A study was conducted to validate a translated version of the Western Ontario an

Question

A study was conducted to validate a translated version of the Western Ontario and McMaster Universities Osteoarthritis 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 (on a scale from 0 to 100 with a higher number indicating less function) 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 less than 75.

Let = .01.

The critical value for the given is ± 2.33.

The z equation is given below:

z= xbar - mue/ sigma upon route of n

5.1 Is this a two-tailed or one-tailed test?

5.2 Draw a graph depicting the critical value, rejection region, and non-rejection region.

5.3. State the null-hypothesis in words and in symbol.

5.4 State the alternative hypothesis in words and in symbol.

5.5 Based on the hypothesis being tested, should you use the critical value of -2.33 or

2.33 to compare against the computed z score to make the decision to reject or retain the

null hypothesis.

5.6 Compute z statistic using the equation given above.

5.7 Comparing z statistics with the critical value, would you reject the null or retain the

null hypothesis? In other words, does the z score lie in the rejection or non-rejection

region?

Explanation / Answer

SOl:5.1 Is this a two-tailed or one-tailed test?

its a left tail z test.

Solution5.3:

Null hypothesis:

Average equals 75

Ho:=75

SOlutin5.4:

Average less than 75

Ha:<75

Solution5.6

5.6 Compute z statistic using the equation given above.

z=70.7-75/14.6/sqrt(76)

z =-2.57

Solution5.7

take |z cal|=|-2.57|=2.57

Z crit=|-2.33|=2.33

Z cal>Z crit

Reject null hypothesis

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