# M-4

7

DBA-830: Module 4 Problem Set

Problem 1

a).You know the population mean and variance

It would be appropriate to use Z-test when the population mean andthe variance are known. Besides, if one has the information on onesample and need to compare it with the population mean to see if itis significant, z-test would be highly preferred. The formula forz-test applicable in this scenario may be as follows

z=(X-μ)

σ

Where σ is standard deviation

μ is the population mean

X is the sample mean

b). You do not know the population mean and variance

In this scenario where one does not know the population mean andvariance, it would be appropriate to use t-test rather than Z-test.Normally, t-test is used when the size of the population (n) isgreater than 30(n&gt30).The formula for t-test may appear as follows

t = (x – μx) / [s/√ (n)].

Where x is the sample mean and μ is population mean

s= sample standard deviation

n=sample size

Problem 2

Unlike the z-test, that uses information about the population’svariance or SD, the formula for the t-statistic uses an estimateof the population’s variance or standard deviation, which iscalled the sample variance or sample standard deviation, whichis computed from the sample population

Problem 3

a).A researcher wants to compare typing speed (the DV) in twoconditions of light (IV: low, high level of light). In order to usea t-test for independent samples, the design had to have participantsthat were assigned to each of the two groups at a random

b). To perform a t-test where we are comparing twoindependent samples (groups), we need three pieces ofinformation. These are:

i) Whether the variables assume a normal distribution

ii).The variables in the two populations must have equal variance

iii). the number of observations in the two samples must beindependent

Problem 4

The degree of freedom (DF) for each samplethat is being compared using the t-test is computed as. Degree offreedom for one sample using t-test is DF=n-1, while degrees offreedom for a two sample being compared:

Problem 5

When we did the z-test, we computed thez-value and then looked at where it would fall in the distribution ofall possible z-scores. Similarly, when we do a t-test, we compute thet-value and then look at where that t-value would fall in thedistribution of student t-test table.The critical values of tcan be found in Table iii_in the appendix of the textbook.

Problem 6:

a). Anondirectional hypothesis predicts there will be a significantdifference between the two means that can beindependent. A directionalhypothesis predicts there will be a significant difference betweenthe two means and adds dependent.

b).A nondirectional hypothesis uses a_two-tailedtest, while a directional hypothesis uses aone-tailed test to evaluate if thedifference between the means is significant.

c).When I set alpha at 5%, and have a nondirectionalhypothesis, the 5% of outcomesthat occur less frequently than my critical t-value

1.Fall in one tail of the distribution OR

2.Are divided between the two tails (2 ½% in one tail and 2 ½ % inthe other tail)? They are dividedbetween the two tails.

Problem7

a).Hypothesis: There will be a significant difference between the means of Group Aand Group Bdf = 25 computed t-value = 2.35.

1.Is this a directional or non-directional hypothesis?_nondirectional hypothesis

2.Given the type of hypothesis, the proportion of extreme values that Iam interested in will be found in one tail or in two tails combined.It will be found in a two-tailed test because the hypothesis givenis nondirectional

3.Given that the df = 25 and alpha is .05, what would be the value of tthat my computed value would have to equal or exceed? t- Criticalvalue will be 1.8 and computed value is 2.35. The critical value doesnot exceed the compute value and hence fail to reject null hypothesis

4.Is my computed value significant at this alpha level?

Mycritical value is not significant at this alpha

a).Hypothesis: There will be a significant difference between the meansof Group A and Group B specifically, the mean of Group B will besignificantly greater than the mean of Group Adf = 25 computedt-value = 2.35.

5.Is this a directional or non-directional hypothesis? it isdirectional hypothesis

6.Given the type of hypothesis, the proportion of extreme values that Iam interested in will be found in one tail or in two tails combined.The proportions of extreme values that I am interested will befound in one tailed

7.Given that the df = 25 and alpha is .05, what would be the value of tthat my computed value would have to equal or exceed? the t-valuecomputed will be approximate 1.325 and does not equal to the t-valuecritical hence reject the null hypothesis

8. Is my computed value significant at this alpha level?

Computedvalue is not significant at this alpha level

Problem 8

a).Is there a significant difference in reported performance between thetwo conditions? Use two-tailed test with alpha set at p &lt.01.

Hypothesis: Ho: μA– μB=0

H1: μA– μB≠0

Theformulas applicable are as follows

S2is the variance while t-test for the two samples means first let usdetermine the sample variance using excel as shown below

 n=9 n=9     Group A, Well Lit Room Group B, Dim Lit Room     X X2 (X1-mean1)^2 (X2-mean2)^2 7 9 0.308641975 1103.716049 8 11 0.197530864 974.8271605 10 13 5.975308642 853.9382716 6 10 2.419753086 1038.271605 8 11 0.197530864 974.8271605 5 9 6.530864198 1103.716049 7 15 0.308641975 741.0493827 12 14 19.75308642 796.4938272 5 10 6.530864198 1038.271605 Mean 1=7.555555556 Mean 2=11.33333333 42.22222222 8625.111111

S2=7.5556+11.3333=18.889=1.1805

9+9-2 16

t=7.5556-11.3333= -3.78

1.1805(1/9+1/9) 0.2623

t =-14.4109

Usingthe two-tailed test at alpha 0.01, it will be 0.01/12=0.005

1-0.005=0.995

Degreeof Freedom (D.F) =n+n-1=9+9-1=16

Tablevalue at D.F=16 at 0.005=0.995 will be 2.921

RejectionRegion

Rejection Region

It canbe observed that t-computed (-14.4109) is less than t-critical(2.921) hence reject null hypothesis and conclude that there asignificant difference in reported performance between the twocondition.

b).Compute Cohen’s d to estimate the size of the treatmenteffect. After you have computed Cohen’s d, how would youinterpret it?

Cohen’sd, may be computed by the formula below where M2 and M1 aredifference between two sample mean and SD is samples standarddeviation

Cohen’s d= (11.33333333-7.555555556)= 3.7778

(42.22222222+8625.111111) /2 8667.3333

Cohen’s d=0.0004358

Cohen’s d show that the standard difference between the two meansabove is 0.0004358