Question

Suppose you have been hired by a pharmaceutical firm to perform a "clinical trial" to test for possible behavioral effects of a new pain reliever.  One of the things you decide to test for is the existence of depression‑producing side effects.  You use a standard instrument (test) for measuring depression.  The overall long‑term average for a normal adult population is 50 with a standard deviation of 10.  Higher numbers mean a greater amount of depression.  You have the resources of testing 18 people on this depression test after they have taken the new pain reliever.  Make up 16 data points such that the mean is between 52 and 56and the standarddeviation is at not too far from 10--say between 7 and 15).  Scores:

_____25_____    ___65_______    _____45_____    _____35_____    ____55______    _____65_____

___65_______    _____30_____    ___65_______    _____55_____    ___49_______    ____59______

_____59_____    ____69______    _____59_____    ____48______    

1.         Confirm that the mean is between 52 and 56.  If it is not, then change your data and start over.

2.         Find s (unbiased) of above data, confirming that s is between 7 and 15. If it is not, then change your data and start over.  Explain how it was obtained or attach work.

3.    State Ho using symbols and also in words.

        

4.    State H₁ using symbols and also in words.

        

      

5.         In everyday language say what a Type 1 error would be for this experiment?  

6.         In everyday language say what a Type 2 error would be for this experiment?  

7.         Which one would be worse?  Why?

      

8.         Based on your answer to #5-7, choose an appropriate alpha level and discuss the choice.

1. alpha = _______  Reasons:

2. What are the consequences of a higher level?  

3. What are the consequences of a lower level?  

        

9.          a.        Compute the obtained z value.  (Show work).  

        

        

           b.            What is the probability of getting a z value as unusual as the one you got simply due to random chance (i.e. add up areas in both tails)?

        

        

        

c.         Compare your answer in 9b to alpha and state your conclusion.

        

                

        

        

        

10.       Pretend that you do not know the population standard deviation

        

1. Write down s (unbiased) from #2.

2. Find .  (Show work).

3. Compute the obtained t‑value  (Show work).

        

        

4. Enter the df.  ______________

5. Find the critical t‑value for your chosen alpha level (two-tailed test).

6. Compare the critical t‑value to the obtained t‑value and state your conclusion.

        

        

        

     

11.         a.       Compute the confidence interval of the population mean based solely on your 18 data points.  Use the same alpha level as above.  (E.g. if your alpha level is 0.1, then this is your 90% confidence interval)

        

        

        

        

        

                

b.         Is the population mean included in your confidence interval?  Is this consistent with your conclusion in 10f?   Explain.

12.       Given your decision in 10f, what are the two possible outcomes of this experiment? (HINT: Consider the 2 x 2 hypothesis outcome table discussed in class)

        

        

        

  

      

        

13.       What are some methodological flaws in this experimental design and how could they be corrected?

Answer 1

1) Mean =

  

Hence the mean is between 52 and 56.

2) s.d. =

= =

= 13.39

Hence s(unbiased) is between 7 and 15.

3) There is no existence of depression producing side effects i.e.

4)   There is a greater amount of depression i.e.

5) Type 1 error is the rejection of a true null hypothesis i.e. There is no depression but still we are rejecting it and thinking that there is a greater amount of depression.

6) Type 2 error is the failure to reject a false null hypothesis i.e. There is a greater amount of depression but we are accepting the fact that there is no existence of depression.

7) Type 2 error is worst because we are not treating the patients with depression as we have accepted the fact that there is no depression.

8) a)Let us take as we have to reduce the probability of type 2 error as it is the worst so by taking higher value of will reduce this probability.

b) Higher values of make it easier to reject the null hypothesis, so choosing higher values for can reduce the probability of a Type II error.

c) Lower values of make it harder to reject the null hypothesis, so choosing lower values for can reduce the probability of a Type I error.

10) Here

  

Hence

Now from the t-distribution table at level of significance = 0.20 and d.f. = 16-1=15 we get

Since , we will accept the null hypothesis that is there is no existence of depression.

11) We need to calculate confidence interval solely on 18 data points i.e.

Also

Now confidence interval:

Yes the population mean is included in the confidence interval.