Question

Check my work The following data show the retail price for 12 randomly selected laptop computers along with their corresponding processor speeds in gigahertz. Computers Speed 2.0 1.6 1.6 1.8 2.0 1.2 2.0 1.6 2.0 NOWN Price $1008.50 461.00 532.00 971.00 1068.50 506.00 1098.50 693.50 1057.00 1001.00 468.50 435.50 Click here for the Excel Data File a. Develop a linear equation that can be used to describe how the price depends on the processor speed. (Negative value should be indicated by a minus sign. Round your answers to the nearest whole number.) Price = speed

Answer 1

a. Using Regression function in Data Analysis package , We get the following results :

SUMMARY OUTPUT Price Speed 1008 461 1.6 532 1.6 1.8 Regression Statistics Multiple R 0.834625309 R Square 0.696599406 Adjusted R Square 0.666259346 Standard Error 161.526905 Observations 2 1.2 2. 12 971 1068.5 506 1098.5 693.5 1057 1001 468.5 435.5 1.6 2 ANOVA df 1.6 1 1.4 Regression Residual Total SS M SF Significance F. 1 599040.8187 599041 22.9597 0.000732793 10 260909.4105 26090.9 11 859950.2292 Intercept Speed Coefficients Standard Error Stat P-value Lower 95% -385.924242 246.736423 -1.5641 0.14886 -935.687253 703.6157025 146.8426409 4.79163 0.00073 376.4299091 Upper 95% Lower 95.0% Upper 95.0% 163.838768 -935.687253 163.838768 1030.8015 376.4299091 1030.8015

Therefore Our Linear Equation is : Price = -385.924242 + 703.6157025*Speed

b. Estimated values calculated from regression equation :

$1587846925006_image.png$

We can say that yes 2nd laptop is underpriced and also the 4th laptop is overpriced .

C-1.  Correlation between price and speed is equal to the sqrt( R^2) in the above ouput .

therefore $\rho$ = 0.834625309

C-2. From hypothesis we ccan see that it is a right tailed test , therefore the decision will be that , we reject our null hypothesis if Test-statistic (t) > right tailed value of t at given level of significaance.

C-3.  Test-statistic (t) = $\rho$ *sqrt(n-2) / (1-$\rho$^2) ~ t(n-2)

= 8.699

C-4. Using T probability point table we get the tabulated T value as 2.2280.

Now since Our t>2.2280 , therefore We have sufficient evidence to say that our correlation coefficient Between the 2 variables is greater than 0 , at 5% level of significance.