++++++++PLEASE SHOW ALL WORK and STEPS++++++ For the data given in DATA FILE , a
ID: 3157314 • Letter: #
Question
++++++++PLEASE SHOW ALL WORK and STEPS++++++
For the data given in DATA FILE, assume a model of the form:
y = 0 + 1x1 + 2x2
a) Find the least-squares prediction equation for y.
b) Conduct a test to determine if x1, is a useful linear predictor of y. Use = .05.
c) Form a 95% confidence interval for 2. Interpret the result.
d) Find and interpret R2 and Ra2.
e) Is the overall model statistically useful for predicting weight change? Test using = .05.
++++++DATA FILE BELOW++++++
Trial
Y
X1
X2
1
-6
0
28.5
2
-5
2.5
27.5
3
-4.5
5
27.5
4
0
0
32.5
5
2
0
32
6
3.5
1
30
7
-2
2.5
34
8
-2.5
10
36.5
9
-3.5
20
28.5
10
-2.5
12.5
29
11
-3
28
28
12
-8.5
30
28
13
-3.5
18
30
14
-3
15
31
15
-2.5
17.5
30
16
-0.5
18
22
17
0
23
22.5
18
1
20
24
19
2
15
23
20
6
31
21
21
2
15
24
22
2
21
23
23
2.5
30
22.5
24
2.5
33
23
25
0
27.5
30.5
26
0.5
29
31
27
-1
32.5
30
28
-3
42
24
29
-2.5
39
25
30
-2
35.5
25
31
0.5
39
20
32
5.5
39
18.5
33
7.5
50
15
34
0
62.5
8
35
0
63
8
36
2
69
7
37
8
42.5
7.5
38
9
59
8.5
39
12
52.5
8
40
8.5
75
6
41
10.5
72.5
6.5
42
14
69
7
Trial
Y
X1
X2
1
-6
0
28.5
2
-5
2.5
27.5
3
-4.5
5
27.5
4
0
0
32.5
5
2
0
32
6
3.5
1
30
7
-2
2.5
34
8
-2.5
10
36.5
9
-3.5
20
28.5
10
-2.5
12.5
29
11
-3
28
28
12
-8.5
30
28
13
-3.5
18
30
14
-3
15
31
15
-2.5
17.5
30
16
-0.5
18
22
17
0
23
22.5
18
1
20
24
19
2
15
23
20
6
31
21
21
2
15
24
22
2
21
23
23
2.5
30
22.5
24
2.5
33
23
25
0
27.5
30.5
26
0.5
29
31
27
-1
32.5
30
28
-3
42
24
29
-2.5
39
25
30
-2
35.5
25
31
0.5
39
20
32
5.5
39
18.5
33
7.5
50
15
34
0
62.5
8
35
0
63
8
36
2
69
7
37
8
42.5
7.5
38
9
59
8.5
39
12
52.5
8
40
8.5
75
6
41
10.5
72.5
6.5
42
14
69
7
Explanation / Answer
For the data given in DATA FILE, assume a model of the form:
y = 0 + 1x1 + 2x2
All the questions we can done using MINITAB.
Steps :
Enter all the data in MINITAB sheet --> Stat --> Regression -->Regression --> Response : Y --> Predictors : X1 and X2 --> Results : select second option --> ok --> ok
a) Find the least-squares prediction equation for y.
The regression equation is
Y = 12.2 - 0.0477 X1 - 0.442 X2
b) Conduct a test to determine if x1, is a useful linear predictor of y. Use = .05.
Predictor Coef SE Coef T P
Constant 12.154 4.173 2.91 0.006
X1 -0.04768 0.05153 -0.93 0.361
X2 -0.4420 0.1217 -3.63 0.001
S = 3.33537 R-Sq = 45.7% R-Sq(adj) = 42.7%
Here we have to test the hypothesis that,
H0 : B = 0 Vs B not= 0
where B is population slope for X1.
The test statistic t =-0.93
P-value = 0.361
P-value > alpha (0.05)
Accept H0 at 5% level of significance.
Conclusion : The population slope for X1 is 0.
c) Form a 95% confidence interval for 2. Interpret the result.
95% confidence interval for B2 is,
b - E < B < b + E
where b is sample slope for X2 = -0.442
E = tc*SEb
where tc is critical value for t-distribution.
tc we can find by using EXCEL.
syntax is,
=TINV(probability, deg_freedom)
probability = 1 - c
c is confidence level = 95% = 0.95
deg_freedom = n-2 = 42-2= 40
tc = 2.021
SEb = 0.1217
E = 2.021*0.1217 = 0.246
lower limit = b - E = -0.442 - 0.246 = -0.688
upper limit = b + E = -0.442 + 0.246 = -0.196
95% confidence interval for population slope is (-0.688, -0.196).
d) Find and interpret R2 and Ra2.
R2 = 45.7%
It expresses the proportion of the variation in y which is explained by variation in x2.
e) Is the overall model statistically useful for predicting weight change? Test using = .05.
Here we have to test the hypothesis that,
H0 : B1 = B2 = 0
H1 : Atleast one of the slope is differ than 0.
Analysis of Variance
Source DF SS MS F P
Regression 2 346.08 173.04 15.55 0.000
Residual Error 37 411.61 11.12
Total 39 757.69
Test statistic F = 15.55
P-value = 0.000
P-value < alpha
Reject H0 at 5% level of significance.
COnclusion : Atleast one of the slope is differ than 0.
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