Answer Using MINITAB with fromulas please Applying the Concepts Companies often

Question

Answer Using MINITAB with fromulas please

Applying the Concepts

Companies often develop and test hypotheses about their products. Regulatory and consumer protection groups must also test claims resulting from hypotheses used to promote products.

Submit by Day 7 a 1- to 2-page paper that addresses the following:

Assume you are working at the Consumer Protection Agency. Recently, you have been getting complaints about the highway gas mileage of a new minivan. The car company agrees to allow you to select randomly 41 of its new minivans to test their highway mileage. The company claims that its minivans get 28 miles per gallon on the highway. Your test results show a sample mean of 26.7 and a sample standard deviation of 4.2.

Part 1 (Confidence Interval):

Calculate a 95% confidence interval around your sample mean.

Is the claimed mean inside your confidence interval?

What does your result mean, in terms of the company's claim?

Part 2 (Two-tail test):

List the null and alternative hypotheses for the appropriate test.

Use alpha = 0.05. Find the critical value(s) and calculate the observed value of the test statistic.

Is the observed test statistic in the critical (rejection) region?

Will the p-value be higher or lower than your alpha? What does this result mean, in terms of the company's claim?

Part 3 (One-tail test):

List the null and alternative hypotheses for the test.

Use alpha = 0.05.

Find the critical value and calculate the observed value of the test statistic.

Is the observed test statistic in the critical region?

Will the p-value be higher or lower than your alpha?

What does this result mean, in terms of the company's claim?

Part 4 (Conclusion):

What conclusions did you reach?

What did you learn from each method of checking the claim for means?

Were there important differences between methods? Which method would you prefer?

Which carries a higher risk of a type I error?

Based on this experience, why do you think it’s important to decide on the method before conducting the test?

Based on your results, do you support the company's claim?

What action, if any, should the company take?

Note: The 1- to 2-page requirement does not include the graphs you prepare for this Assignment. Embed the graphs you create in the Word document with your written responses to the questions. Drawings may be hand-drawn and either scanned or photographed, or they may be drawn using Minitab or other statistical drawing packages.

Explanation / Answer

a.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=26.7
Standard deviation( sd )=4.2
Sample Size(n)=41
Confidence Interval = [ 26.7 ± t a/2 ( 4.2/ Sqrt ( 41) ) ]
= [ 26.7 - 2.021 * (0.656) , 26.7 + 2.021 * (0.656) ]
= [ 25.374,28.026 ]

b.
Given that,
population mean(u)=28
sample mean, x =26.7
standard deviation, s =4.2
number (n)=41
null, Ho: =28
alternate, H1: !=28
level of significance, = 0.025
from standard normal table, two tailed t /2 =2.329
since our test is two-tailed
reject Ho, if to < -2.329 OR if to > 2.329
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =26.7-28/(4.2/sqrt(41))
to =-1.982
| to | =1.982
critical value
the value of |t | with n-1 = 40 d.f is 2.329
we got |to| =1.982 & | t | =2.329
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.9819 ) = 0.0544
hence value of p0.025 < 0.0544,here we do not reject Ho
ANSWERS
---------------
null, Ho: =28
alternate, H1: !=28
test statistic: -1.982
critical value: -2.329 , 2.329
decision: do not reject Ho
p-value: 0.0544

c.
Given that,
population mean(u)=28
sample mean, x =26.7
standard deviation, s =4.2
number (n)=41
null, Ho: >28
alternate, H1: <28
level of significance, = 0.025
from standard normal table,left tailed t /2 =2.021
since our test is left-tailed
reject Ho, if to < -2.021
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =26.7-28/(4.2/sqrt(41))
to =-1.982
| to | =1.982
critical value
the value of |t | with n-1 = 40 d.f is 2.021
we got |to| =1.982 & | t | =2.021
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -1.9819 ) = 0.02719
hence value of p0.025 < 0.02719,here we do not reject Ho
ANSWERS
---------------
null, Ho: >28
alternate, H1: <28
test statistic: -1.982
critical value: -2.021
decision: do not reject Ho
p-value: 0.02719

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