Question

A recent study assessed the long-term quality of life of oral/oropharyngeal cancer survivors who received a new type of treatment. A total of 135 survivors completed a quality of life questionnaire called the EORTC QLQ-C30. Their mean social functioning score was 75.0 with a standard deviation of 23.2. For this scale, a score of 82.6 is what we expect from the population of people who receive standard treatment. Perform a test to determine whether the mean score after the new treatment is significantly different than 82.6. You must specify the null and alternative hypotheses, hand calculate all steps of the test (but may use Stata to calculate a p-value or critical value) and write a sentence summarizing the result of the test in the context of the problem. (Source: Ann Maxillofac Surg. 2015 Jan-Jun;5(1):26-31)

Hand calculate a confidence interval for the mean global functioning score after the new treatment. Write a one-sentence interpretation of the confidence interval.

Explain the relationship between the result of the hypothesis test in Question 1 and the confidence interval in Question 2. In other words, what does the confidence interval tell you about the hypothesis test result?

Answer 1

H0:Null Hypothesis: = 82.6

HA: Alternative Hypothesis: 82.6

SE = s/

= 23.2/ = 1.9967

Test statistic is given by:

t = (75.0 - 82.6)/1.9967 = - 3.81

ndf = n - 1 = 135 - 1 = 134

Take = 0.05

From Table, critical values of t = 1.9778

Since the calculated value of t = - 3.81 is less than critical value of t = - 1.9778, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that mean score after the new treatment is significantly different from 82.6.

t score = - 3.81

ndf = 134

By Technology, P - Value = 0.0002

Since P - value = 0.0002 is less than = 0.05, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that mean score after the new treatment is significantly different from 82.6.

Confidence Interval:

75.0 (1.9778 X 1.9967)

= 75.0 3.9491

= ( 71.0509 ,78.9491)

So,

Confidence Interval:

71.0509 < < 78.9491

Since 82.6 is not included in the Confidence Interval: ( 71.0509 ,78.9491), it confirms the conclusion in the hypothesis test result that mean score after the new treatment is significantly different from 82.6.