Answer

One Sample T-Test

Population Mean (): 21

Sample Mean (): 13.15

Sample Standard Deviation (s): 7.160577171

Sample size (n): 72

t=  = -9.302247179

1.994

BCR: Reject  given that |t| >

|t| = 9.302247179

9.302247179 > 1.994

In locating the critical value of t, I used the student-t distribution table. I majorly used two metrics (the degree of freedom of the sample and  which is the significance level) from Engineering Statistics Handbook (u.d). Given that this type of test was two-tailed-given the hypothesis, I used the two-tailed tests. The degree of freedom was gotten using n-1 (72-1) which gave 71. So, in the t-table, I chose 70 as the closest degrees of freedom and checked vertically along the 5% significance level. This gave me the value 1.994. To assess whether my calculated t was significant, I applied the Basic Critical Region (BCR) Test which states that Reject  given that |t| >  according to Test (2018)

The obtained t (9.3) falls within the rejection region given by the critical t (1.994)Therefore, at 5% level of significance, we reject the null hypothesis and conclude that there is enough evidence to show that stress scores reported in the Week 6 sample are significantly different from those of the population of online students.