A dean of a business school claims that the GMAT scores of applicants to the sch

Question

A dean of a business school claims that the GMAT scores of applicants to the school's M.B.A. program have increased during the past five years. Five years ago, the mean of GMAT scores of M.B.A. applicants was 560. Twenty applications for this year's program were randomly selected and the GMAT scores recorded. If we were to assume that the distribution of GMAT scores of this year's applicants is the same as that of five years ago, with the possible exception of the mean, can we conclude at the 5% significance level that the dean's claim is true?

GMAT Scores
597,606,504,584,603,607,494,560,549,606,585,587,535,539,637,540,625,500,545,577

c) Create a 95% confidence interval and perform a hypothesis test using the p-value. If outliers were found, remove them, rerun the analysis, and compare the results.

d) Compare the results from the hypothesis test and the 95% confidence interval. Discuss what information each method gives and also talk about any limitations of each method.

Explanation / Answer

HYPOTHESIS TEST
Given that,
population mean(u)=560
sample mean, x =569
standard deviation, s =41.8745
number (n)=20
null, Ho: =560
alternate, H1: !=560
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.093
since our test is two-tailed
reject Ho, if to < -2.093 OR if to > 2.093
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =569-560/(41.8745/sqrt(20))
to =0.961
| to | =0.961
critical value
the value of |t | with n-1 = 19 d.f is 2.093
we got |to| =0.961 & | t | =2.093
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 0.9612 ) = 0.3485
hence value of p0.05 < 0.3485,here we do not reject Ho
ANSWERS
---------------
null, Ho: =560
alternate, H1: !=560
test statistic: 0.961
critical value: -2.093 , 2.093
decision: do not reject Ho
p-value: 0.3485

---------------------------------------------------------------------------------------------

TRADITIONAL METHOD
given that,
sample mean, x =569
standard deviation, s =41.8745
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 41.8745/ sqrt ( 20) )
= 9.363
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
margin of error = 2.093 * 9.363
= 19.598
III.
CI = x ± margin of error
confidence interval = [ 569 ± 19.598 ]
= [ 549.402 , 588.598 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =569
standard deviation, s =41.8745
sample size, n =20
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
we use 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
confidence interval = [ 569 ± Z a/2 ( 41.8745/ Sqrt ( 20) ]
= [ 569-(2.093 * 9.363) , 569+(2.093 * 9.363) ]
= [ 549.402 , 588.598 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 549.402 , 588.598 ] contains the true population mean
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 mean

PART D.
We come at similar situation in both confidence, hypothesis. We have true sitatuion that GMAT
score values are 560

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