Question

Economists wish to look at the relationship between the price of regular unleaded gasoline in the U.S. (units=$/gallon) and the acquisition cost of OPEC crude oil (units=$10/barrel). Summary measures for a random sample of 13 years are given below.

n=	13
?yi=	31.5
?y2i=	80.95
?xi=	60.5
?x2i=	329.67
?xiyi=	156.98

What is the Lower and upper end points for 95% confidence interval for the mean value of y when x0=2.9 (with appropriate units)?

. If x0=1.7, what is the predicted value for y (with appropriate units)?

. If x0=1.7, what is 80% prediction interval for an individual value of y (with appropriate units)?

What is the estimated regression equation? (Specify which variable is y and which is x)

If you will only answer the first part don’t bother answering the question please.

Please give detailed step by step solution

Answer 1

The simple linear regression model to define the relationship between price of regular unleaded gasoline which is the pedicted variable and the acquisition cost of OPEC crude oil which is the predicter variable is obtained as follow;

The simple linear regression model is;

$Y=a+bX$

The value of coefficient can be estimate using the formula;

$\widehat{b}=\frac{N\sum X_{i}Y_{i}-\sum X_{i}\sum Y_{i}}{N\sum X_{i}^{2}-\left ( \sum X_{i} \right )^{2}}=\frac{13\times 156.98-60.5\times 31.5}{13\times 329.67- 60.5^{2}}=0.2158$

$\widehat{a}=\frac{\sum Y_{i}}{N}-\widehat{b}\ \frac{\sum X_{i}}{N}=\frac{31.5}{13}-0.2158\times \frac{60.5}{13}=1.41877$

The equation is obtained as;

$\widehat{Y}=1.4187+0.2158X$

The estimated value for X=1.7 is obtained as;

$\widehat{Y}=1.4187+0.2158\times 1.7=1.78556$

To calculate the confidence interval, the estimated value for X=2.9 is obtained as;

$\widehat{Y}=1.4187+0.2158\times 2.9=2.04452$

Standard error for estimate at X=2.9 is obtained as;

$S_{YX}=\sqrt{\frac{\sum \left ( Y-\widehat{Y} \right )^{2}}{n-2}}=\sqrt{\frac{\sum \left ( Y^{2}+\widehat{Y}^{2} -2Y\widehat{Y}\right )}{n-2}}=\sqrt{\frac{\sum Y^{2}+\sum\widehat{Y}^{2} -\sum2Y\widehat{Y}}{n-2}}$

$S_{YX}=\sqrt{\frac{80.95+13\times 2.04452^{2}-2\times 31.5\times 2.04452}{11}}=0.76788$

The confidence interval for the mean is calculated using the formula;

$95\%\ CI=\widehat{Y}\pm t_{n-2}s_{YX}\sqrt{\frac{1}{n}+\frac{\left ( Y-\overline{Y} \right )^{2}}{\sum \left (Y-\overline{Y} \right )^{2}}}$

$95\%\ CI=\widehat{Y}\pm t_{n-2}s_{YX}\sqrt{\frac{1}{n}+\frac{\left ( Y-\overline{Y} \right )^{2}}{\sum Y^{2}-n\overline{Y}^{2}}}$

$Where\ \ \boldsymbol{t_{n-2}=1.7958}\ for\ 95\%\ confidence Interval$

$95\%\ CI=2.04452\pm 1.7958\times 0.76788\sqrt{\frac{1}{13}+\frac{\left ( 2.9-\frac{60.5}{13} \right )^{2}}{329.67-13\times \left (\frac{60.5}{13} \right )^{2}}}$

$95\%\ CI=2.04452\pm 0.51754=\left [1.52698, 2.56205 \right ]$

80% prediction interval for an individual value of y is calculated as;

$80\%\ CI=\widehat{Y}\pm t_{n-2}s_{YX}\sqrt{1+\frac{1}{n}+\frac{\left ( Y-\overline{Y} \right )^{2}}{\sum \left (Y-\overline{Y} \right )^{2}}}$

$80\%\ CI=\widehat{Y}\pm t_{n-2}s_{YX}\sqrt{1+\frac{1}{n}+\frac{\left ( Y-\overline{Y} \right )^{2}}{\sum Y^{2}-n\overline{Y}^{2}}}$

$Where\ \ \boldsymbol{t_{n-2}=0.8755}\ for\ 80\%\ confidence Interval$

$80\%\ CI=1.78556\pm 0.8755\times 0.76788\sqrt{1+\frac{1}{13}+\frac{\left ( 1.7-\frac{60.5}{13} \right )^{2}}{329.67-13\times \left (\frac{60.5}{13} \right )^{2}}}$

$80\%\ CI=1.78556\pm 0.98788=\left [0.80348, 2.76763 \right ]$