Question

Te data shown below indicate the average tuition Cim doners) over a b year period tur private and Public colleges in a resion, use a=002 were needed Ar conduct a simple linear regression analysis of these data in which the average tuition for private colleges is predicted 44 te average public college tuition. Test using انا رانا

Answer 1

a)

Answer:

The estimated regression slope coefficient, b1 = 2.69

The estimated regression intercept, b0 = 3326

Hypothesis: C)

$H_0:\beta_1=0$

07 lg :

Rejection Region

t < -4.60 or t 4.60

Test statistic

t = 24.23

Conclusion

Since the test statistic is in the rejection region, reject the null hypothesis. Conclude that the regression model is statistically significant.

Explanation:

The simple linear regression model

The linear regression model is defined as,

Y = bo + b1X

where the dependent variable, Y = Average tuition fee for the private colleges and the independent variable, X = Average tuition fee for the public colleges

The regression analysis is done in excel by following these steps,

Step 1: Write the data values in excel. The screenshot is shown below,

A B C X 1 y 2 8752 9082 3 4 2010 2308 3169 3649 4616 5652 12277 13379 15627 18398 5 6 7 8

Step 2: DATA > Data Analysis > Regression > OK. The screenshot is shown below,

IE INSERT PAGE LAYOUT FORMULAS DATA REVIEW 000 - Data Analysis Х ? OK Cancel AZA Data Analysis Analysis Tools Fourier Analysis Histogram Moving Average Random Number Generation Rank and Percentile Regression Sampling t-Test: Paired Two Sample for Means t-Test: Two-Sample Assuming Equal Variances t-Test: Two-Sample Assuming Unequal Variances Help 36.6

Step 3: Select Input Y Range: 'Y' column, Input X Range: 'X' column then OK. The screenshot is shown below,

А B с D E F G 1 Y IX 2 3 ? X 87521 20101 Regression Input Input Y Range: 4 OK 5 SA$1:SAS7 TO 6 Cancel Input X Range: SBS1:SBS7 7 Help 8 Labels Confidence Level: Constant is Zero 95 % 9 10 11 SGS1 12 13 Output options Output Range: New Worksheet Ply: New Workbook Residuals Residuals Standardized Residuals 14 15 16 17 Residual Plots Line Fit Plots 18 Normal Probability Normal Probability Plots

The result is obtained. The screenshot is shown below,

G Н. K M 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.996612 5 R Square 0.993235 6 Adjusted R Square 0.991543 7 Standard Error 344.2407 8 Observations 6 9 10 ANOVA df F gnificance F 587.24 1.72E-05 11 12 Regression 13 Residual 14 Total SS MS 1 69588920 69588920 4 474006.7 118501.7 5 70062927 15 16 17 Intercept 18 X Coefficientsandard Err t Stat P-value Lower 95%Upper 95% 3325.62 420.0915 7.916418 0.001378 2159.259 4491.981 2.689277 0.110976 24.23303 1.72E-05 2.381159 2.997395

From the regression output summary,

The estimated regression slope coefficient, b1 = 2.69

The estimated regression intercept, b0 = 3326

Slope hypothesis test

Hypothesis

$H_0:\beta_1=0$

07 lg :

This is a two-tailed test.

Rejection Region

The t critical value is obtained from the t critical value table for the significance level = 0.01, the degree of freedom = n-2=4, and for a two-tailed distribution.

$\text{For left-tail}\Rightarrow t<-4.60$

$\text{For Right-tail}\Rightarrow t>4.60$

Test statistic

From the regression output summary,

t = 24.23

Conclusion

Since the test statistic is in the rejection region, reject the null hypothesis. Conclude that the regression model is statistically significant.

b)

Answer:

$269

Explanation:

From the regression output summary,

The regression equation is,

$\widehat{Y}=3325.62+2.6893X$

$\Delta \widehat{Y}=\Delta\left (3325.62+2.6893\times X \right )$

$\Delta \widehat{Y}=2.6893\times \Delta X$

For X = $100 increase in X variable,

$\Delta \widehat{Y}=2.6893\times 100=268.93\approx 269$

c)

Answer:

$20806

Explanation:

The regression equation is,

$\widehat{Y}=3325.62+2.6893X$

For X = $6500

$\widehat{Y}=3325.62+2.6893\times 6500=20805.92$