Math / Data & Statistics

Questionand after the diesel exhaust exposure. The resulting data are given in the accompanying table. For purposes of this exercise, assume that the sample of 10 men is representative of healthy adult males. \begin{tabular}{|c|c|c|c|c|} \hline \multirow{2}{*}{ Subject } & \multicolumn{4}{|c|}{ MPF (in Hz) } \\ \cline { 2 - 5 } & Location 1 before & Location 1 after & Location 2 before & Location 2 after \\ \hline 1 & 6.4 & 8.0 & 6.8 & 9.4 \\ \hline 2 & 8.6 & 12.7 & 9.5 & 11.2 \\ \hline 3 & 7.4 & 8.4 & 6.6 & 10.2 \\ \hline 4 & 8.6 & 9.0 & 9.0 & 9.7 \\ \hline 5 & 9.9 & 8.4 & 9.6 & 9.2 \\ \hline 6 & 7.8 & 11.0 & 9.0 & 11.8 \\ \hline 7 & 7.4 & 14.4 & 7.8 & 9.3 \\ \hline 8 & 6.7 & 7.3 & 7.1 & 8.0 \\ \hline 10 & 8.8 & 11.2 & 7.4 & 9.3 \\ \hline \end{tabular}
Use a table or technology. Round your answers to two decimal places.) ( 081 , 315 ) Hz

Studdy Solution

STEP 1

1. The sample consists of 10 healthy adult males.
2. We are constructing a 90% confidence interval for the difference in mean MPF at brain location 1 before and after exposure.
3. The data is paired, meaning each subject has two measurements: one before and one after exposure.
4. The differences in MPF are normally distributed or the sample size is large enough to apply the Central Limit Theorem.

STEP 2

1. Calculate the differences in MPF for each subject at location 1.
2. Compute the mean and standard deviation of these differences.
3. Determine the critical value for a 90% confidence interval.
4. Calculate the confidence interval for the mean difference.

STEP 3

Calculate the differences in MPF for each subject at location 1:
\begin{align} \text{Subject 1:} & \quad 8.0 - 6.4 = 1.6 \\ \text{Subject 2:} & \quad 12.7 - 8.6 = 4.1 \\ \text{Subject 3:} & \quad 8.4 - 7.4 = 1.0 \\ \text{Subject 4:} & \quad 9.0 - 8.6 = 0.4 \\ \text{Subject 5:} & \quad 8.4 - 9.9 = -1.5 \\ \text{Subject 6:} & \quad 11.0 - 7.8 = 3.2 \\ \text{Subject 7:} & \quad 14.4 - 7.4 = 7.0 \\ \text{Subject 8:} & \quad 7.3 - 6.7 = 0.6 \\ \text{Subject 9:} & & \\ \text{Subject 10:} & \quad 11.2 - 8.8 = 2.4 \\ \end{align}

STEP 4

Compute the mean of these differences:
$\text{Mean difference} = \frac{1.6 + 4.1 + 1.0 + 0.4 - 1.5 + 3.2 + 7.0 + 0.6 + 2.4}{9}$

STEP 5

Compute the standard deviation of these differences. First, find the squared differences from the mean, then take the square root of the variance:
$\text{Standard deviation} = \sqrt{\frac{\sum (\text{difference} - \text{mean difference})^2}{n-1}}$

STEP 6

Determine the critical value for a 90% confidence interval using a t-distribution with $n-1 = 9$ degrees of freedom.

STEP 7

Calculate the confidence interval for the mean difference:
$\text{Confidence Interval} = \text{Mean difference} \pm (\text{Critical value} \times \text{Standard deviation})$
The 90% confidence interval for the difference in mean MPF at brain location 1 before and after exposure is:
$(0.81, 3.15) \, \text{Hz}$

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Studdy Solution

STEP 1

STEP 2

1. Calculate the differences in MPF for each subject at location 1.
2. Compute the mean and standard deviation of these differences.
3. Determine the critical value for a 90% confidence interval.
4. Calculate the confidence interval for the mean difference.

STEP 3

STEP 4

Compute the mean of these differences:
$\text{Mean difference} = \frac{1.6 + 4.1 + 1.0 + 0.4 - 1.5 + 3.2 + 7.0 + 0.6 + 2.4}{9}$

STEP 5

Compute the standard deviation of these differences. First, find the squared differences from the mean, then take the square root of the variance:
$\text{Standard deviation} = \sqrt{\frac{\sum (\text{difference} - \text{mean difference})^2}{n-1}}$

STEP 6

Determine the critical value for a 90% confidence interval using a t-distribution with $n-1 = 9$ degrees of freedom.

STEP 7

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Studdy Solution

STEP 1

STEP 2

1. Calculate the differences in MPF for each subject at location 1.2. Compute the mean and standard deviation of these differences.3. Determine the critical value for a 90% confidence interval.4. Calculate the confidence interval for the mean difference.

STEP 3

STEP 4

Compute the mean of these differences:Mean difference=1.6+4.1+1.0+0.4−1.5+3.2+7.0+0.6+2.49\text{Mean difference} = \frac{1.6 + 4.1 + 1.0 + 0.4 - 1.5 + 3.2 + 7.0 + 0.6 + 2.4}{9}Mean difference=91.6+4.1+1.0+0.4−1.5+3.2+7.0+0.6+2.4​

STEP 5

STEP 6

Determine the critical value for a 90% confidence interval using a t-distribution with n−1=9 n-1 = 9 n−1=9 degrees of freedom.

STEP 7

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1. Calculate the differences in MPF for each subject at location 1.
2. Compute the mean and standard deviation of these differences.
3. Determine the critical value for a 90% confidence interval.
4. Calculate the confidence interval for the mean difference.

Compute the mean of these differences:
$\text{Mean difference} = \frac{1.6 + 4.1 + 1.0 + 0.4 - 1.5 + 3.2 + 7.0 + 0.6 + 2.4}{9}$

Determine the critical value for a 90% confidence interval using a t-distribution with $n-1 = 9$ degrees of freedom.