Question

1. ) Suppose a researcher is interested in the effectiveness of a new drug that is intended to naturally increase hours of sleep for elderly patients experiencing insomnia above 5 hours. Suppose he collects information from a group of 64 patients, and calculates a sample mean of 6.4 hours of sleep and a standard deviation of s=0.8.  

1. Write out your null and alternative hypotheses.
2. Carry out the appropriate statistical test to determine if the new treatment is effective (One sided)?

2. ) A drug company is investigating a new pain relief medicine for migraine. To test the effectiveness of the new drug, 10 volunteers were selected and randomly assigned to receive either the new medication or a placebo. The subjects were instructed to take the drug or placebo during their next migraine headache episode and to report their pain on a scale of 1 to 10 (10 being most pain). The following table summarize the results:

Placebo Group	Treatment Group
9	4
8	5
7	3
7	7
6	5

1. Write out your null and alternative hypotheses.
2. Carry out the appropriate statistical test to determine if the new drug is effective (Two sided).

3.) A psychiatrist is interested in the relationship between sleep and depression. In order to test this relationship, the psychiatrist groups participants in her study into quartiles of sleep and then reports each participant’s score on the geriatric depression scale (GDS). The following table summarizes the GDS scores by sleep category.

Sleep Group I (<Quartile1)	Sleep Group II (Between Quartile I and Quartile II)	Sleep Group III (Between Quartile II and Quartile III)	Sleep Group IV (Greater than Quartile III)
4	5	6	8
4	6	7	4
5	8	8	3
3	6	9	9
2	7	8	11
6	5	3	8

           

Carry out the appropriate statistical test to determine if there is a difference in mean score on the GDS among the groups categorized by sleep.

Answer 1

1)

Answer)

As the population s.d is unknown here and we are using sample s.d as the best estimate, we will use t distribution table to conduct the test.

A)

Null hypothesis Ho : u = 5

Alternate hypothesis Ha : u > 5

B)

Test statistics t = (sample .mean - claimed mean)/(s.d/√n)

t = (6.4 - 5)/(0.8/√64)

t = 14

Degrees of freedom is = n-1 = 63

For 63 dof and 14 test statistics, P-value from t distribution is 0.

As the obtained P-value is extremely small.

We reject the null hypothesis Ho.

So, we have enough evidence to conclude that new drug that is intended to naturally increase hours of sleep for elderly patients experiencing insomnia above 5 hours.