Please can someone answer this problem for me. I am having problems with questio
ID: 3073626 • Letter: P
Question
Please can someone answer this problem for me. I am having problems with questions 2 through 5. euestions 2-6 A career counselor wishes to estimate, based on the following sample data, the mean increase in the annual salary of people (in thousands of dollars) per additional year of education pursued. Years, r12 Salary 13 141516 17 25 31 4352 58 60 2. Which of the following is the null and the alternative hypotheses to test whether the number of years of education pursued is useful for predicting the annual salary?(1 point) c. Ho: B10 vs. Ha: Bi>0 e. Ho: B10 vs. Ha: Bi0 3. The value of the test statistic is: (1 point) a. -0.099. b. 9.064. c. -9.064 d. 10.064. e. 0.099. 4. The rejection region at the 1% significance level is: (1 point) a. [4.604, oo) b. (-oo,-3.747] U [3.747, 0o) c. (-0,-4.604] U [4.604, o) d. (-oo,-3.747] e. [3.747, 0o) 5. Which of the following is the correct decision regarding the rejection of the null hypothesis? (1 point) a. The decision is not to reject Ho, because the test statistic falls in the rejection region (-o0,-3.747] U [3.747,oo). The decision is to reject Ho, because the test statistic falls in the rejection region [4.604, 0o). The decision is not to reject Ho, because the test statistic falls in the rejection region [3.747, ). b. c. The decision is to reject Ho, because the test statistic falls in the rejection region (-oo,-3.747]. The decision is to reject Ho, because the test statistic falls in the rejection region (-o0,-4.604] U [4.604, oo). d. e.Explanation / Answer
Line of Regression Y on X i.e Y = bo + b1 X
calculation procedure for regression
mean of X = X / n = 14.5
mean of Y = Y / n = 44.8333
(Xi - Mean)^2 = 17.5
(Yi - Mean)^2 = 1042.83
(Xi-Mean)*(Yi-Mean) = 132.5003
1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2
= 132.5003 / 17.5
= 7.5714
o = Y / n - b1 * X / n
o = 44.8333 - 7.5714*14.5 = -64.9527
value of regression equation is, Y = o +1 X
Y'=-64.9527+7.5714* X
Standard Error of Y on X i.e Y = bo + b1 X
Standard error of 1 = Sqrt( ( Y -Yi )^2/ n-2 * (Xi - Mean)^2 ))
Y -Yi )^2 = 39.619 ; n-2 = 6 - 2 = 4 ; (Xi - Mean)^2 =17.5
Standard Error = Sqrt( 39.619/ 4 * 17.5)
Standard Error = Sqrt(0.566)
Standard Error = 0.7523
calculated/given are
1= 7.5714
0= -64.9527
standard error of 1= 0.7523
number (n)=6
null, Ho: 1 =0
alternate, H1: 1 !=0
level of significance, alpha = 0.01
from standard normal table, two tailed t alpha/2 =4.6041
since our test is two-tailed
reject Ho, if to < -4.6041 OR if to >
we use test statistic (t) = 1/standard error pf (1)
to =7.5714/ 0.7523
to =10.0641
| to | =10.0641
critical value
the value of |t alpha| with n-2 = 4 d.f is 4.6041
we got |to| =10.0641 & | t alpha | =4.6041
make decision
hence value of | to | > | t alpha| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 10.0641 ) = 0.0005
hence value of p0.01 > 0.0005,here we reject Ho
ANSWERS
---------------
null, Ho: 1 =0
alternate, H1: 1 !=0
test statistic: 10.0641
critical value: ( - Infinity, -4.6041 ] U [4.6041, Infinity)
decision: reject Ho
p-value: 0.0005
e. the decision is to reject Ho, because the test statistic falls in the rejection region
falls in the region ( - Infinity, -4.604 ] Union [ 4.604, Infinity)
X Y (Xi - Mean)^2 (Yi - Mean)^2 (Xi-Mean)*(Yi-Mean) 12 25 6.25 393.3598 49.5833 13 31 2.25 191.3602 20.75 14 43 0.25 3.361 0.9167 15 52 0.25 51.3616 3.5834 16 58 2.25 173.362 19.7501 17 60 6.25 230.0288 37.9168