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.

Respuesta :

Answer:

[tex]t = 0.875[/tex]

Step-by-step explanation:

Given

Brand A            Brand B

[tex]n_ 1= 12[/tex]               [tex]n_2 = 12[/tex]

[tex]\bar x_1 = 21.8[/tex]            [tex]\bar x_2 = 18.9[/tex]

[tex]\sigma_1 = 8.7[/tex]              [tex]\sigma_2 = 7.5[/tex]

Required

Determine the test statistic (t)

This is calculated as:

[tex]t = \frac{\bar x_1 - \bar x_2}{s\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}[/tex]

Calculate s using:

[tex]s = \sqrt{\frac{(n_1-1)*\sigma_1^2+(n_2-1)*\sigma_2^2}{n_1+n_2-2}}[/tex]

The equation becomes:

[tex]s = \sqrt{\frac{(12-1)*8.7^2+(12-1)*7.5^2}{12+12-2}}[/tex]

[tex]s = \sqrt{\frac{1451.34}{22}}[/tex]

[tex]s = \sqrt{65.97}[/tex]

[tex]s = 8.12[/tex]

So:

[tex]t = \frac{\bar x_1 - \bar x_2}{s\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}[/tex]

[tex]t = \frac{21.8 - 18.9}{8.12 * \sqrt{\frac{1}{12} + \frac{1}{12}}}[/tex]

[tex]t = \frac{21.8 - 18.9}{8.12 * \sqrt{\frac{1}{6}}}[/tex]

[tex]t = \frac{21.8 - 18.9}{8.12 * 0.408}}[/tex]

[tex]t = \frac{2.9}{3.31296}}[/tex]

[tex]t = 0.875[/tex]