Question

You work as a cheeseburger sales person and believe that eating cheeseburgers will increase walking speeds in adults. In order to test the idea you randomly select 6 adults from your town and you measure their walking speed again. For this experiment you decide your critical value of 0.05 to be your cutoff for statistical significance. Your data is

Data as follows:

Subject/pre cb walk speed/ post cab walk speed

1. 4. 2

2. 5. 6

3. 7. 5

4. 3. 1

5. 6. 3

6. 8. 4

1. Which statistical test is appropriate for the data set? Why

2. What are your hypotheses for this experiment?

3. Describe the characteristics of your comparison distribution.

4. What is the cutoff critical score for this experiment?

5. What is your sample t score

6. What is your decision regarding the null hypothesis?

7. What is the effect size of this study?

8. Describe the overall conclusion you can draw from this study. Interpret and explain your findings.

Answer 1

(1)

Dependent 2 Samples t test.

REASON: Here the 2 samples: pre cb walk spped and post cb walk speed are dependent.

(2) H0: Null Hypothesis: 0

HA:Altrnative Hypothesis: 0

(3)

Pre cb walk speed (X)	Post cb walk speed (Y)	X-Y=d
4	2	2
5	6	- 1
7	5	2
3	1	2
6	3	3
8	4	4

From d values, the following statistics are calculated:

n = 6

= 12/6 = 2

s_d= 1.6733

SE = s_d /

= 1.6733/ = 0.6831

The comparison distribution is a t distribution with mean = = 2 and SD = 0.6831

(4)

= 0.05

One Tail - Left Side Test

ndf = n - 1 = - 1 = 5

From Table, critical value of t = - 2.0150

(5)

t = /SE

= 2/0.6831 = 2.9278

(6)

Since calculated value of t = - 2.9268 is less than critical value of t = - 2.0150, the difference is significant. Reject null hypothesis.

(7)

Effect Size = (M1 - M2)/SD

                = 2/1.6733 = 1.1952

Since Effect Size> 1, the effect size is Large.

(8) Conclusion:
The data support the claim that eating cheeseburger will increase walking speeds in adults.