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Sati [7]
3 years ago
13

Two competing headache remedies claim to give fast-acting relief. An experiment was performed to compare the mean lengths of tim

e 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.
Mathematics
1 answer:
DaniilM [7]3 years ago
3 0

Answer:

t = 0.875

Step-by-step explanation:

Given

<u>Brand A</u>            <u>Brand B</u>

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

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