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DBA830: Module 4 Problem Set
Problem 1
a).You know the population mean and variance
It would be appropriate to use Ztest 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, ztest would be highly preferred. The formula forztest applicable in this scenario may be as follows
z=(Xμ)
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 ttest rather than Ztest.Normally, ttest is used when the size of the population (n) isgreater than 30(n>30).The formula for ttest 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 ztest, that uses information about the population’svariance or SD, the formula for the tstatistic 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 ttest for independent samples, the design had to have participantsthat were assigned to each of the two groups at a random
b). To perform a ttest 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 ttest is computed as. Degree offreedom for one sample using ttest is DF=n1, while degrees offreedom for a two sample being compared:
Problem 5
When we did the ztest, we computed thezvalue and then looked at where it would fall in the distribution ofall possible zscores. Similarly, when we do a ttest, we compute thetvalue and then look at where that tvalue would fall in thedistribution of student ttest 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_twotailedtest, while a directional hypothesis uses aonetailed 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 tvalue
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 tvalue = 2.35.
1.Is this a directional or nondirectional 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 twotailed 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 computedtvalue = 2.35.
5.Is this a directional or nondirectional 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 tvaluecomputed will be approximate 1.325 and does not equal to the tvaluecritical 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 twotailed test with alpha set at p <.01.
Hypothesis: Ho: μ_{A}– μ_{B}=0
_{ }H1: μ_{A}– μ_{B}≠0
Theformulas applicable are as follows
S^{2}is the variance while ttest 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 
(X1mean1)^2 
(X2mean2)^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 
S^{2}=7.5556+11.3333=18.889=1.1805
9+92 16
t=7.555611.3333= 3.78
1.1805(1/9+1/9) 0.2623
t =14.4109
Usingthe twotailed test at alpha 0.01, it will be 0.01/12=0.005
10.005=0.995
Degreeof Freedom (D.F) =n+n1=9+91=16
Tablevalue at D.F=16 at 0.005=0.995 will be 2.921
RejectionRegion
Rejection Region
Comments
It canbe observed that tcomputed (14.4109) is less than tcritical(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= (11.333333337.555555556)= 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