Question

A drug manufacturer claims that its new drug causes faster red-cell buildup in anemic people than the drug currently used. A team of doctors tests the drug on 6 randomly selected people and compares the results with the current buildup factor of 7.1 (this is the buildup factor for the current drug). The results of the 6 peoples are (theses are red-cell buildup factors):

6.3 7.8 8.1 8.3 8.7 9.4.

Test the manufacturer’s claim by stating the null and alternative hypothesis, then make your decision and draw your conclusions using the following methods. (Note: since this is a medical study, researchers often want to be very certain that a new drug does what it states it does. As a result, the test is often performed with a low level of significance, as is the case here, meaning that you require very strong evidence against the null hypothesis before you will reject it)

(a) Classical testing at α = 1%

(b) The p-value method.

Answer 1

Part a

Here, we have to use one sample t test for the population mean. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: The new drug do not causes faster red-cell buildup in anemic people.

Alternative hypothesis: H_a: The new drug causes faster red-cell buildup in anemic people.

H₀: µ ≤ 7.1 versus H_a: µ > 7.1

This is an upper tailed or right tailed (one tailed) test.

The test statistic formula for this test is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

Xbar = 8.1

S = 1.041153207

n = 6

df = n – 1 = 6 – 1 = 5

α = 0.01

Critical value = 3.3649

(by using t-table)

t = (8.1 – 7.1)/[ 1.041153207/sqrt(6)]

t = 2.3527

Test statistic = t = 2.3527 < Critical value = 3.3649

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that the new drug causes faster red-cell buildup in anemic people.

Part b

Here, we have to use one sample t test for the population mean. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: The new drug do not causes faster red-cell buildup in anemic people.

Alternative hypothesis: H_a: The new drug causes faster red-cell buildup in anemic people.

H₀: µ ≤ 7.1 versus H_a: µ > 7.1

This is an upper tailed or right tailed (one tailed) test.

The test statistic formula for this test is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

Xbar = 8.1

S = 1.041153207

n = 6

df = n – 1 = 6 – 1 = 5

α = 0.01

t = (8.1 – 7.1)/[ 1.041153207/sqrt(6)]

t = 2.3527

P-value = 0.0327

(by using t-table)

P-value > α = 0.01

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that the new drug causes faster red-cell buildup in anemic people.