COMM 250 Research & T Test Practice

COMM 250 Research & t test Practice

FBK provide an example of using a t test to analyze an hypothesis and a set of data on pp. 344-347. You will need to know this for both RAT6 and the final exam. Here are two more examples, the one I did in class, and the one from ITE 12 in class.

Experimental Research Example # 1

(Presented in Class)

Research Scenario

A researcher notices that people who interrupt others frequently do not appear to be very good listeners. She devises an experiment to test this.

a) A pretest is used to divide participants into two groups – six “Low-Interrupters” (Group A) and six “High-Interrupters” (Group B)

b) Each participant engages in a 30 minute conversation about a controversial topic in which they have expressed an interest.

c) A “listening rating” was calculated based on a subjects’ ability to answer questions about the person with whom they talked (1 point for each of 16 questions). The following listening scores were recorded:

a) the A scores = 4, 6, 10, 10, 14, 16

b) the B scores = 1, 3, 4, 5, 5, 6

Compute a t statistic and determine whether the data supports the research hypothesis. Set the probability level – also called the significance level - at .05 (pronounced “point oh five”).

Use this exercise as a way to practice . . .

1) Write out this investigator’s research hypothesis.

2) Write out the null hypothesis.

3) Does the research hypothesis call for a one-tailed or two-tailed test?

4) What is the calculated value of t for this set of data?

5) How many “degrees of freedom” are there in this example?

6) With the alpha level set at .05, what is the critical value of “t” needed to rejecting the null hypothesis?

7) Do the results of this statistical test allow you to reject the null hypothesis that there is no difference between the two groups?

8) What conclusion can the researcher draw about the research hypothesis from this study? (Practice by writing it out as a sentence.)

9) Describe at least one extraneous variable that should be considered as a possible mitigating factor (that is, a variable that the researcher should mention when “qualifying” the results or the generalizability of the results).

The Answers

1) Write out this investigator’s research hypothesis.

The more a person interrupts another in conversation, the poorer a listener they are.

2) Write out the null hypothesis.

The number of one’s conversational interruptions is unrelated to one’s level of listening skills.

3) Does the research hypothesis call for a one-tailed or two-tailed test?

Since this is a directional hypothesis, it calls for a one-tailed (directional) test.

“A” Scores / Minus the Mean / Diff from Mean / Diff 2 / Diff Squared / “B” Scores / Minus the Mean / Diff from Mean / Diff 2 / Diff Squared
4 / -10 / - 6 / - 6 2 / 36 / 1 / - 4 / - 3 / - 3 2 / 9
6 / -10 / - 4 / - 4 2 / 16 / 3 / - 4 / - 1 / - 1 2 / 1
10 / -10 / 0 / 0 2 / 0 / 4 / - 4 / 0 / 0 2 / 0
10 / -10 / 0 / 0 2 / 0 / 5 / - 4 / 1 / 1 2 / 1
14 / -10 / 4 / 4 2 / 16 / 5 / - 4 / 1 / 1 2 / 1
16 / -10 / 6 / 6 2 / 36 / 6 / - 4 / 2 / 2 2 / 4
60 / Sum Sq = / 104 / 24 / Sum Sq = / 16
/ 6 = / 10 / (Mean) / / 6 = / 4 / (Mean)

M1 - M2

t = ------

------

½ ∑ d1 2 + d22 n1 + n2

½ ______

√ n1 + n2 - 2 n1 x n2

10 - 4

t = ------

------

½ 104 + 16 6 + 6

½ ______

√ 6 + 6 – 2 6 x 6

t = ------

------

½ 120 12

½ ______X ______

√ 10 36

t = ------

------

√ 12 ( .33)

t = 6 / 1.99 = 3.02

4) What is the calculated value of t for this set of data?

t = 3.02

5) How many “degrees of freedom” are there in this example?

(The degrees of freedom equals the number of observations (n1) in sample 1 plus the number in sample 2 (n2) minus the number of groups. In other words:)

df = n1 + n2 – 2, which equals: 6 + 6 – 2 = 10

6) With the alpha level set at .05, what is the critical value of “t” needed to rejecting the null hypothesis?

According to the t table, for an alpha level of .05, with a one-tailed test and 10 df, the critical value of t is = 1.812. (This means that if the calculated value of t is greater than or equal to 1.812, the null hypothesis can be rejected and the research hypothesis is supported.)

7) Do the results of this statistical test allow you to reject the null hypothesis that there is no difference between the two groups?

Yes, because the calculated value of t (3.02) is greater than the critical value required (1.812). What we say is that “there is a statistically significant difference between the groups.”

8) What conclusion can the researcher draw about the research hypothesis from this study? (Practice by writing it out as a sentence.)

This study supports the hypothesis that the more people interrupt during conversation, the poorer are their listening skills.

It is unclear whether the same results would be found with a very large sample. Also, it is possible that another variable affected both interrupting and listening. For example, perhaps using a “controversial” topic of interest to the participants is a mistake. The study could be replicated using a more neutral topic.

Note – you can create your own research examples and a set of numbers and practice your own t-test until you are confident you have it down. Get your teammates to do it as well and you can compare answers.

Experimental Research Example # 2

(The scores used below were given out in class and are part of lecture notes for week 12)

Research Scenario

A professor is concerned about low student scores on quizzes. He figures that while most of them are doing the readings, they do not know how to study for his quizzes and may be unfamiliar with the style of question that he writes. Therefore, he has reason to believe that providing “practice quizzes” in advance of in-class quizzes will improve student performance. He decides to test this. He divides the class in two groups – eight students are given a practice quiz (Group A) and eight are not (Group B)

· the A scores = 12, 12, 12, 11, 11, 11, 10, 9

· the B scores = 9, 8, 8, 7, 7, 7, 6, 4

Compute a t statistic and determine whether the data supports the research hypothesis. Set the probability level – also called the significance level - at .05 (pronounced “point oh five”).

Use this exercise as a way to practice . . .

1) Write out this investigator’s research hypothesis.

2) Write out the null hypothesis.

3) Does the research hypothesis call for a one-tailed or two-tailed test?

4) What is the calculated value of t for this set of data?

5) How many “degrees of freedom” are there in this example?

6) With the alpha level set at .05, what is the critical value of “t” needed to rejecting the null hypothesis?

7) Do the results of this statistical test allow you to reject the null hypothesis that there is no difference between the two groups?

8) What conclusion can the researcher draw about the research hypothesis from this study? (Practice by writing it out as a sentence.)

The Answers

1) Write out this investigator’s research hypothesis.

Experience with practice quizzes will improve in-class student quiz scores.

2) Write out the null hypothesis.

Taking practice quizzes do not improve student in-class quiz scores.

3) Does the research hypothesis call for a one-tailed or two-tailed test?

Since this is a directional hypothesis, it calls for a one-tailed (directional) test.

“A” Scores / Minus the Mean / Diff from Mean / Diff 2 / Diff Squared / “B” Scores / Minus the Mean / Diff from Mean / Diff 2 / Diff Squared
12 / -11 / 1 / 1 2 / 1 / 9 / - 7 / 2 / 2 2 / 4
12 / -11 / 1 / 1 2 / 1 / 8 / - 7 / 1 / 1 2 / 1
12 / -11 / 1 / 1 2 / 1 / 8 / - 7 / 1 / 1 2 / 1
11 / -11 / 0 / 0 2 / 0 / 7 / - 7 / 0 / 0 2 / 0
11 / -11 / 0 / 0 2 / 0 / 7 / - 7 / 0 / 0 2 / 0
11 / -11 / 0 / 0 2 / 0 / 7 / - 7 / 0 / 0 2 / 0
10 / -11 / - 1 / - 1 2 / 1 / 6 / - 7 / - 1 / - 1 2 / 1
9 / -11 / - 2 / - 2 2 / 4 / 4 / - 7 / - 3 / - 3 2 / 9
88 / Sum Sq = / 8 / 56 / Sum Sq = / 16
/ 8 = / 11 / (Mean) / / 8 = / 7 / (Mean)

M1 - M2

t = ------

------

½ ∑ d1 2 + d22 n1 + n2

½ ______X ______

√ n1 + n2 - 2 n1 x n2

11 - 7

t = ------

------

½ 8 + 16 8 + 8

½ ______X ______

√ 8 + 8 – 2 8 x 8

t = ------

------

½ 24 16

½ ______X ______

√ 14 64

t = ------

------

√ 1.71 ( .25)

t = 4 / 0.65 = 6.15

4) What is the calculated value of t for this set of data?

t = 6.15

5) How many “degrees of freedom” are there in this example?

(The degrees of freedom equals the number of observations (n1) in sample 1 plus the number in sample 2 (n2) minus the number of groups. In other words:)

df = n1 + n2 – 2, which equals: 8 + 8 – 2 = 14

6) With the alpha level set at .05, what is the critical value of “t” needed to rejecting the null hypothesis?

According to the t table, for an alpha level of .05, with a one-tailed test and 14 df, the critical value of t is = 1.761. (This means that if the calculated value of t is greater than or equal to 1.761, the null hypothesis can be rejected and the research hypothesis is supported.)

7) Do the results of this statistical test allow you to reject the null hypothesis that there is no difference between the two groups?

Yes, because the calculated value of t (6.15) is greater than the critical value required (1.761). What we say is that “there is a statistically significant difference between the groups.”

8) What conclusion can the researcher draw about the research hypothesis from this study? (Practice by writing it out as a sentence.)

This study supports the hypothesis - practice quizzes do, in fact, help students score higher on in-class quizzes.

It is unclear whether the same results would be found with a very large sample. Also, it is possible that .