Posted by lola on .
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,
Use a onesample ztest.
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 ztable at .05 level of significance for a onetailed test.
If the ztest statistic exceeds the critical value from the ztable, reject the null. If the ztest statistic does not exceed the critical value from the ztable, do not reject the null.
I hope this will help get you started.