As one part of a study of commercial bank branches, data are obtained on the num
ID: 3314099 • Letter: A
Question
As one part of a study of commercial bank branches, data are obtained on the number of independent business (r) located in sample ZIP code areas and the number of bank branches (v) located in these areas. x:92116124210 216 267306 378 415 502 615 703 y- 4 a) b) c) d) e) Check whether a simple linear regression model is reasonable for this study Fit a regression line and state the estimates of intercept and slope, Assuming a linear regression model is constructed, what is the interpretation of H, :B-O? What is the natural research hypothesis H, for the problem in part (c)? Calculate a 90% confidence interval for B. By using this confidence interval test the hypothesis in part (c).Explanation / Answer
a.
calculation procedure for regression
mean of X = X / n = 328.6667
mean of Y = Y / n = 5.4167
(Xi - Mean)^2 = 436262.666
(Yi - Mean)^2 = 56.92
(Xi-Mean)*(Yi-Mean) = 4844.667
b1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2
= 4844.667 / 436262.666
= 0.011
bo = Y / n - b1 * X / n
bo = 5.4167 - 0.011*328.6667 = 1.767
value of regression equation is, Y = bo + b1 X
Y'=1.767+0.011* X
b.
Y'=1.767+0.011* X
intercept Y = 1.767
slope m = 0.011
c.
linear regression line Y'=1.767+0.011* X and hypothesis test for given data
d.
Given that,
mean(x)=328.6667
standard deviation , s.d1=199.1488
number(n1)=12
y(mean)=5.4167
standard deviation, s.d2 =2.2747
number(n2)=12
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.1
from standard normal table, two tailed t /2 =1.8
since our test is two-tailed
reject Ho, if to < -1.8 OR if to > 1.8
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =328.6667-5.4167/sqrt((39660.24454/12)+(5.17426/12))
to =5.622
| to | =5.622
critical value
the value of |t | with min (n1-1, n2-1) i.e 11 d.f is 1.8
we got |to| = 5.62242 & | t | = 1.8
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 5.6224 ) = 0
hence value of p0.1 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 5.622
critical value: -1.8 , 1.8
decision: reject Ho
p-value: 0
e.
TRADITIONAL METHOD
given that,
mean(x)=328.6667
standard deviation , s.d1=199.1488
number(n1)=12
y(mean)=5.4167
standard deviation, s.d2 =2.2747
number(n2)=12
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((39660.245/12)+(5.174/12))
= 57.493
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 11 d.f is 1.796
margin of error = 1.796 * 57.493
= 103.258
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (328.6667-5.4167) ± 103.258 ]
= [219.992 , 426.508]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=328.6667
standard deviation , s.d1=199.1488
sample size, n1=12
y(mean)=5.4167
standard deviation, s.d2 =2.2747
sample size,n2 =12
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 328.6667-5.4167) ± t a/2 * sqrt((39660.245/12)+(5.174/12)]
= [ (323.25) ± t a/2 * 57.493]
= [219.992 , 426.508]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [219.992 , 426.508] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
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