they need to increase or decrease collection frequency. During the last visit, suppose that 5 phones were 50%, 40%, 70%, 75%, and 45% full, respectively. Do you think the frequency of visits needs to be changed, or is this just chance variation?

It is important to note that an action will be made (i.e. increase or decrease visits) if the average fill shifts away from in either direction. Since we are looking for statistical evidence of an average shift in either direction, the appropriate hypotheses and test procedure are:

1.
Hypotheses: versus
2.
Test Statistic:
3.
P-value: Presuming H0 is true, the likelihood of chance variation yielding a t-statistic more extreme than -2.01 on either side of 0 (since H1 direction is both high and low) is .11.

4.
Conclusion: Since P-value > .05, we do not reject H0. The sample does not provide enough evidence that the mean fill has shifted from 70%.

Summary of the two-tailed  t-test for :

versus
Test statistic:
P-value: Total area greater than |t| and less than -|t| under t-curve with n-1 degrees of freedom If t is far enough from 0 on either side (the direction of H1), the P-value will be small.
Conclusion: If P-value .05, we reject H0 with statistical significance. If P-value .01, we reject H0 with high statistical significance. If P-value>.05, we do not reject H0.

Next: The t-test is not Up: Test of Significance involving Previous: Upper-tailed t-test

2003-09-08