Answer to Question #155773 in Statistics and Probability for kamrul

Question

The density of molten salt mixtures, y g/cm3 , was measured at various temperatures x°C. The results were:

xi 250 270 290 310 330 350

yi 1.955 1.935 1.890 1.920 1.895 1.865.

Assume that the graphical checks of problem above are satisfactory.

a. Calculate the estimated variance around the regression line, �)|+ &

b. Is the estimated regression coefficient b, significantly different from zero at the 1% level of significance? Can we conclude that temperature has a significant effect on density in this case?

c. What is the 95% confidence interval for ß, the true slope or regression coefficient?

d. Calculate the 95% confidence interval for the mean value of y at each of x = 250, 300, 350.

e. Calculate the correlation coefficient for this data set.

f. Calculate the coefficient of determination. 

Answer 1

Using Excel to fit the regression model

Steps:

1. Click of "data" tab and select "data analysis"

2. From the pop up, select "regression" and click "ok"

3. Under "input_y_range" select the range of cells with y values.

Under "input_x_range" select the range of cells with x values.

4. Check the box for "confidence level" and input 99%

5. Check the box "residuals". Select the output range and click ok

The output from excel is shown below

The estimated model is given as

$\hat y_i=2.1414-0.00077143x_i$

The residuals are obtained by

$e_i=observed (y_i)-estimate(\hat y_i)$

Part a)

variance around the regression line is the variance of the residuals which is obtained in from excel as

$Var(e_i) =0.0002469$

Part b)

The coefficient

$b=-0.00077143$. The test of significance for the coefficient returns a t statistic

$t=-3.67423$ and a corresponding $P\ value= 0.0213$

The p value is greater than 0.01 therefore the coefficient "b" is not significant at a 1% significance level. We can conclude that temperature has no significant effect on density

Part c)

The 95% confidence interval for the coefficient $\beta$ is $[-0.001354, -0.0001885]$

Part d)

At the $95\%$ confidence level, the t critical value is given as $t=+/-2.200$

$x= t*\sigma +\mu$

The standard deviation of the dependent variable y is given by

$=[\frac{1}{6-1} *[(1.955-1.91)^2+(1.935-1.91)^2+(1.89-1.91)^2+(1.92-1.91)^2+(1.895-1.91)^2+(1.865-1.91)^2]]^{0.5}$

$\sigma=0.03286$

- When x= 250

$\hat y_i=2.1414-0.00077143(250)=2.3343$

The confidence interval is given as

$+/-2.200*0.03286+2.3343=[2.2620 , 2.4066]$

- when x=300

$\hat y=2.1414-0.00077143(300)= 1.90997$

The confidence interval is given as

$+/-2.200*0.03286+1.90997=[1.83768 , 1.98226]$

- when x= 350

$\hat y= 2.1414-0.00077143(350)=[1.79911,1.94369]$

Part e)

The correlation coefficient is the square root of the r square value. From the Excel output

$r^2=0.77143$

Therefore

$r=\sqrt{0.77143}=0.8783$

Part f)

The coefficient of determination is the $r^2$ Value.

$r^2=0.77143$