Question 1

An article in the Journal of Sound and Vibration [“Measurement of Noise-Evoked Blood Pressure by Means of Averaging Method: Relation between Blood Pressure Rise and PSL” (1991, Vol. 151(3), pp. 383-394)] described a study investigating the relationship between noise exposure and hypertension. The following data are representative of those reported in the article.

A. Draw a scatter diagram of y (blood pressure rise in millimeters of mercury) versus x (sound pressure level in decibels). Does a simple linear regression model seem reasonable in this situation?

Looking at the scatter diagram with a fitted regression line, we can see that there is a linear relationship between the values. Hence, a simple regression model in this situation seems to be reasonable.

B. Fit the simple linear regression model using least squares. Find an estimate of \(\sigma^2\)

From the data, we get n = 20, \(\sum{X_i}\) = 1656, \(\sum{Y_i}\) = 86, \(\bar{x}\) = 82.8, \(\bar{y}\) = 4.3, \(\sum{X_i^2}\) = 140,176, \(\sum{Y_i^2}\) = 49, \(\sum{X_iY_i}\) = 7654

Using these data, we first get \(S_{xx}\) using the formula: \[ S_{xx} ~ = ~ \sum{X_i^2}~-~\frac{(\sum{X_i})^2}{n} \] So: \[ S_{xx} ~ = ~ 140,176~-~\frac{(1656)^2}{20}~= ~ 3,059.2 \]

We then get \(S_{xy}\) using the formula: \[ S_{xy}~=~\sum{X_iY_i}~-~\frac{(\sum{Y_i})(\sum{X_i})}{n} \]

So, \[ S_{xy}~=~7,654~-~\frac{(86)(1656)}{20}~=~533.2 \]

Using the values obtained for \(S_{xx}\) and \(S_{xy}\) we can get the coefficient \(\hat{\beta_1}\) with \[ \hat{\beta_1} ~=~\frac{S_{xy}}{S_{xx}}~=\frac{533.2}{3059.2}~=~0.17429 \] So \(\hat{\beta_1}\) = 0.17429

Now with \(\hat{\beta_1}\), we can solve for the intercept \(\hat{\beta_0}\) with the values of \(\bar{y}\) and \(\bar{x}\): \[ \hat{\beta_0}~=~\bar{y}~-~\hat{\beta_1}\bar{x}~=~4.3 ~-~(0.17429)(82.8)~=~-10.13121 \] So \(\hat{\beta_0}\) = -10.13121

Using the values of \(\hat{\beta_0}\) and \(\hat{\beta_1}\), Our simple regression model is: \[ \hat{y}~= \hat{\beta_0}~+~\hat{\beta_1}x ~:~ \hat{y}~=-10.13121~+~0.17429x \] So our regression model is:

\(\hat{y}\) = -10.13121 + 0.17429x

Estimating \(\sigma^2\): \[ \sigma^2~=~\frac{SS_E}{n-2}~where~SS_E~=~SS_T~-~\hat{\beta_1}S_{xy} \] \[ SS_T~=~\sum{Y_i^2~-~n\bar{y}^2}~=~494~-(20)(4.3)^2~=~124.2 \] \[ SS_E~=~124.2~-~(0.17429)(533.2)~=~31.268572 \]

Getting the estimated \(\sigma^2\): \[ \sigma^2~=~\frac{SS_E}{n-2}~=~\frac{31.268572}{20-2}~=~1.73714 \]

Our estimated \(\sigma^2\) is 1.73714

C.Find the predicted mean rise in blood pressure level associated with a sound pressure level of 85 decibels.

Letting x = 85 and using the simple regression model \[ \hat{y}~=-10.13121~+~0.17429x \] Substituting the value of x we get: \[ \hat{y}~=-10.13121~+~0.17429(85)~=~4.68255 \]

So, The predicted mean rise in blood pressure level associated with a sound pressure level of 85 decibels is 4.68255