9.41 Single-sample t test, military training, and anger: Bardwell, Ensign, and M

Question

    9.41 Single-sample t test, military training, and anger: Bardwell, Ensign, and Mills (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 three groups: college men, adult men, and male psychiatric outpatients. Let’s examine anger scores for six 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.

        The population mean anger score for college men is 8.90. Conduct all six steps of a single-sample t test. Report the statistics as you would in a journal article.

        Now calculate the test statistic to compare this sample mean to the population mean anger score for adult men (M = 9.20). You do not have to repeat all the steps from part (a), but conduct step 6 of hypothesis testing and report the statistics as you would in a journal article.

        Now calculate the test statistic to compare this sample mean to the population mean anger score for male psychiatric outpatients (M = 13.5). Do not repeat all the steps from part (a), but conduct step 6 of hypothesis testing and report the statistics as you would in a journal article.

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

Explanation / Answer

we know that the t stat is given as

t = (xbar-mu)/(sd/sqrt(n))

now in the first case mu = 8.90

Hypothesis is

H0 : There is no difference between the mean values

H1 : There is a statistical difference between the mean values

we quickly calculate the mean , sd and sample size for the given data points in excel as

t = (13.33- 8.9)/(1.105/sqrt(6))
= 9.82

lets consider the mu as 9.20

t= (13.33- 9.20)/(1.105/sqrt(6))
=9.15

lets consider the mu = 13.5

t = (13.33- 13.5)/(1.105/sqrt(6))

= -0.37

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