Question

15 patients have been examined after intake for Glucose levels of a drug. If mean difference= 2.55 and standard deviation difference= 2.8 of glucose level, was observed between before and after intake of this drug.

Does that prove the new drug is effective in the treatment of diabetes at 95% CL? please explain

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 0
Alternative Hypothesis, Ha: μ ≠ 0

Rejection Region
This is two tailed test, for α = 0.05 and df = 14
Critical value of t are -2.145 and 2.145.
Hence reject H0 if t < -2.145 or t > 2.145

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (2.55 - 0)/(2.8/sqrt(15))
t = 3.527

P-value Approach
P-value = 0.0034
As P-value < 0.05, reject the null hypothesis.

sample mean, xbar = 2.55
sample standard deviation, s = 2.8
sample size, n = 15
degrees of freedom, df = n - 1 = 14

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.145

ME = tc * s/sqrt(n)
ME = 2.145 * 2.8/sqrt(15)
ME = 1.551

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (2.55 - 2.145 * 2.8/sqrt(15) , 2.55 + 2.145 * 2.8/sqrt(15))
CI = (1 , 4.1)

as the confidenc einterval does not contain 0 , so, prove the new drug is effective in the treatment of diabetes at 95% CL