The moods of U. S. Marines following a month-long training exercise conducted at cold temperature and at high altitudes were assessed. Negative moods, including fatigue and anger, increased substantially during the training and lasted up to three month after the training ended. The scores for 5 of the Marines were 14, 10, 13, 10, 11. The mean mood score was compared to population norms for college men; the population mean anger score for college men is 8.90. a) Test the null hypothesis that the population mean is 8.90 against the alternative that the population mean is greater than 8.90 at α=.05. Show all 6 steps. b) Interpret the results. What did we learn about the Marines' negative moods?

Step-by-step explanation:

Sample mean = (14 + 10 + 13 + 10 + 11) / 5 = 11.6

Sample standard deviation, s = √ (summation (x - mean) / n

Summation (x - mean) = (14 - 11.6) ^2 + (10 - 11.6) ^2 + (13 - 11.6) ^2 + (10 - 11.6) ^2 + (11 - 11.6) ^2 = 13.2

s = √ (13.2/5) = 1.62

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 8.9

For the alternative hypothesis,

µ > 8.9

it is a right tailed test because of >.

Since the number of samples is 5 and no population standard deviation is given, the distribution is a student's t.

Since n = 5,

Degrees of freedom, df = n - 1 = 5 - 1 = 4

t = (x - µ) / (s/√n)

Where

x = sample mean = 11.6

µ = population mean = 8.9

s = samples standard deviation = 1.62

t = (11.6 - 8.9) / (1.62/√5) = 3.73

We would determine the p value using the t test calculator. It becomes

p = 0.01

Since alpha, 0.05 > than the p value, 0.01, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that mean anger score of the marines is greater than that of college men.

Therefore, the marines's negative moods increased and it is higher than that of college men.