Automatic Zoom : POM 212 Business Statistics Homework Assignment-4 Note: The hom

Question

Automatic Zoom : POM 212 Business Statistics Homework Assignment-4 Note: The homework is due on the day of Test-4 Chapter-10 Problems: Problem-1 In order to estimate the difference between the average Miles per Gallon of two different models of automobiles, samples are taken and the following information is collected. Model A Model B Sample Size 50 32 9 35 10 Sample Mean Sample Variance At 95% confidence develop an interval estimate for the difference between the average Miles per Gallon for the two models. b. Is there conclusive evidence to indicate that one model gets a higher MPG than the other? Why or why not? Explain. Problem-2 8

Explanation / Answer

1.

a.
TRADITIONAL METHOD
given that,
mean(x)=32
standard deviation , s.d1=3
number(n1)=50
y(mean)=35
standard deviation, s.d2 =3.1622
number(n2)=55
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((9/50)+(10/55))
= 0.602
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 49 d.f is 2.01
margin of error = 2.01 * 0.602
= 1.209
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (32-35) ± 1.209 ]
= [-4.209 , -1.791]
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DIRECT METHOD
given that,
mean(x)=32
standard deviation , s.d1=3
sample size, n1=50
y(mean)=35
standard deviation, s.d2 =3.1622
sample size,n2 =55
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 = [( 32-35) ± t a/2 * sqrt((9/50)+(10/55)]
= [ (-3) ± t a/2 * 0.602]
= [-4.209 , -1.791]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-4.209 , -1.791] 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

b.
Given that,
mean(x)=32
standard deviation , s.d1=3
number(n1)=50
y(mean)=35
standard deviation, s.d2 =3.1622
number(n2)=55
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.677
since our test is left-tailed
reject Ho, if to < -1.677
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =32-35/sqrt((9/50)+(9.99951/55))
to =-4.987
| to | =4.987
critical value
the value of |t | with min (n1-1, n2-1) i.e 49 d.f is 1.677
we got |to| = 4.98748 & | t | = 1.677
make decision
hence value of | to | > | t | and here we reject Ho
p-value:left tail - Ha : ( p < -4.9875 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -4.987
critical value: -1.677
decision: reject Ho
p-value: 0

there is evidence to support the claim that one model is not greater than other model because its reject the null hypothesis at one model is lower than the other

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