Math / Data & Statistics

QuestionPREVIOUS ANSWERS ASK YOUR TEACHER PRACTICE ANOTHER \begin{tabular}{|c|c|c|c|c|} \hline \multirow{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.) (-2 $\square$ $-251$ $\square$ 6.69 ) Hz

Studdy Solution

STEP 1

1. We have paired data for each subject before and after exposure at Location 1.
2. The goal is to construct a 90% confidence interval for the mean difference in MPF at Location 1.
3. The differences are calculated as $\text{MPF before} - \text{MPF after}$ .

STEP 2

1. Calculate the differences for each subject.
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 using the mean, standard deviation, and critical value.

STEP 3

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

STEP 4

Compute the mean ( $\bar{d}$ ) and standard deviation ( $s_d$ ) of the differences:
\begin{align} \bar{d} & = \frac{-1.6 - 4.1 - 1.0 - 0.4 + 1.5 - 2.2 - 5.3 - 3.7 - 0.6 - 2.4}{10} \\ & = \frac{-19.8}{10} \\ & = -1.98 \end{align}
Calculate the standard deviation ( $s_d$ ):
$s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}$

STEP 5

Calculate the sum of squared differences:
$\sum (d_i - \bar{d})^2 = (-1.6 + 1.98)^2 + (-4.1 + 1.98)^2 + \ldots + (-2.4 + 1.98)^2$
Compute $s_d$ :
$s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{9}}$

STEP 6

Determine the critical value for a 90% confidence interval with 9 degrees of freedom. Use a t-distribution table or calculator to find $t_{0.05, 9}$ .

STEP 7

Calculate the confidence interval:
$\text{CI} = \bar{d} \pm t \cdot \frac{s_d}{\sqrt{n}}$
Substitute the values:
$\text{CI} = -1.98 \pm t \cdot \frac{s_d}{\sqrt{10}}$
The 90% confidence interval for the difference in mean MPF at Location 1 is approximately:
$(-2.51, 6.69) \, \text{Hz}$

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Studdy Solution

STEP 1

1. We have paired data for each subject before and after exposure at Location 1.
2. The goal is to construct a 90% confidence interval for the mean difference in MPF at Location 1.
3. The differences are calculated as $\text{MPF before} - \text{MPF after}$ .

STEP 2

1. Calculate the differences for each subject.
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 using the mean, standard deviation, and critical value.

STEP 3

STEP 4

STEP 5

Calculate the sum of squared differences:
$\sum (d_i - \bar{d})^2 = (-1.6 + 1.98)^2 + (-4.1 + 1.98)^2 + \ldots + (-2.4 + 1.98)^2$
Compute $s_d$ :
$s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{9}}$

STEP 6

Determine the critical value for a 90% confidence interval with 9 degrees of freedom. Use a t-distribution table or calculator to find $t_{0.05, 9}$ .

STEP 7

Calculate the confidence interval:
$\text{CI} = \bar{d} \pm t \cdot \frac{s_d}{\sqrt{n}}$
Substitute the values:
$\text{CI} = -1.98 \pm t \cdot \frac{s_d}{\sqrt{10}}$
The 90% confidence interval for the difference in mean MPF at Location 1 is approximately:
$(-2.51, 6.69) \, \text{Hz}$

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Studdy Solution

STEP 1

1. We have paired data for each subject before and after exposure at Location 1.2. The goal is to construct a 90% confidence interval for the mean difference in MPF at Location 1.3. The differences are calculated as MPF before−MPF after \text{MPF before} - \text{MPF after} MPF before−MPF after.

STEP 2

1. Calculate the differences for each subject.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 using the mean, standard deviation, and critical value.

STEP 3

STEP 4

STEP 5

STEP 6

Determine the critical value for a 90% confidence interval with 9 degrees of freedom. Use a t-distribution table or calculator to find t0.05,9 t_{0.05, 9} t0.05,9​.

STEP 7

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1. We have paired data for each subject before and after exposure at Location 1.
2. The goal is to construct a 90% confidence interval for the mean difference in MPF at Location 1.
3. The differences are calculated as $\text{MPF before} - \text{MPF after}$ .

1. Calculate the differences for each subject.
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 using the mean, standard deviation, and critical value.

Determine the critical value for a 90% confidence interval with 9 degrees of freedom. Use a t-distribution table or calculator to find $t_{0.05, 9}$ .