A lead inspector at ElectroTech, an electronics assembly shop, wants to convince
ID: 3177494 • Letter: A
Question
A lead inspector at ElectroTech, an electronics assembly shop, wants to convince management that it takes longer, on a per-component basis, to inspect large devices with many components than it does to inspect small devices because it is difficult to keep track of which components have already been inspected. To prove her point, she has collected data from the last 25 devices. The data are shown in the accompanying table Number of Components Inspection Time on Device (seconds) 31 83 13 30 17 59 15 51 11 40 24 71 98 21 42 11 18 62 25 30 79 12 49 10 32 19 63 17 53 18 60 24 71 102 44 17 58 14 44 21 68 13 46 23 69 Click here for the Excel Data File a. A scatterplot of the above data is shown below. Does the lead inspector's claim seem credible? 120 100 80Explanation / Answer
Result:
a). Yes, there is positive association, the lead inspectors claim seem credible.
B1).
Linear : Adjusted R2 =0.9247
Quadratic : Adjusted R2 = 0.9704
Cubic : Adjusted R2 =0.9795
B2).
Best: cubic model
c).
Predicted time = 90.23
Predicted values for: y
95% Confidence Interval
95% Prediction Interval
x
xx
xxx
Predicted
lower
upper
lower
upper
Leverage
38
1,444
54872.0000
90.226
86.370
94.083
82.963
97.490
0.393
Regression Analysis
r²
0.9247
n
25
r
0.962
k
1
Std. Error
5.791
Dep. Var.
y
ANOVA table
Source
SS
df
MS
F
p-value
Regression
9,469.6719
1
9,469.6719
282.39
2.07E-14
Residual
771.2881
23
33.5343
Total
10,240.9600
24
Regression output
confidence interval
variables
coefficients
std. error
t (df=23)
p-value
95% lower
95% upper
Intercept
18.8457
2.5509
7.388
1.64E-07
13.5686
24.1227
x
2.0359
0.1212
16.804
2.07E-14
1.7853
2.2866
Regression Analysis
R²
0.973
Adjusted R²
0.9704
n
25
R
0.986
k
2
Std. Error
3.554
Dep. Var.
y
ANOVA table
Source
SS
df
MS
F
p-value
Regression
9,963.1135
2
4,981.5567
394.44
5.86E-18
Residual
277.8465
22
12.6294
Total
10,240.9600
24
Regression output
confidence interval
variables
coefficients
std. error
t (df=22)
p-value
95% lower
95% upper
Intercept
-0.7059
3.4978
-0.202
.8419
-7.9599
6.5481
x
4.0253
0.3268
12.316
2.41E-11
3.3475
4.7031
xx
-0.0401
0.0064
-6.251
2.73E-06
-0.0534
-0.0268
Regression Analysis
R²
0.982
Adjusted R²
0.9795
n
25
R
0.991
k
3
Std. Error
2.960
Dep. Var.
y
ANOVA table
Source
SS
df
MS
F
p-value
Regression
10,056.9774
3
3,352.3258
382.64
1.76E-18
Residual
183.9826
21
8.7611
Total
10,240.9600
24
Regression output
confidence interval
variables
coefficients
std. error
t (df=21)
p-value
95% lower
95% upper
Intercept
-20.8727
6.8153
-3.063
.0059
-35.0458
-6.6996
x
7.3482
1.0510
6.991
6.66E-07
5.1624
9.5339
xx
-0.1956
0.0478
-4.091
.0005
-0.2950
-0.0962
xxx
0.0021
0.0006
3.273
.0036
0.0008
0.0034
Predicted values for: y
95% Confidence Interval
95% Prediction Interval
x
xx
xxx
Predicted
lower
upper
lower
upper
Leverage
38
1,444
54872.0000
90.226
86.370
94.083
82.963
97.490
0.393
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.