Can you please use Excel. I don\'t have the software used to solve this problem.

Question

Can you please use Excel. I don't have the software used to solve this problem. Can also solve by hand. This is a resubmission I did not provide the right data previously.

A traffic engineer wants to determine if there is any difference in the volume of traffic at two different intersections, and assigns an employee to make the traffic count. Each day, the engineer tosses a coin to determine which intersection the employee will be sent to. After 20 days, the traffic has been counted at location 1 on nine days, and at location 2 on 11 days.
a. Create stem-and-leaf plots of the data from each of the two locations, side-by-side box-and-whisker plots, and stacked histograms.
b. Determine if the variances of the traffic counts from the two locations are equal. Use =.01.
c. Assess the normality of data for each location.
d. Based on your results so far, test whether the average counts of the two locations are equal. Justify your selection of test method.
e. If you decide that the average counts of the two locations are not equal, and you have done the parametric test in part (d.), estimate this difference with a 95% confidence interval.

This is a test question can you please show as much work as possible? The data is below:

Traffic Traffic
Count Count
at at
Location 1

Location 2

Explanation / Answer

Given that,
mean(x)=656.1
standard deviation , s.d1=93.2588
number(n1)=10
y(mean)=689.6
standard deviation, s.d2 =96.8426
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.878
since our test is two-tailed
reject Ho, if to < -2.878 OR if to > 2.878
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (9*8697.203777 + 9*9378.489175) / (20- 2 )
s^2 = 9037.846476
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=656.1-689.6/sqrt((9037.846476( 1 /10+ 1/10 ))
to=-33.5/42.515518
to=-0.787948
| to | =0.787948
critical value
the value of |t | with (n1+n2-2) i.e 18 d.f is 2.878
we got |to| = 0.787948 & | t | = 2.878
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -0.7879 ) = 0.4404
hence value of p0.01 < 0.4404,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.787948
critical value: -2.878 , 2.878
decision: do not reject Ho
p-value: 0.4404

TRADITIONAL METHOD
given that,
mean(x)=656.1
standard deviation , s.d1=93.2588
number(n1)=10
y(mean)=689.6
standard deviation, s.d2 =96.8426
number(n2)=10
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (9*8697.204 + 9*9378.489) / (20- 2 )
s^2 = 9037.846
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 9037.846 * (1/10+1/10) )
=42.516
III.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.01
from standard normal table, two tailed and value of |t | with (n1+n2-2) i.e 18 d.f is 2.878
margin of error = 2.878 * 42.516
= 122.36
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (656.1-689.6) ± 122.36 ]
= [-155.86 , 88.86]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=656.1
standard deviation , s.d1=93.2588
sample size, n1=10
y(mean)=689.6
standard deviation, s.d2 =96.8426
sample size,n2 =10
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 656.1-689.6) ± t a/2 * sqrt( 9037.846 * (1/10+1/10) ]
= [ (-33.5) ± 122.36 ]
= [-155.86 , 88.86]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-155.86 , 88.86]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion

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