Brand name producers of aspirin claim that one advantage of their aspirin over generic aspirin is that brand name aspirin is much more consistent in the amount of active ingredient used. This in turn means that users can expect the same results each time they use the brand name aspirin, while the effects of the generic aspirin can be a lot more variable. A random sample of 200 brand name aspirin tablets had a mean and standard deviation of active ingredient of 325.01 and 10.12 mg. A second independent sample of 180 generic aspirin tablets was measured for the amount of active ingredient, and the mean standard deviation were 323.47 and 11.43 mg. Given that the amount of active ingredient is normally distributed for both the brand name and the generic aspirin, do these data support the brand name producers claim? Let alpha = 0.025.

Step-by-step explanation:

The claim here is that the brand name aspirin is more consistent in the amount of active ingredient used than the generic aspirin.

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean amount of active ingredients in brand name aspirin and μ2 be the mean amount of active ingredients in generic name aspirin

The random variable is μ1 - μ2 = difference in the mean amount of active ingredients between the brand name and generic aspirin

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 ≥ μ2 H0 : μ1 - μ2 ≥ 0

The alternative hypothesis is

H1 : μ1 < μ2 H1 : μ1 - μ2 < 0

This is a left tailed test

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

x1 = 325.01

x2 = 323.47

s1 = 10.12

s2 = 11.43

n1 = 200

n2 = 180

t = (325.01 - 323.47)/√(10.12²/200 + 11.43²/180)

t = 1.24

1.237877

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [10.12²/200 + 11.43²/180]²/[(1/200 - 1)(10.12²/200)² + (1/180 - 1)(11.43²/180)²] = 1.53233946713/0.00537245359

df = 285

We would determine the probability value from the t test calculator. It becomes

p value = 0.108

Since alpha, 0.025 < than the p value, 0.108, then we would fail to reject the null hypothesis. Therefore, at 2.5% level of significance, these data support the brand name producers claim

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