Question

Q1) Independent-Samples t Test (25 points total)

A clinic specializing in eating disorders would like to evaluate the effectiveness of a nutritional training program on binge eating disorder (BED). A psychologist was hired to conduct a small study to examine the potential effect of nutritional training on number of binges per month. The psychologist recruited 10 patients in the clinic to participate in a nutritional training program and 12 patients in the clinic with similar demographics who do not participate in the nutritional program. Afterward, all participants reported the number of binges from the previous month. The data are listed in the table below. The psychologist is not predicting a particular direction of the potential differences between the two groups and she sets the alpha level at .05 for the hypothesis test.

Nutritional training			No Nutritional training
Subject ID #	# of binges		Subject ID #	# of binges
1	6		11	7
2	4		12	8
3	2		13	6
4	5		14	4
5	3		15	7
6	5		16	6
7	5		17	3
8	3		18	8
9	7		19	9
10	3		20	6
			21	7
			22	5

1. What is the dependent (outcome) variable? What is the independent (grouping) variable? (1 point total: .5 for each variable)

Dependent variable is number of doctors visits while the independent variable is potential effect of cat ownership.

2. Create the null and alternative hypotheses for this study, using both words and symbol notation (2 points total: 1 per hypothesis- .5 for written format, .5 for notation)

H₀:µ_d =0 .There is no potential effect of cat ownership on a senior resident’s number of visits to the doctor’s office.

H_a:µ_d ≠ 0. There is a potential effect of cat ownership on senior resident’s visits to the doctor’s office.

3. Calculate M₁and M₂(2 points total: 1 point per sample mean- .5 for work, .5 for answer)

Owning a cat = 7.7

Not owning a cat = 8.5

4. Calculate df₁ , df₂, and df_total (2 points total: 1 for df1 and df2, 1 for total df, including work)

df₁=9

df₂=11

df_total=20

5. Calculate ²₁ and ²₂ (estimated variance for population 1 and variance for population 2) (4 points total: 2 points for each variance. 1 for work and 1 for result)

s²₁=4.6778

s²₂=4.2727

6. Calculate the pooled variance for the two populations (2 points total: 1 point for work, 1 point for result)

S²_p=4.455

7. Use the pooled variance to calculate the variance for sampling distribution 1 and the variance for sampling distribution 2 (2 points total: 1 for work and 1 for result)

Hint: Sampling distribution derives from the original population and it consists of means of all possible samples drawn from the original population.

8. Calculate standard deviation of the comparison distribution (2 points total: 1 for work and 1 for result)

Hint: This is the distribution consisting of differences between means and its standard deviation is the denominator of the t statistic formula.

9. Calculate the t statistic (2 points total: 1 for work and 1 for result)

10. Considering the research question and the hypotheses, should the test be one-tailed or two-tailed? Why? (2 points total: 1 for each answer)

The test should be two tailed as the researcher is not predicting a particular direction of the potential differences between the two groups.

11. Determine the critical t value(s) for this hypothesis test based on the degree of freedom, from (d), and the preset alpha level. (1 point total)

12. Compare the calculated t statistic with the critical t value by stating which is more “extreme”, and then make a decision about the hypothesis test by stating clearly “reject” or “fail to reject” the null hypothesis. (1 point total: .5 for comparison, .5 for decision)

13. Calculate the pooled standard deviation for the populations and then use it to calculate the standardized effect size of this test. (2 point total: 1 for pooled standard deviation, 1 for effect size. Deduct .5 if a result is wrong but the process is correct.)

Answer 1

The hypothesis being tested is:

H₀: µ₁ = µ₂

H₁: µ₁ < µ₂

Nutritional training	No Nutritional training
4.30	6.33	mean
1.57	1.72	std. dev.
10	12	n
	20	df
	-2.033	difference (# of binges - # of binges)
	2.738	pooled variance
	1.655	pooled std. dev.
	0.709	standard error of difference
	1.725	critical value
	-2.870	t-value
	.0047	p-value (one-tailed, lower)

Since 2.870 > 1.725, we can reject the null hypothesis.

Since the p-value (0.0047) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the nutritional training program on binge eating disorder (BED) is effective.

Effect size = -2.033/1.655 = 1.228

This is large effect size.