Question

Q2) Independent-Samples t-Test (15 points total)

In a research project, researchers collected demographic and health data from a sample of elderly residents in the community. To examine any possible gender differences in their sample, they want to see if the females and the males differ significantly on the education level (number of years of formal schooling). The researchers are not predicting any direction in the possible gender differences so the hypotheses should be non-directional. They would like to run a two-tailed test with α = .10.

Male Subject ID	Education		Female Subject ID	Education
1	12		13	16
2	12		14	18
3	14		15	18
4	12		16	16
5	16		17	16
6	16		18	14
7	12		19	16
8	14		20	12
9	16		21	18
10	16		22	18
11	15		23	16
12	13		24	16
			25	18
			26	12

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

DV: Education

IV: Sex

b. Create the null and alternative hypotheses (non-directional) for this study, using both words and symbol notation (1 point total: .5 for each hypothesis, both written and symbol notation need to be correct to earn the credit for each hypothesi

for null hypothesis: Ho: uMale=Ufemale
against the alternative hypothesis H1: uMale=/uFemale

c. Calculate M₁ andM₂(1 point total: .5 point per sample mean, both process and result must be correct to earn credit for each mean)

d. Calculate df₁ , df₂, and df_total (1 point total: deduct .5 for each error)

e. Calculateestimated variance for population 1 (^s₁²) and estimated variance for population 2 (s₂²) (2 points total: 1 point for each variance, .5 if the process is correct but the answer is wrong)

f. Calculate the pooled variance (S_pooled²) from the two population variances (from question e above) (1 point total: .5 if the process is correct but the answer is wrong)

g. Use the pooled variance (from question f above) to calculate the variance for sampling distribution 1 (S_M1²) and the variance for sampling distribution 2 (S_M2²) (2 points total: 1 for each variance, .5 if the process is correct but the result is calculated incorrectly)

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

h. Calculate standard deviation (S_diffmean)of the comparison distribution (1 point total: .5 if the process is correct but the answer is wrong)

Hint: This comparison distribution consists of differences between all possible sample means drawn from the two sampling distributions. Its standard deviation is the denominator of the t statistic formula.

i. Calculate the t statistic (1 point total: .5 if the process is correct but the answer is wrong)

j. For the two-tailed test, find the critical t values for this hypothesis test based on the total degree of freedom (from question d above), and the preset alpha level. (1 point total)

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

l. Calculate the pooled standard deviation for the populations (use the pooled variance calculated in question f); and then calculate the standardized effect size of this test. (1 point total: .5 for pooled standard deviation, .5 for effect size. Both process and result must be correct to earn the credit for each item.)

Answer 1

B)

Ho :   µ1 - µ2 =   0
Ha :   µ1-µ2 ╪   0

Ho : Difference between mean education between male and female is zero.
Ha :    Difference between mean education between male and female is not zero.

c)

Sample #1   ---->   1
mean of sample 1,    x̅1=   14.000

      
Sample #2   ---->   2
mean of sample 2,    x̅2=   16.000

d)

df1 = 14

df2 =16

Degree of freedom, DF=   n1+n2-2 =    24

e)

standard deviation of sample 1,   s1 =    1.758

standard deviation of sample 2,   s2 =    2.075

f)

pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.9365

Fpr variance we can square above values

g)

std error , SE =    Sp*√(1/n1+1/n2) =    0.7618                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -2.0000   -   0   ) /    0.76   =   -2.6253

t-critical value , t* =        1.711   (excel formula =t.inv(α/2,df)              
Decision:   | t-stat | > | critical value |, so, Reject Ho                      

Please let me know in case of any doubt.

Thanks in advance!

Please upvote!