In: Statistics and Probability
I am curious about how many hours per week my students spend on their online coursework. So, I surveyed 9 of my students on how many hours per week they spend on my online PSY 230 course. According to a survey of online college students (in all courses), students studied an average of 9 hours a week for a 3-credit course. I assume that the national study time scores are normally distributed, and I set the significance level at α = .05.
Student |
Study hours |
1 |
12 |
2 |
19 |
3 |
21 |
4 |
14 |
5 |
11 |
6 |
9 |
7 |
12 |
8 |
10 |
9 |
7 |
(a)
Question:
Calculate the t statistic for the sample
Answer :
From the given data, the following statistics are calculated:
n = 9
= 12.7778
s = 4.5765
Test Statistic is given by:
(b)
Question:
Specify whether the test is a one-tailed or two-tailed test based on the hypotheses form (b)
two-tailed test
(c)
Question:
determine the critical t value(s) based on the type of test and the preset alpha level
df = 9 - 1 = 8
= 0.05
From Table, critical values of t =
2.306
(d)
Question:
Compare the t statistic with the critical t value. Is the calculated t statistic more extreme or less extreme than the critical t value? Then make a decision by stating whether we “reject” or “fail to reject” the null hypothesis
Since calculated value of t = 2.476 is greater than critical value of t = 2.306, the calculated t statistic more extreme than the critical t value. Rejectnull hypothesis.
(e)
Question:
Interpret the result in 1-2 sentences (you may restate the hypothesis accepted or explain it in your own words)
The null hypothesis that the number of hours spent by the students of the Professor do not differ from the national study time scores is rejected. So, we conclude that the number of hours spent by the students of the Professor differ from the national study time scores
(f)
(i)
Raw Effect size =
= 12.7778 - 9 = 3.7778
(ii)
Standardized Effect Size is given by: