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[t]{2}{*}{Subject} & \multicolumn{4}{|c|}{MPF (in Hz)} \\ \hline & 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 & 8.8 & 11.0 & 9.0 & 11.8 \\ \hline 7 & 9.1 & 14.4 & 7.8 & 9.3 \\ \hline 8 & 7.4 & 11.1 & 8.1 & 9.1 \\ \hline 9 & 6.7 & 7.3 & 7.2 & 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.) ( $\square$ , $\square$ ) Hz

Studdy Solution

STEP 1

1. The sample consists of 10 men, which is representative of 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 approximately normally distributed, or the sample size is large enough for the Central Limit Theorem to apply.

STEP 2

1. Calculate the differences in MPF for each subject at location 1 before and after exposure.
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 before and after exposure:
\begin{align} \text{Subject 1: } & 8.0 - 6.4 = 1.6 \\ \text{Subject 2: } & 12.7 - 8.6 = 4.1 \\ \text{Subject 3: } & 8.4 - 7.4 = 1.0 \\ \text{Subject 4: } & 9.0 - 8.6 = 0.4 \\ \text{Subject 5: } & 8.4 - 9.9 = -1.5 \\ \text{Subject 6: } & 11.0 - 8.8 = 2.2 \\ \text{Subject 7: } & 14.4 - 9.1 = 5.3 \\ \text{Subject 8: } & 11.1 - 7.4 = 3.7 \\ \text{Subject 9: } & 7.3 - 6.7 = 0.6 \\ \text{Subject 10: } & 11.2 - 8.8 = 2.4 \\ \end{align}

STEP 4

Compute the mean and standard deviation of these differences:
\begin{align} \text{Mean difference: } & \frac{1.6 + 4.1 + 1.0 + 0.4 - 1.5 + 2.2 + 5.3 + 3.7 + 0.6 + 2.4}{10} = 2.08 \\ \text{Standard deviation: } & s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \end{align}

STEP 5

Calculate the standard deviation:
$s = \sqrt{\frac{(1.6-2.08)^2 + (4.1-2.08)^2 + (1.0-2.08)^2 + (0.4-2.08)^2 + (-1.5-2.08)^2 + (2.2-2.08)^2 + (5.3-2.08)^2 + (3.7-2.08)^2 + (0.6-2.08)^2 + (2.4-2.08)^2}{9}}$

STEP 6

Determine the critical value for a 90% confidence interval. For a t-distribution with 9 degrees of freedom, the critical value $t^*$ can be found using a t-table or technology.

STEP 7

Calculate the confidence interval:
$\text{Confidence interval: } \bar{x} \pm t^* \left(\frac{s}{\sqrt{n}}\right)$
Substitute the values:
$2.08 \pm t^* \left(\frac{s}{\sqrt{10}}\right)$
The 90% confidence interval for the difference in mean MPF at brain location 1 before and after exposure is:
$(\square, \square) \text{ Hz}$

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Studdy Solution

STEP 1

STEP 2

1. Calculate the differences in MPF for each subject at location 1 before and after exposure.
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 and standard deviation of these differences:
\begin{align} \text{Mean difference: } & \frac{1.6 + 4.1 + 1.0 + 0.4 - 1.5 + 2.2 + 5.3 + 3.7 + 0.6 + 2.4}{10} = 2.08 \\ \text{Standard deviation: } & s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \end{align}

STEP 5

Calculate the standard deviation:
$s = \sqrt{\frac{(1.6-2.08)^2 + (4.1-2.08)^2 + (1.0-2.08)^2 + (0.4-2.08)^2 + (-1.5-2.08)^2 + (2.2-2.08)^2 + (5.3-2.08)^2 + (3.7-2.08)^2 + (0.6-2.08)^2 + (2.4-2.08)^2}{9}}$

STEP 6

Determine the critical value for a 90% confidence interval. For a t-distribution with 9 degrees of freedom, the critical value $t^*$ can be found using a t-table or technology.

STEP 7

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Studdy Solution

STEP 1

STEP 2

1. Calculate the differences in MPF for each subject at location 1 before and after exposure.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 and standard deviation of these differences:\begin{align*} \text{Mean difference: } & \frac{1.6 + 4.1 + 1.0 + 0.4 - 1.5 + 2.2 + 5.3 + 3.7 + 0.6 + 2.4}{10} = 2.08 \\ \text{Standard deviation: } & s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \end{align*}

STEP 5

STEP 6

Determine the critical value for a 90% confidence interval. For a t-distribution with 9 degrees of freedom, the critical value t∗ t^* t∗ can be found using a t-table or technology.

STEP 7

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1. Calculate the differences in MPF for each subject at location 1 before and after exposure.
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 and standard deviation of these differences:
\begin{align} \text{Mean difference: } & \frac{1.6 + 4.1 + 1.0 + 0.4 - 1.5 + 2.2 + 5.3 + 3.7 + 0.6 + 2.4}{10} = 2.08 \\ \text{Standard deviation: } & s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \end{align}

Determine the critical value for a 90% confidence interval. For a t-distribution with 9 degrees of freedom, the critical value $t^*$ can be found using a t-table or technology.