SOLUTION i: Let
be the coefficient of YEARS OF EXPERIENCE
NULL HYPOTHESIS Ha:
ALTERNATIVE HYPOTHESIS Ha:
ii) Under null hypothesis test statistic is
Test statistic t= coefficient of b1/s.e(b1)
Since standard error of b1 is not given we will make use of TABLE 2
From table 2 we have
F=622.51
As we know that
F= t^2
622.51= t^2
t= sqrt(622.51)
t= 24.95
Degrees of freedom= 30-2=28
P value= 0.0000 (By using Excel's function TDIST)
iii) Since P value is SMALLER than the level of significance therefore SIGNIFICANT.
Decision: REJECT H0.
Conclusion: We have sufficient evidence to show that more years of working experience lead to higher salaries at 0.05 level of significance.
A company manager is interested in analyzing the relationship between years of working experience and the...
Analysis of Variance Table Response: Price Df Sum Sq Mean Sq F value Pr(>F) Living.Area 1 1.3501e+12 1.3501e+12 362.0394 < 2e-16 *** Bedrooms 1 2.3642e+10 2.3642e+10 6.3394 0.01241 * Fireplaces 1 7.6232e+07 7.6232e+07 0.0204 0.88642 Residuals 259 9.6588e+11 3.7293e+09 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 > Using= 0.05, perform an F test of overall linear relationship. State the hypotheses, the value of F-test statistic, p-value, and your conclusion.
Using R output provided
1). Perform hypothesis testing for B(beta)1=2 using
A(alpha)=0.05
> summary(ls) Call: Residuals: Min 1Q Median 3Q Max 0.20283 -0.14691 -0.02255 0.06655 0.44541 Coefficients: (Intercept) 0.365100.099043.686 0.003586 ** Signif. codes: 0 '***' 0.001 '0.01 '*'0.05 '.' 0.1''1 Estimate Std. Error t value Pr>Itl) 0.96683 0.18292 5.286 0.000258** Residual standard error: 0.1932 on 11 degrees of freedom Multiple R-squared 0.7175, Adjusted R-squared: 0.6918 F-statistic: 27.94 on 1 and 11 DF, p-value: 0.0002581 anovaCLs) Analysis of Variance Table Response:...
> summaryCls) Call: Lm(formula y X) Residuals: -0.20283 -0.146910.02255 0.06655 0.44541 Coefficients: (Intercept) 0.36510 0.09904 3.686 0.003586 ** Min 1Q Median 3Q Max Estimate Std. Error t value Pr(>ltl) 0.96683 0.18292 5.286 0.000258*** Signif. codes: 00.001*0.010.050.11 Residual standard error: 0.1932 on 11 degrees of freedom Multiple R-squared 0.7175, Adjusted R-squared: 0.6918 F-statistic: 27.94 on 1 and 11 DF, p-value: 0.0002581 > anovaCls) Analysis of Variance Table Response : y Df Sum Sq Mean Sq F value PrOF) 1 1.04275 1.04275...
(a) Using the above t-test data to determine whether or not there
is a linear relationship between the two variables.
(b) Using the above ANOVA F-test data to determine whether or not
there is a linear relationship between the two variables.
(c) How do the results in (a) compare to those in (b)?
We were unable to transcribe this imageAnalysis of Variance Table Response: DatSGPA Dat $ACT 1 3. 588 3. 5878 9. 2402 0.002917 Df Sum Sq Mean sq...
What are the implications of predictability results in Part 2 and 3 for investment decisions? Part 2 use log dividend-price ratio to predict the 5-year stock market excess log returns: lm(formula = lnexret[2:t] ~ dp[1:t - 1]) Residuals: Min 1Q Median 3Q Max -0.54389 -0.07305 0.01977 0.10712 0.34107 Coefficients: Estimate Std. Error t value (Intercept) 0.58469 0.12768 4.579 dp[1:t - 1] 0.13510 0.03771 3.582 Pr(>|t|) (Intercept) 1.58e-05 *** dp[1:t - 1] 0.000567 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’...
How do I interpret the p-values in terms of rejecting or failing
to reject H0 at a 95% confidence level? What does the intercept
column mean in terms of p-value? How does the p-value of the F test
compare and what does it mean? In the simple linear regression I'd
conclude age isn't related to pulmonary disease (what does
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2. 2. After we fit the model, the R commander output is provided below. Coefficients: (Intercept) -5.128e+03 1.103e+02 46.49 2e-16** Estimate std. Brror t value Pr(lt|) TEMP PERT TEM: FERT 1.45se-01 9.692e-03 -15.01 1.06e-12 3.110e+01 1.344e+00 23.13 2e-16* 1.397e+02 3.140e+00 44.51 < 2e-16** TEMPSQ FERTSO -1.334e-01 6.853e-03 19.46 6.46e-15 -1.144e+00 2.741e-02 41.74 <2e-16 signif. codes: 00.001 0.01 0.05 011 Residual standard error: 1.679 on 21 degrees of freedom Multiple R-squared: 0.993, F-statistic: 596.3 on 5 and 21 DF, p-value: 2.2e-16...
Question 1 1 pts Consider the following model for estimating the salary of employees at a company (in $) by the number of years employed at the company. technitron.1m-(salary- yrs.empl, data technitron.af) sumary(technitron.1n ss Call: # In(formula . salary ~ yrs.enp1. dat. . technitron.df) # Residuals: Medianax Min -12854.7-4188.9 281.5 3254.4 16493. a# as Coefficients Estimate Std. Error t value Pr( ltl) < 2e-16 5.93e-10 (Intercept) 28394.2 1107 .2 1794.อ 140.4 15.828 7.884 a: yrs . enp1 ㆅ signif. codes...
The R code will help to answer
the question.
8. DeGroot&Shervish (2002) consider an experiment to study the combined effects of taking a stimulant and a tranquilizer. In this experiment three types of stimulant and four types of tranquilizer are administered to a group of rabbits. Each rabbit received one of the stimulants, then 20 minutes later, one of the tranquilizers. One hour later their response time (in microseconds) to a stimulus was measured. The results were: Tranquilizer Stimulant 1...
write answer step by step on the paper
What would you expect to see if you were to run the following code? Please describe th analysis and the result briefly. (20 pts) 5- > myANOVA <- aov(Learning" Group Condition) >summary(myANOVA) Group Condition Group:Condition Residuals Df Sum Sq 1.8454 1 0.1591 0.3164 59 1.3325 Mean Sq 1.84537 0.15910 0.31640 0.02258 F value 81.7106 7.0448 14.0100 Pr(>F) 9.822e-13** 0.0102017 0.000414*** Signif. codes:0.001*0.01' 0.05'0.1'"'1 > boxplot(Learning"Group"condition,col:c("#ffdddd","#ddddff"))