Posted by **lola** on Thursday, April 19, 2012 at 5:13pm.

A large television repair service claims that its average repair charge is $124. Suspecting that the average charge is higher, a consumer group randomly choose a sample of 35 statements for tv repairs done by the repair service and found the mean was $125.50 and s = 2.25. Test the claim at the .05 level of significance

- Statistics -
**MathGuru**, Thursday, April 19, 2012 at 6:18pm
Use a one-sample z-test.

Hypotheses:

Ho: µ = 124 -->null hypothesis

Ha: µ > 124 -->alternate hypothesis

z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

With your data:

z = (125.50 - 124)/(2.25/√35) = ?

Finish the calculation.

Check a z-table at .05 level of significance for a one-tailed test.

If the z-test statistic exceeds the critical value from the z-table, reject the null. If the z-test statistic does not exceed the critical value from the z-table, do not reject the null.

I hope this will help get you started.

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