Question

Two competing headache remedies claim to give fast-acting relief. An experiment was performed to compare the mean lengths of time required for bodily absorption of brand A and brand B headache remedies. Twelve people were randomly selected and given an oral dose of brand A and another 12 people were randomly selected and given an oral dose of brand B. The lengths of time in minutes for the drugs to reach a specified level in the blood were recorded. The mean and standard deviation for brand A was 21.8 and 8.7 minutes, respectively. The mean and standard deviation for brand B was 18.9 and 7.5 minutes, respectively. Past experience with the drug composition of the two remedies permits researchers to assume that both distributions are approximately Normal. Let us use a 5% level of significance to test the claim that there is no difference in the mean time required for bodily absorption.
Required:
Find or estimate the p — value of the sample test statistic.

Answer 1

Answer:

$t = 0.875$

Step-by-step explanation:

Given

Brand A            Brand B

$n_ 1= 12$               $n_2 = 12$

$\bar x_1 = 21.8$            $\bar x_2 = 18.9$

$\sigma_1 = 8.7$              $\sigma_2 = 7.5$

Required

Determine the test statistic (t)

This is calculated as:

$t = \frac{\bar x_1 - \bar x_2}{s\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

Calculate s using:

$s = \sqrt{\frac{(n_1-1)*\sigma_1^2+(n_2-1)*\sigma_2^2}{n_1+n_2-2}}$

The equation becomes:

$s = \sqrt{\frac{(12-1)*8.7^2+(12-1)*7.5^2}{12+12-2}}$

$s = \sqrt{\frac{1451.34}{22}}$

$s = \sqrt{65.97}$

$s = 8.12$

So:

$t = \frac{\bar x_1 - \bar x_2}{s\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

$t = \frac{21.8 - 18.9}{8.12 * \sqrt{\frac{1}{12} + \frac{1}{12}}}$

$t = \frac{21.8 - 18.9}{8.12 * \sqrt{\frac{1}{6}}}$

$t = \frac{21.8 - 18.9}{8.12 * 0.408}}$

$t = \frac{2.9}{3.31296}}$

$t = 0.875$