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

`t = 0.875`

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{(1)/(n_1) + (1)/(n_2)}}`

Calculate s using:

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

The equation becomes:

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

`s = \sqrt{(1451.34)/(22)}`

`s = √(65.97)`

`s = 8.12`

So:

`t = \frac{\bar x_1 - \bar x_2}{s\sqrt{(1)/(n_1) + (1)/(n_2)}}`

`t = \frac{21.8 - 18.9}{8.12 * \sqrt{(1)/(12) + (1)/(12)}}`

`t = \frac{21.8 - 18.9}{8.12 * \sqrt{(1)/(6)}}`

`t = (21.8 - 18.9)/(8.12 * 0.408)}`

`t = (2.9)/(3.31296)}`

`t = 0.875`