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Upper-tailed t-test

Cheap N Fresh food stores advertise that their average checkout waiting time is 4 minutes or less. A random selection of shoppers at peak time yielded the following waiting times: 3.5, 5.0, 3.8, 4.5, 7.2, 5.0. Does this evidence disprove the store's claim?

1.
Hypotheses: $H_0: \mu\leq 4.0$ versus $H_1: \mu > 4.0$
2.
Test Statistic: $t = \frac{\overline{X}-4.0}{S/\sqrt{n}} =
\frac{ 4.83 -4.0}{1.31/\sqrt{6}} = 1.55$
3.
P-value: Presuming H0 is true, the likelihood of chance variation yielding a t-statistic higher than 1.55 (this is the H1 direction) is .09.

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4.
Conclusion: Since P-value > .05, we do not reject H0. The sample does not provide enough evidence to disprove the store's claim.

Summary of the upper-tailed  t-test for $\mu$:

$H_0: \mu \leq c$ versus $H_1: \mu > c$
Test statistic: $t= (\overline{X}- c)/(S/\sqrt{n})$
P-value: Total area greater than t (the direction of H1) under t-curve with n-1 degrees of freedom. If t is far enough above 0 (the direction of H1), the P-value will be small.
Conclusion: If P-value $\leq$ .05, we reject H0 with statistical significance. If P-value $\leq$ .01, we reject H0 with high statistical significance. If P-value>.05, we do not reject H0.


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Next: Two-tailed t-test Up: Test of Significance involving Previous: Test of Significance involving

2003-09-08