Question

Medical professionals are interested in knowing if overall quality of life among chronic pain sufferers is improved for people who take medication versus those that do not. They administer a quality of life survey that reports a score between 1 (very low quality of life) and 10 (very high quality of life). Based on the information from these samples, should doctors prescribe medication to improve the lives of chronic pain sufferers?

Medicated Group

Mean = 7.6

Sd= 1.2

N=175

Non-Medicated Group

Mean = 6.5

Sd=2.0

N=200

• Use either a one-sample t-test, independent-sample t-test, or a 95% Confidence Interval to answer each of the questions.

• For large samples one sample t-tests (n >120) use critical t scores of ±1.96 (for 95% confidence level two tailed test) or ±1.65 (for 95% confidence level one tailed test).
• For small samples (n<120) use critical t score obtained from t-distribution table. You will need to calculate degrees of freedom, which is simply the sample size minus 1 (df = n-1) and use an alpha value of .05.
• For comparing means between two samples (regardless of sample sizes), use an independent samples t-test, with a critical t of ±1.96.

Answer 1

SOLUTION-

LET BE THE MEAN SCORE FOR MEDICATED SCORE AND BE THE MEAN SCORE FOR NON MEDICATED SCORE.

WE WANT TO TEST THE CLAIM THAT MEDICATION IMPROVES LIVES OF PATIENTS. SO THE HYPOTHESIS IS,

WE PERFORM A TWO SAMPLE T-TEST TO TEST THE HYPOTHESIS STATED ABOVE. ALSO WE USE MINITAB-16 FOR COMPUTATION:

STEPS- STAT> BASIC STATISTICS> TWO SAMPLE-T> ENTER THE SUMMARIZED DATA( SET THE MEDICATION AS SAMPLE-1)> UNDER 'OPTIONS', ENTER THE CONFIDENCE LEVEL 95.0 AND ALTERNATE AS 'GREATER THAN'> OK.

OBSERVATIONS-

THE TEST STATISTIC IS, T= 6.55 AND THE P-VALUE IS 0.000

AS P-VALUE< LEVEL OF SIGNIFICANCE, WE REJECT THE NULL HYPOTHESIS AND CONCLUDE THAT MEDICATION IMPROVES THE THE LIFE OF SUFFERERS.

****IN CASE OF DOUBT, COMMENT BELOW. ALSO LIKE THE SOLUTION, IF POSSIBLE.