A researcher performed a study to validate a translated version of the Western Ontario and McMaster University index (WOMAC) questionnaire used with spanish-speaking patients with hip or knee osteoarthritis. For the 76 women classified with severe hip pain, the WOMAC mean function score was 70.7 with a standard deviation of 14.6. We wish to know if we may conclude that the mean function score for a population of similar women subjects with severe hip pain is different from 75. Let α =0.01
The null and alternative hypothesis is ,
The test is two-tailed test.
Since , the population standard deviation is not known.
Therefore , use t-test.
Now , df=degrees of freedom=n-1=76-1=75
Teh critical values are ,
; The Excel function is , =TINV(0.01,75)
The test statistic is ,
Decision : Here , the value of the test statistic does not lies in the rejection region.
Therefore , fail to reject Ho.
Conclusion : Hence , there is not sufficient evidence to conclude that the mean function score for a population of similar women subjects with severe hip pain is different from 75.
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