Question

The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures the motivation, attitudes, and study habits of college students. Scores range from 0 to 200 and follow, approximately, a Normal distribution, with mean 115 and standard deviation ?=25σ=25 . You suspect that incoming freshman have a mean ?μ that is different from 115, since they are often excited yet anxious about entering college. To verify your suspicion, you test the hypotheses ?0:?=115H0:μ=115 , ??:?≠115Hα:μ≠115 . You give the SSHA to 25 incoming freshman and find their mean score. Based on this, you reject ?0H0 at significance level ?=0.01α=0.01 .

Which of these choices would be most helpful in assessing the practical significance of your results?

construct a 99% confidence interval for ?μ to see the magnitude of the difference between 115 and your sample results

report the ?P‑value of your test

take another sample and retest, just to make sure the results are not due to chance

test the hypotheses again, this time using significance level ?=0.001

Answer 1

a.
Given that,
population mean(u)=115
standard deviation, σ =25
Assume, sample mean, x =110 because not given in the data
number (n)=25
null, Ho: μ=115
alternate, H1: μ!=115
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 110-115/(25/sqrt(25)
zo = -1
| zo | = 1
critical value
the value of |z α| at los 1% is 2.576
we got |zo| =1 & | z α | = 2.576
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1 ) = 0.317
hence value of p0.01 < 0.317, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=115
alternate, H1: μ!=115
test statistic: -1
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 0.317
we donot have enough evidence to support the claim that incoming freshman have a mean μ that is different from 115.
b.
Given that,
population mean(u)=115
standard deviation, σ =25
sample mean, x =110
number (n)=25
null, Ho: μ=115
alternate, H1: μ!=115
level of significance, α = 0.001
from standard normal table, two tailed z α/2 =3.291
since our test is two-tailed
reject Ho, if zo < -3.291 OR if zo > 3.291
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 110-115/(25/sqrt(25)
zo = -1
| zo | = 1
critical value
the value of |z α| at los 0.1% is 3.291
we got |zo| =1 & | z α | = 3.291
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1 ) = 0.317
hence value of p0.001 < 0.317, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=115
alternate, H1: μ!=115
test statistic: -1
critical value: -3.291 , 3.291
decision: do not reject Ho
p-value: 0.317
we donot have enough evidence to support the claim that incoming freshman have a mean μ that is different from 115.
c.
TRADITIONAL METHOD
given that,
standard deviation, σ =25
sample mean, x =110
population size (n)=25
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 25/ sqrt ( 25) )
= 5
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 5
= 12.88
III.
CI = x ± margin of error
confidence interval = [ 110 ± 12.88 ]
= [ 97.12,122.88 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =25
sample mean, x =110
population size (n)=25
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 110 ± Z a/2 ( 25/ Sqrt ( 25) ) ]
= [ 110 - 2.576 * (5) , 110 + 2.576 * (5) ]
= [ 97.12,122.88 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [97.12 , 122.88 ] contains the true population mean
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 mean