Two new drugs are being compared for effectiveness in lowering total cholesterol
ID: 3294058 • Letter: T
Question
Two new drugs are being compared for effectiveness in lowering total cholesterol levels. On a sample of 10 subjects, drug A is found to lower cholesterol by an average of 65 mg/dl with SS=2025.On a sample of 15 subjects, drug B lowered the mean cholesterol level by 72 mg/dl with SS=2525. Use a 0.05 level of significance (2 tail test) to test the claim that the two drugs are equally effective. Assume that the population is normally distributed.
THE FIRST USE OF THE TERM SS IS TO DETERMINE THE VARIANCE
a. The null hypotheses (H0) _______________
b. The alternative hypothesis (H1) ____________
c. The critical value _______________
d. The test statistic __________________
e. Your decision about the claim ________________
f. Your conclusion about the claim
Explanation / Answer
(a) The null hypothesis : H0 : Two drugs are equally effective. 1 = 2
(b) The alternative hypothesis : H1 : Two drugs are not equally effective. 1 2
(c) Mean value of drag A effectiveness to lower cholestrol level xA = 65 mg/dl
Variance of drug A = SS/ (n-1) = 2025/ (10-1) = 225
Standard Deviation for drug A = sqrt (225) = 15 mg/dl
Mean value of drug Beffectiveness to lower cholestrol level xA = 72 mg/dl
Variance of drug B = SS/ (n-1) = 2525/ (15-1) = 180.357
Standard Deviation for drug B = sqrt (180.357) = 13.43 mg/dl
Here Degrees of Freedom = 10 + 15 - 2 = 23
so, critical value of t for dF = 23 and alpha = 0.05; tcr = 2.0686
(d) Test Statistic
Pooled Variance sp2 = (SSA + SSB)/ dF = (2025 + 2525)/23 = 197.826
so pooled standard deviation sp = 14.065 mg/dl
t = (xA - xB) / sp * sqrt (1/nA + 1/nB ) = (72 - 65)/ [14.065 * sqrt(1/15 + 1/10)] = 7/ 5.742 = 1.22
(e) SO here t < tcritical , we can not reject the null hypothesis.
(f) We can conclude that both of the drugs are effective in the same manner. There is no statistically difference between both.
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