Question

M5 Assignment (30 points)

Help with J, K, I and M please!

General instruction:

1. Complete the guided Practice Problem first, if you have not done so. It will help you become proficient in answering the questions in this assignment.
2. Check out the module, Assignment Guides, which demonstrates how to show the math/process work and provides instruction on essential procedures in Word and Excel, and APA formatting examples.
3. Use this Word document to fill in the answers to the questions. You must type out a clear answer to each question, even if the answer is also contained in the Excel file submitted to show work.

Q1) Independent-Samples t Test (15 points total)

In a cognitive psychology experiment, the researcher is interested in whether encoding condition has a significant effect on memory for a set of drawings. In encoding condition A, subjects are asked to name the object in each drawing; in encoding condition B, subjects are asked to name the color of each drawing. Then all subjects are tested on their memory of the drawings. Each recruited subject is randomly assigned to either encoding condition A or encoding condition B.

Based on memory theories, the researcher hypothesized that condition A (object naming) would lead to significantly better memory compared to condition B (color naming), and she sets the significance level at α = .05 for a one-tailed test.

Hint: You may remember a similar scenario in the previous assignment, but with a crucial difference. In the previous assignment, each subject is tested in BOTH condition A and condition B, while each subject here is tested in only ONE condition (either A or B).

Subject ID	Encoding A		Subject ID	Encoding B
1	87		13	85
2	80		14	80
3	78		15	76
4	76		16	77
5	86		17	86
6	77		18	68
7	83		19	85
8	82		20	79
9	91		21	89
10	90		22	85
11	95		23	82
12	83		24	80

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

DV: Memory

IV: encoding

b. Create the null and alternative hypotheses for this study, using both words and symbol notation.

Note that the hypotheses should be directional. (1 point total: .5 for each hypothesis, both written and symbol notation need to be correct to earn the credit for each hypothesis)

Ho: uA=uB ( There is no significant difference between each condition)

H1: uA>uB (Condition A would lead to more money than condition B)

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

M1 = Avg. of A = (87+80+(etc…)+83)/12 =

1008/12 = 84

M2 = Avg of B = (85+80+(etc…)+80)/12 =

972/12 = 81

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

df= 12-1=11

df₂=12-1=11

total: 11+11=22

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)

s12 =

((87-84)2+(80-84)2+..(etc)..+(83-84)2)/(12-1) = 35.45

s22 =

((85-81)2+(80-81)2+..(etc)..+(80-81)2}/(12-1))= 32.18

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)

{11x35.45 + 11x32.18}/{22} = 33.82

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.

S_M1² = (12 - 1)x35.45/12 + 12 - 2 = 17.725

S_M2² = (12 - 1)x32.18/12 + 12 - 2 = 16.09

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.

= 17.725 + 16.09 = 5.815

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

|3 | / 5.8153 =  0.516

j. For the one-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.)

m. Draw a conclusion based on the hypothesis test result and the effect size. In other words, did encoding condition have a significant effect on memory score? Was the effect small, medium, or large? (1 point total: .5 for each element).

Answer 1

The provided sample means are shown below: X1 = 84 X2 = 81 Also, the provided sample standard deviations are: S1 = 5.9518 S2 = 5.6729 12 and n2 = = and the sample sizes are ni 12. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = M2 Ha: U1 > U2 This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region Based on the information provided, the significance level is a 0.05, and the degrees of freedom are df = 22. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this right-tailed test is te = 1.717, for a = 0.05 and df = 22. The rejection region for this right-tailed test is R= {t:t > 1.717}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X1 - X2 (n1-1) sî+(12-1)s (m + ni+n2-2 n2 84 – 81 = 1.264 (12-1)5.95182+(12-1)5.67292 12+12-2 (12 + 12

(4) Decision about the null hypothesis Since it is observed that t = 1.264 <te = 1.717, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.1097, and since p 0.1097 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean ui is greater than um, at the 0.05 significance level. Confidence Interval The 95% confidence interval is – 1.922 < M1 – M2 < 7.922.

Thus there is not sufficient evidence to conclude that encoding conditions have effect on memory score.

I have given my best to solve your problem. Please like the answer if you are satisfied with it. ?