Question

Assignment 10 1. A random sample of 37 drivers insured with a company having similar auto insurance policies) was selected The data table lists each of the driver's driving experience in years (DU), and their monthly auto insurance pre- miums in dollars (CU). A researcher wished to determine the following: Correlation: Experience (yrs) 1. On an a priori basis, would the researcher expect a positive or negative inverse) relationship between years driving experience and monthly insurance pre- miums? 4 4 S 5 2. Calculate the sums-of-squares. 6 7 3. To determine the strength of the bivariate relation- ship, calculate the (zero-order, bivariate) correlation coefficient and coefficient-of-determination. 7 7 174 356 316 320 400 336 525 595 455 639 583 660 600 520 806 806 770 660 9 11 4. Calculate the percentage of the variation in month- ly insurance premiums (the DV) that is attributable to years driving experience (the IV). 5. At the 5% significance level, determine whether or or not p (tho) is significant (i... #0, to determine whether or not a correlation exists between years driving experience and monthly insurance premiums). 960 Premium (5) y 87 7.569 89 7,921 79 6,241 64 4,096 80 6,400 S6 3.136 75 5,625 85 7.225 65 4,225 71 5,041 53 2,809 60 3,600 50 2.500 40 1,600 62 3,844 62 3,844 SS 3,025 44 1,936 60 3,600 50 2,500 42 1,764 45 2,025 41 1,681 961 45 2,025 30 SO 2.500 39 1.521 70 4,900 35 1.225 70 4,900 40 1,600 33 1,089 79 6,241 78 6,084 81 6,561 19 361 2,115 133,075 * = 12.92 11 12 13 13 13 14 15 16 18 18 20 20 21 23 18 8 24 12 25 15 12 14 9 11 Regression: 1. Calculate the slope () for the BLR equation. 2. Calculate the y-intercept for the regression model equation. 3. Write the equation for the linear regression model. 4 16 16 25 25 36 49 49 49 81 121 121 144 169 169 169 196 225 256 324 324 400 400 441 529 324 64 576 144 625 225 144 196 81 121 196 484 7,518 900 756 900 820 651 1,035 31 900 540 4. For a person with 10 years of driving experience, what is the predicted monthly insurance premium? 400 936 840 875 1.050 480 462 5. Calculate the Standard Error Estimate. 6. Calculate the Standard Error of b, 711 7. Ata - .05, determine whether or not B -0. 8. Construct the 95% confidence interval for B. 858 1,134 418 24,247 22 478 y = 57.16 n = 37 kiv = 1

Answer 1

Correlation:

1. The researcher will expect a negative relationship between years driving experience and insurance premium

2. Sum of squares and cross products:  $\sum x_{i}^{2}=7518\: \: \: \sum y_{i}^{2}=133075\: \: \sum x_{i}*y_{i}=24247$

3. Correlation coefficient r=-0.7608. The coefficient of determination

R2 = 0.5789

4.The percentage of variation in monthly insurance premiums that is attributable to years driving experience=57.89%

5. The hypothesis test indicates that correlation exists between years driving experience and the insurance premium(t=6.9354, critical t=2.0301).

Regression:

1. Slope for BLR:

Bi SSry SS

  

B1 -3076.5135 1342.7568

  $\hat{\beta }_{1}=-2.2912$

2. y-intercept : $\hat{\beta }_{0}=\bar{y}-\hat{\beta }_{1}*\bar{x}$

$\hat{\beta }_{0}=57.16-(-2.2912)*12.92$

Bo= 57.16 +29.6022

  $\hat{\beta }_{0}=86.7619$

3. The equation for the linear model: premium=86.7619-2.2912*years driving

4. For a person with 10 years experience the premium= 86.7619-2.2912*10

=63.8499

5. The standard error estimate =12.1045

6. The standard error of b1=0.3303

7. The test statistic : $t=\frac{\hat{\beta }_{1}}{SE(\hat{\beta }_{1})}=\frac{-2.2912}{0.3303}=-6.9361$ . The critical value is 2.03010. Since the $\left | t \right |=6.9361>2.0310$ , we reject the null hypothesis and conclude that the regression slope cannot be assumed to be zero.

8. 95% confidence interval for b1,  $CI=\hat{\beta }_{1}\mp t_{35,0.95}*SE(\hat{\beta }_{1})$

  $CI=-2.2912\mp 2.0310*0.3303$

  $CI=-2.2912\mp 0.6708$

$CI=(-2.9620,-1.6204)$

----------------------------------------------------Calculations-------------------------------------------------------------------------------------------

η 1 SSxx = Σ? Ti n i=1 i=1

$SS_{XX}=7518-\frac{1}{37}\left ( 478 \right )^{2}$

$SS_{XX}=7518-\frac{228484}{37}$

$SS_{XX}=7518-6175.2432$

$SS_{XX}=1342.7568$

$SS_{YY}=\sum_{i=1}^{n}y_{i}^{2}-\frac{1}{n}\left ( \sum_{i=1}^{n}y_{i} \right )^{2}$

SSYY 133075 - (2115) 37

$SS_{YY}=133075-\frac{4473225}{37}$

$SS_{YY}=133075-120897.973$

$SS_{YY}=12177.0270$

$SS_{XY}=\sum_{i=1}^{n}x_{i}*y_{i}-\frac{1}{n}\left ( \sum_{i=1}^{n}x_{i} \right )\left ( \sum_{i=1}^{n}y_{i} \right )$

SSXY 24247 – (478) (2115) 37

$SS_{XY}=24247-\frac{1010970}{37}$

$SS_{XY}=24247-27323.5135$

$SS_{XY}=-3076.5135$

Correlation coefficient $r=\frac{SS_{xy}}{\sqrt{SS_{xx}*SS_{yy}}}$

  $r=\frac{-3076.5135}{\sqrt{1342.7568*12177.0270}}$

$r=\frac{-3076.5135}{\sqrt{16350785.81}}$

$r=\frac{-3076.5135}{4043.6105}$

$r=-0.7608$

5. Null hypothesis of correlation test:

$H_{0}:\rho =0$

$H_{1}:\rho \ne0$

Test statistic: $t=\frac{r*\sqrt{n-2}}{\sqrt{1-r^2}}$ will be distributed as a t with 37-2=35 df.

  $t=\frac{-0.7608*\sqrt{37-2}}{\sqrt{1-(-0.7608)^2}}$

  $t=\frac{-0.7608*\sqrt{35}}{\sqrt{0.4211}}$

$t=\frac{-0.7608*5.9161}{0.6489}$

$t=\frac{-4.5010}{0.6489}$

$t=-6.9354$

The critical value of t at 35 df at 5% level is 2.0301. Since, t calculated value >the critical value , we reject the null hypotesis and conclude that the correlation is significant and cannot be assumed to be 0.

5. The regression sum of squares: $SSR=\hat{\beta }_{1}*SS_{XY}$

  $SSR=-2.2912*-3076.5135=7048.9077$

Total sum of squares : $SST=SS_{YY}=12177.0270$

Error SS ESS=SST-SSR=5128.1439

Mean square error=5128.1439/35=146.5184

Therefore the standard error =$SE=\sqrt{MSe}=\sqrt{146.5184}=12.1045$

6. Standard error of b1: $\frac{SE}{\sqrt{SS_{XX}}}$

$=\frac{12.1045}{\sqrt{1342.7568}}=\frac{12.1045}{36.6436}=0.3303$

7. Test statistic: $t=\frac{\hat{\beta }_{1}}{SE(\hat{\beta }_{1})}=\frac{-2.2912}{0.3303}=-6.9361$