Math  /  Data & Statistics

QuestionQuestion 9 of 10 (1 point) I Question Attempt 3 of Unlimited
Fast reactions: In a study of reaction times, the time to respond to a visual stimulus (x)(x) and the time to respond to an auditory stimulus ( yy ) were recorded for each of 7 subjects. Times were measured in thousandths of a second. The results are presented in the following table. \begin{tabular}{cc} \hline Visual & Auditory \\ \hline 161 & 159 \\ 176 & 163 \\ 178 & 201 \\ 188 & 193 \\ 201 & 197 \\ 203 & 206 \\ 211 & 189 \\ \hline \end{tabular} Send data to Excel
The least-squares regression line y^=b0+b1x=51.2341+0.7203x\hat{y}=b_{0}+b_{1} x=51.2341+0.7203 x and Σ(xxˉ)2=1895.4286\Sigma(x-\bar{x})^{2}=1895.4286 are known for this data. Construct a 95%95 \% confidence interval for the slope. Round the answers to at least four decimal places.
The 95%95 \% confidence interval is \square <β1<<\beta_{1}< \square 1. Save For Later Submit Assig

Studdy Solution

STEP 1

1. The data consists of paired observations for visual and auditory reaction times.
2. The least-squares regression line is given by y^=51.2341+0.7203x\hat{y} = 51.2341 + 0.7203x.
3. The sum of squares of the deviations from the mean for xx is Σ(xxˉ)2=1895.4286\Sigma(x-\bar{x})^{2} = 1895.4286.
4. We need to construct a 95% confidence interval for the slope β1\beta_1.

STEP 2

1. Calculate the standard error of the slope.
2. Determine the critical value for the t-distribution.
3. Construct the confidence interval for the slope.

STEP 3

Calculate the standard error of the slope (SEb1SE_{b_1}) using the formula:
SEb1=Σ(yiy^i)2(n2)Σ(xxˉ)2SE_{b_1} = \sqrt{\frac{\Sigma(y_i - \hat{y}_i)^2}{(n-2) \cdot \Sigma(x-\bar{x})^2}}
We need to calculate Σ(yiy^i)2\Sigma(y_i - \hat{y}_i)^2. First, compute the predicted values y^i\hat{y}_i for each xix_i, and then find the residuals yiy^iy_i - \hat{y}_i.

STEP 4

Calculate the predicted values y^i\hat{y}_i using the regression equation for each xix_i:
\begin{align*} \hat{y}_1 &= 51.2341 + 0.7203 \times 161 \\ \hat{y}_2 &= 51.2341 + 0.7203 \times 176 \\ \hat{y}_3 &= 51.2341 + 0.7203 \times 178 \\ \hat{y}_4 &= 51.2341 + 0.7203 \times 188 \\ \hat{y}_5 &= 51.2341 + 0.7203 \times 201 \\ \hat{y}_6 &= 51.2341 + 0.7203 \times 203 \\ \hat{y}_7 &= 51.2341 + 0.7203 \times 211 \\ \end{align*}
Calculate each y^i\hat{y}_i.

STEP 5

Calculate the residuals yiy^iy_i - \hat{y}_i and then Σ(yiy^i)2\Sigma(y_i - \hat{y}_i)^2.

STEP 6

Substitute Σ(yiy^i)2\Sigma(y_i - \hat{y}_i)^2 and Σ(xxˉ)2\Sigma(x-\bar{x})^2 into the formula for SEb1SE_{b_1}.

STEP 7

Determine the critical value tt^* for a 95% confidence interval with n2n-2 degrees of freedom (where n=7n = 7).

STEP 8

Construct the confidence interval for the slope β1\beta_1 using:
b1±tSEb1b_1 \pm t^* \cdot SE_{b_1}
Substitute the values for b1b_1, tt^*, and SEb1SE_{b_1}.
The 95% confidence interval for the slope β1\beta_1 is:
Lower bound<β1<Upper bound\boxed{\text{Lower bound}} < \beta_1 < \boxed{\text{Upper bound}}

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