Question

A medical study was conducted to compare the difference in effectiveness of two drugs (A and B) in lowering cholesterol levels. A random sample of 8 participants was recruited to try one of the drugs for one month and the other drug - during another month (order of taking A and B was randomly). Based on observations collected for both drug A and B results researchers want to compare effectiveness of these two drugs. If the mean difference is 1.63 and the standard deviation of that difference is 3.78, what is the 99% confidence interval that corresponds to this case?

Select one:

a. (-2.38; 5.64)

b. (-1.82; 5.08)

c. (-3.04; 6.31)

d. (-2.86; 6.12)

Answer 1

As the mean difference is 1.63 and standard deviation is 3.78

so if we calcuate the standard error = standard deviation / sqrt(number of sample)

=3.78 /sqrt(8) = 1.3364

If we calculate t critical value of 99% at degree of freedom 7 (8-1) is 3.4994

so 3.4994 so we need to have the interval 3.4994*1.3364 =4.6767

So we need to have the inteeval of 4.6767 on either side of the mean

so lower limit = 1.63 -4.6767 = -3.046

so upper limit = 1.63 +4.6767 = 6.3067

So range of the same is (-3.04,6.31)

So correct answe is option C(-3.04,6.31)