Hypothesis Testing SHEILA LIBOON

 

SHEILA REYES-LIBOON REYES-LIBOON Master in Social Work Major in Social Work Practice in Health Care MSWSSW STAT 218 Statistics in Research  Thursday 530-830pm Problem 1: A psychologist proposed a new form of psychotherapy for patients suffering from depression, claiming that on the average a patient would regain a normal outlook and attitude in two months (about 8 weeks). The hospital decided to adopt the new treatment on a trial basis, and to test the psychologist’s claim on a sample of 25 depression patients, at a t the stated risk of 5 percent probability of rejecting the claim it was true. Final results showed that the mean time required for the patients to return to normal was 9.5 weeks, with a standard deviation of 3.2 weeks. What would the decision of the hospital administration be? Given: n=(sample size) 25 depression patients population mean=8weeks sample mean= 9.5 weeks (return to normal) standard deviation=3.2 weeks level of significance=(5%) .05 probability of rejecting the claim Step 1: Ho: Patients suffering from depression would regain a normal outlook and attitude to equal to 9.5 weeks Ha: Patients suffering from depression would regain a normal outlook and attitude to not equal/ at least to 9.5 weeks Step 2: Test Statistics = one tailed t-test Step 3:

 

Step 4:

Accept Ho. Reject Ha

Step 5: After testing the psychologist’s proposed new form of psychotherapy on a sample of 25 patients, it was found out that the average time it took for the patients to return to normal was 9.5 weeks, with a standard deviation of 3.2 weeks. The test made use of the one tailed t-test which yielded a value of 2.34. The t-value of 2.34 is below the given Critical Value of 2.92 thus giving us enough evidence to say that the null hypothesis is valid and should be  by the hospital administration. accepted by accepted

 

Problem 2: The researcher wishes to study the comparative ability of boys and girls to learn nonsense syllables. Ten boys and girls are chosen at random from grade six elementary class. The boy-girl pairing then made on the basis of IQ and age. As a result of the test, boys had a mean of 7 while girls had a mean of 10 with a population standard deviation of 4.5. 4. 5. Use 5% significance level. Given: n1= 10 boys mean :7 (boys, test result) n2 = 10 girls mean 10: (girls, test result) population standard deviation = 4.5 level of significance = (5%).05 Step 1: Ho: There is no significant difference between boys and girls ability to learn nonsense syllables. Ha: There is a significant difference differen ce between boys and girls ability to learn nonsense syllables. Step 2: Test Statistics = two tailed z test Step 3:

 

Step 4: 4:  

Accept Ho. Reject Ha Step 5: After comparing the abilities of boys and girls to learn nonsense syllables, it was found out that the boys got an average test result of 7 while the girls got a 10, with a population standard deviation of 3.2 weeks. The test made use of the two tailed z-test which yielded a value of -1.49. The z-value of -1.49 is within the given Critical Values of +1.96 and -1.96 which gives us enough evidence to say that the null hypothesis is valid and accepted. Thus, there is no significant difference between difference between the ability boys and girls to learn nonsense syllables.

 

Problem 3: The time span between test and retest is to be tested for its effect on the scores of a particular type of examination. Twelve examinees took the retest one week after they had taken the initial test. The results of the test-retest of the 12 examinees are as follows: mean(test) = 89.5, sd = 10.3 mean (retest) = 95.8, sd = 7.5. One may wish to find out if in retest after a one-week time span the mean score of the examinees showed a significant increase. Use 0.05 significance level.  level.  Given: n=12 examinees took (mean test)  x    1=89.5     ̅  x    2=     ̅ 95.8 (mean test) level of significance =

retest standard deviation = 10.3 standard deviation deviation = 7.5 0.05

Step 1: Ho: There is no significant difference between test and retest results of the 12 examinees Ha: There is a significant difference between test and retest results of the 12 examinees Step 2: Test Statistics = two tailed t-test Step 3:

 

Step 4:

Accept Ho. Reject Ha Step 5: Comparing the effect of a test and a retest given a week after showed that there is an increase in the average score. This was examined with the use of the two tailed t-test which yielded a value of -1.71. The t-value of -1.71 is within the given Critical Values of -+4.303 and -4.303 thus giving us enough evidence to say that there is no significant difference between difference between the test and retest results of the 12 examinees.

 

Problem 4: A bus company advertised a mean time of 150 minutes for a trip between two cities. A consumer group had a reason to believe that the mean time was more than 150 minutes. A sample of 40 trips showed a mean of 153 minutes and a standard deviation of 7.5 minutes. Using a 5% level of significance, is there sufficient evidence to support the consumer group’s contention? Given: mean time= 150 minutes trip between two cities sample= 40 trips mean= 153 minutes standard deviation= 7.5 minutes level of significance= (5%) .05 Step 1 Ho: A trip between two cities has a mean time of equal to 150 minutes Ha: A trip between two cities has a mean time of more than 150 minutes Step 2: Test Statistics = one tailed t-test Step 3:

Step 4:

Accept Ho. Reject Ha

 

  Step 5: After examining the consumer group’s claim that on average it takes more than 150 minutes for a trip between two cities, the study showed a mean of 153 minutes and a standard deviation of 7.5 minutes. This study was tested with the use of the one tailed t-test which yielded a value of 2.52. The t-value of 2.52 is below the given Critical Value of 2.92 thus giving us enough evidence to say that the null hypothesis must be accepted. In conclusion, there is no sufficient evidence  evidence to support the consumer group’s contention.  contention.