For chart A2 recalculate the correlation, confidence interval, and P-value (95%
ID: 3183460 • Letter: F
Question
For chart A2 recalculate the correlation, confidence interval, and P-value (95% two-sided). Discuss how this differs from A1 analysis with all 14 cases.
Dataset A1
county
dist
phome
tran_dist
Zphome
Ztran_dist
Zx*Zy
BERKSHIRE
97
77
0.9897
FRANKLIN
62
81
0.9839
HAMPSHIRE
54
75
0.9815
HAMPDEN
52
69
0.9808
WORCESTER
20
64
0.9500
MIDDLESEX
14
47
0.9286
ESSEX
10
47
0.9000
SUFFOLK
4
6
0.7500
NORFOLK
14
49
0.9286
BRISTOL
14
60
0.9286
PLYMOUTH
16
68
0.9375
BARNSTABLE
44
76
0.9773
NANTUCKET
77
25
0.9870
DUKES
52
79
0.9808
N = 14
58.8(22.04)
0.9432(0.0627)
Dataset A2
county
dist
phome
tran_dist
Zphome
Ztran_dist
Zx*Zy
BERKSHIRE
97
77
0.9897
FRANKLIN
62
81
0.9839
HAMPSHIRE
54
75
0.9815
HAMPDEN
52
69
0.9808
WORCESTER
20
64
0.9500
MIDDLESEX
14
47
0.9286
ESSEX
10
47
0.9000
NORFOLK
14
49
0.9286
BRISTOL
14
60
0.9286
PLYMOUTH
16
68
0.9375
BARNSTABLE
44
76
0.9773
DUKES
52
79
0.9808
N = 12
66.0(12.65)
0.9556(0.0302)
For chart A2 recalculate the correlation, confidence interval, and P-value (95% two-sided). Discuss how this differs from A1 analysis with all 14 cases.
Dataset A1
county
dist
phome
tran_dist
Zphome
Ztran_dist
Zx*Zy
BERKSHIRE
97
77
0.9897
FRANKLIN
62
81
0.9839
HAMPSHIRE
54
75
0.9815
HAMPDEN
52
69
0.9808
WORCESTER
20
64
0.9500
MIDDLESEX
14
47
0.9286
ESSEX
10
47
0.9000
SUFFOLK
4
6
0.7500
NORFOLK
14
49
0.9286
BRISTOL
14
60
0.9286
PLYMOUTH
16
68
0.9375
BARNSTABLE
44
76
0.9773
NANTUCKET
77
25
0.9870
DUKES
52
79
0.9808
N = 14
58.8(22.04)
0.9432(0.0627)
Dataset A2
county
dist
phome
tran_dist
Zphome
Ztran_dist
Zx*Zy
BERKSHIRE
97
77
0.9897
FRANKLIN
62
81
0.9839
HAMPSHIRE
54
75
0.9815
HAMPDEN
52
69
0.9808
WORCESTER
20
64
0.9500
MIDDLESEX
14
47
0.9286
ESSEX
10
47
0.9000
NORFOLK
14
49
0.9286
BRISTOL
14
60
0.9286
PLYMOUTH
16
68
0.9375
BARNSTABLE
44
76
0.9773
DUKES
52
79
0.9808
N = 12
66.0(12.65)
0.9556(0.0302)
Explanation / Answer
There are different data points in the datasets A1 and A2. For example, the cities SUFFOLK and NANTUCKET of A1 are not available in A2. We first import the data to R and then calculate the correlation, confidence inteval, and p-value for the two datasets.
> tt <- read.csv("clipboard",sep=" ")
> head(tt)
county dist phome tran_dist
1 BERKSHIRE 97 77 0.9897
2 FRANKLIN 62 81 0.9839
3 HAMPSHIRE 54 75 0.9815
4 HAMPDEN 52 69 0.9808
5 WORCESTER 20 64 0.9500
6 MIDDLESEX 14 47 0.9286
> cor.test(tt$dist,tt$phome)
Pearson's product-moment correlation
data: tt$dist and tt$phome
t = 1.5683, df = 12, p-value = 0.1428
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
-0.1512340 0.7737076
sample estimates:
cor
0.4124404
> ttB <- read.csv("clipboard",sep=" ")
> head(ttB)
county dist phome tran_dist
1 BERKSHIRE 97 77 0.9897
2 FRANKLIN 62 81 0.9839
3 HAMPSHIRE 54 75 0.9815
4 HAMPDEN 52 69 0.9808
5 WORCESTER 20 64 0.9500
6 MIDDLESEX 14 47 0.9286
> cor.test(ttB$dist,ttB$phome)
Pearson's product-moment correlation
data: ttB$dist and ttB$phome
t = 4.0534, df = 10, p-value = 0.002311
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
0.3918448 0.9379377
sample estimates:
cor
0.7884414
Removing the two datapoints from the first set A makes the correlation significant.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.