Question

Q18. Health Department, NYC plans to release a new drug commercially that is expected to lower

        blood pressure effectively. The drug was administered to a random sample of 20 patients.

        Among these patients the decrease in the blood pressure of 7 patients was recorded and no

        change in blood pressure of other patients was recorded . The decrease in the blood pressure of

        the 7 patients was 21, 22, 16, 19, 20, 23, 19.  At 1% level of significance, to support the

        release of drug, Health Department used

        a. mean decrease in blood pressure =20for test of hypothesis

         b. Critical value t_α/2= 3.707 was valid value to compare t_STAT

         c. Degree of freedom = 6 were used to get t_α/2for the conclusion                      

         d none of the above a, b, c are correct for conclusion        

Answer 1

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Lets see:

C. df = n-1 = 20-1 = 19. So option C is wrong

B. Critical value of t can be found by using the formula or looking up the t-tables for df = 6, and alpha = .01

This is a one tailed test( left-tailed test to be precise, as we are testing for the claim that it is LESS than the population parameter)

Excel formula if you want use Excel to dervive the critival t-value =T.INV(0.99, 19)= -2.54

So, B is incorrect

A. Mean decrease is the average of the decrease of all 20 data points, not just 7. Lets calculate the mean and check that' 20.

(21+22+16+19+20+23+19)/20 = 140/20 = 7

So, A is also incorrect

Answer: D is correct. None of the above a, b and c are correct