5.Bardwell et al. (2005) assessed the moods of 60 male U.S. Marines following a

Question

5.Bardwell et al. (2005) assessed the moods of 60 male U.S. Marines following a month-long training exercise conducted in cold temperatures and at high altitudes. Negative moods, including fatigue and anger, increased substantially during the training and lasted up to 3 months after the training ended. Mean mood scores were compared to population norms for 3 groups: college men, adult men, and male psychiatric outpatients. Let’s examine anger scores for 6 Marines at the end of training; these scores are fictional, but their mean and standard deviation are very close to the actual descriptive statistics for the sample: 14, 12, 13, 12, 14, 15. (show your work)

A)The population mean anger score for college men is 8.90. Conduct all 6 steps of a single-sample t test.

B)Now calculate the test statistic to compare this sample mean to the population mean anger score for adult men (M=9.20). You don’t need to repeat all steps from part (a), but conduct step 6 of hypothesis testing and report the statistics academically (i.e., explain what it means for people who do not understand stats).

C)Now calculate the test statistic to compare this sample mean to the population mean anger score for male psychiatric outpatients (M = 13.5). Again, only conduct step 6 of hypothesis testing and report the stats academically.

D)What can we conclude overall about Marines’ moods following high-altitude, cold-weather training?

Explanation / Answer

a.
When population mean anger score for college men (M=8.90)
population mean(u)=8.9
sample mean, x =13.3333
standard deviation, s =1.2111
number (n)=6
null, Ho: =8.9
alternate, H1: !=8.9
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.571
since our test is two-tailed
reject Ho, if to < -2.571 OR if to > 2.571
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =13.3333-8.9/(1.2111/sqrt(6))
to =8.9665
| to | =8.9665
critical value
the value of |t | with n-1 = 5 d.f is 2.571
we got |to| =8.9665 & | t | =2.571
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 8.9665 ) = 0.0003
hence value of p0.05 > 0.0003,here we reject Ho
ANSWERS
---------------
null, Ho: =8.9
alternate, H1: !=8.9
test statistic: 8.9665
critical value: -2.571 , 2.571
decision: reject Ho
p-value: 0.0003

population mean anger score for is diffrent from college men

b.
When population mean anger score for adult men (M=9.20)
to =13.3333-9.2/(1.2111/sqrt(6))
to =8.3597
| to | =8.3597
hence value of | to | > | t | and here we reject Ho
population mean anger score for Marines is diffrent from adult men

c.
mean anger score for male psychiatric outpatients (M = 13.5)
to =13.3333-13.5/(1.2111/sqrt(6))
to =-0.3372
| to | =0.3372
hence value of |to | < | t | and here we do not reject Ho
population mean anger score for Marines is as similar to psychiatric outpatients

d.
by observing above 3 diffrent behaviour we conclude that, overall Marines’moods
are following high-altitude since the reason behind this, the anger score for them are
as similar to psychiatric outpatients and neither in range to college students and
nor to adult men

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