Can you please answer this using excel. I submitted this question before but the
ID: 2930165 • Letter: C
Question
Can you please answer this using excel. I submitted this question before but the answer was provided in software I do not have.
A shoe manufacturer is field testing the durability of a leather sole made by a new process, and is comparing it with the type of leather sole currently in use. Fifteen pairs of shoes are constructed, with one shoe in each pair having the new type of leather sole. After six months of daily use by 15 randomly selected supermarket employees, the shoes are examined and measured for the percentage of wear remaining in each sole.
a. Compute the difference scores on the wear between the two types of soles, and create a stem-and-leaf plot for the difference scores.
b. Assess the normality of these difference scores.
c. Based upon these results, use either the Wilcoxon Signed Ranks test or the t-test to determine whether there is ant difference in the wear between the two types of soles. Justify your selection of test method.
d. If you decide that the wear of the two types of soles are indeed different, and you have done the parametric test in part (c.), 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:
New Old
73 64
43 41
47 43
53 41
58 47
47 32
52 24
38 43
61 53
56 52
56 57
34 44
55 57
65 40
75 68
Explanation / Answer
Given that,
mean(x)=54.2
standard deviation , s.d1=11.5771
number(n1)=15
y(mean)=47.0667
standard deviation, s.d2 =11.6709
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.145
since our test is two-tailed
reject Ho, if to < -2.145 OR if to > 2.145
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =54.2-47.0667/sqrt((134.02924/15)+(136.20991/15))
to =1.681
| to | =1.681
critical value
the value of |t | with min (n1-1, n2-1) i.e 14 d.f is 2.145
we got |to| = 1.68059 & | t | = 2.145
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.6806 ) = 0.115
hence value of p0.05 < 0.115,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.681
critical value: -2.145 , 2.145
decision: do not reject Ho
p-value: 0.115
TRADITIONAL METHOD
given that,
mean(x)=54.2
standard deviation , s.d1=11.5771
number(n1)=15
y(mean)=47.0667
standard deviation, s.d2 =11.6709
number(n2)=15
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((134.029/15)+(136.21/15))
= 4.245
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 14 d.f is 2.145
margin of error = 2.145 * 4.245
= 9.104
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (54.2-47.0667) ± 9.104 ]
= [-1.971 , 16.238]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=54.2
standard deviation , s.d1=11.5771
sample size, n1=15
y(mean)=47.0667
standard deviation, s.d2 =11.6709
sample size,n2 =15
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 = [( 54.2-47.0667) ± t a/2 * sqrt((134.029/15)+(136.21/15)]
= [ (7.133) ± t a/2 * 4.245]
= [-1.971 , 16.238]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-1.971 , 16.238] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion