Chapter 12 Homework 6 Your Name

  • Uncategorized

Chapter12: Homework 6

YourName

7.Whatdoes the confidence interval for µdindicate?

&nbspItindicates themean of the differences d for the population of paired data. Itcontains a range of values of mean differences any value in therange can be represented by the samples.

10.(a)What is the final task after completing the analysis of significantresults?

Afteranalyzing the significant results, the researcher completes theexperiment by computing the effect size.

(b)Why is effect size useful at this stage?

Itis useful because it is the only way one can determine whether theindependent variable is essential in influencing a behavior. Whendetermining this, it is essential to note that as the proportionbecomes larger, the prediction of the individual scores by the meanbecomes more accurate. Measuring effect size helps in addressing suchissues.

12.Forthe following, which type of t-testis required?

(a)Studying the effects of a memory drug on Alzheimer’s patients,testing a group of patients before and after administration of thedrug.

Therelated-samples t-test is required. This is because we shall use onegroup of patients to test the memory under condition before and afteradministration of the drug.

(b)Studying whether men and women rate the persuasiveness of an argumentdelivered by a female speaker differently.

Theappropriate t-test for this is the Independent samples t-test. Inthis case, the persuasiveness of an argument delivery of male andfemale is the key thing to be tested. Looking at the groups, yourealize that they are independent hence the t-test.

(c)The study described in part (b), but with the added requirement thatfor each man of a particular age, there is a woman of the same age.

Related-samplest-test is also required here since both the groups are dependent onone factor- age.

13.Wesolicit two groups of professors at our school: One group agrees tocheck their e-mail once per day the other group can check it asoften as they wish. After two weeks, we give everyone a standard testof productivity to determine who has the higher (more productive)scores. The productivity scores from two independent samples are

Once:Ẋ=43, s2x=22.79, n=15

Often:Ẋ=39, s2x=24.6, n=15

  1. What are H0 and Ha?

Inthis case, our interes lies in studying the relationship betweenchecking email once per day and checking it more often.

Letmhsignify mean productivity score once per day and mcsignify mean productivity score often times per day

ThereforeH0= mh–mc= 0

AndHa = mh– mcnot equal to zero

  1. Compute tobt

Pooledvariance = S2p= {(n1 – 1)s2×1+ (n2 – 1)s2×2}/n1 + n2 – 2

={(15-1)(22.79)+ (15-1)(24.6)}/28

=663.46/28 = 23.695

Standarderror of the difference is given by { S2p(1/n1+ 1/n2)}0.5

={(23.695 (1/15 + 1/15)}0.5

=(3.1593)0.5= 1.7775

Havingboth the pooled variance and the standard error, we can now computethe tobtwhich is

={(43 -39) – (15 – 15)}/ 1.7775 = 4/1.7775 which is approximately2.25.

  1. With α= .05, what is tcrit?

0.05is the level of significance

Toget the degrees of freedom we add the two n and subtract 2

Thatis 15 + 15 – 2 = 28

Wetherefore read the value of tcritfrom the t-table with 0.05 as the level of significance and degreesof freedom of 28

Fromthe table we get tcrit= +/- 2.048

  1. What should we conclude about this relationship?

Inorder to conclude on the relationship, we compare tobtandtcrit. We therefore reject the null hypothesis since the tobt(2.25) is larger than tcrit.(2.048).in conclusion, we expect the mean of checking the email once to beabot 43 and that of checking it more often to be around 39.Therefore, checking email once per day produce different productivityscores than checking it more often.

  1. Compute the confidence interval for the difference between the µs.

Thisis given by

(Sx1– x2)(-tcrit)+ (X1bar – X2bar) &lt= mh– mc&lt= (Sx1– x2)(+tcrit)+ (X1bar – X2bar)

=(1.7775) (-2.048)+ 4 &lt= mh– mc&lt=(1.7775)(2.048)+4

-3.6402 +4 &lt= mh– mc&lt= 3.6402 + 4

Theconfidence interval is 0.3598 &lt= mh– mc&lt=7.6402

  1. Using our two approaches, how big is the effect of checking e-mail on productivity?

D= (X1bar – X2bar)/sqr S2p = 4/ 23.6950.5

=4/4.86775 = 0.8217

Squaredpoint-biserial correlation coefficient = (tobt)2/(tobt)2+ df

=2.252/2.252+ 28 = 5.0625/ 33.0625 = 0.1531

Thisis a moderate to large effect

  1. Describe how you would graph these results

Takenumber of times to check email on x axis and mean productivity scoreson y axis. Plot the data point of once per day check at 43 and themore often check at 39. Connect the two data points with a straightline.

15.Aresearcher asks if people score higher or lower on a questionnairemeasuring their emotional well-being when they are exposed to muchsunshine compared to when they’re exposed to little sunshine. Asample of 8 people is measured under both levels of sunshine andproduces these well-being scores:

Low:14 13 17 15 18 17 14 16

High:18 12 20 19 22 19 19 16

  1. Subtracting low from high, what are H0 and Ha?

Nullhypothesis H0: µD= 0

Alternativehypothesis Ha:µDnot equal to zero

WhereµDis the average difference in the population

  1. Compute tobt.

Low

High

Difference (D)

D2

14

18

4

16

13

12

-1

1

17

20

3

9

15

19

4

16

18

22

4

16

17

19

2

4

14

19

5

25

16

16

0

0

Total 124

145

21

87

Dbar = 21/8 = 2.625

Theestimated population variance of the difference scores = {87 –(212)/8}/8-1

=31.875/7 = 4.5536

Standarderror of the mean difference = (4.5536/ 8)1/2= 0.56921/2= 0.7545

Thereforetobt= (2.625 -0)/ 0.7545 = 3.4791

  1. With α= .05, what do you conclude about this study?

Degreesof freedom = 8 -1 = 7

Fromthe table tcrit with alpha = 0.05 and df = 7 is 2.365

Sincethe calculated value (3.4791) is larger than the tabulated value(2.635, we reject the null hypothesis. Therefore tobtis significant. We therefore conclude that people exposed to muchsunshine exhibit a significantly higher well-being score than whenexposed to little sunshine.

  1. Compute the appropriate confidence interval.

(Sx1– x2)(-tcrit)+ (X1bar – X2bar) &lt= mh– mc&lt= (Sx1– x2)(+tcrit)+ (X1bar – X2bar)

(0.7545)(-2.365)+ 2.625 &lt= µD(0.7545)(2.365) + 2.625

0.5806&lt= µD&lt=4.4094

(e)Whatis the predicted well-being score for someone when tested under lowsunshine? Under high sunshine?

Xbarunder little sunshine = 124/8 = 15.5

Xbarunder high sunshine = 145/8 = 18.125

f).Onaverage, how much more accurate are these predictions than if you didnot know how much sunshine people experience?

r2pb= (tobt)2/(tobt)2+df

=(3.4791)2/3.47912+ 7 = 12.1041/19.1041

=0.6336

Onaverage, 63.36% predictions are more accurate

(g)How scientifically important are these results?

Fromf, you note that r2pb= 63.36%. this implies that the results are very much important.

17.Wepredict that children exhibit more aggressive acts after watching aviolent television show. The scores for ten participants before andafter watching the show are

Sample1 (After) Sample 2 (Before)

5 4

6 6

4 3

4 2

7 4

3 1

2 0

1 0

45

32

  1. Subtracting before from after, what are H0 and Ha?

H0: µD= 0 there is no change in the scores before and after watching aviolent television show

Ha: µD&gt0 the acts of the children are more aggressive sfter watching aviolent television show

  1. Compute tobt

After

Before

Difference

D2

5

4

1

1

6

6

0

0

4

3

1

1

4

2

2

4

7

4

3

9

3

1

2

4

2

0

2

4

1

0

1

1

4

5

-1

1

3

2

1

1

Total

12

36

Mean(Dbar) = 12/10 = 1.2

Estimatedpopulation of the difference scores = (26 – 144/10)/10 – 1

=11.6/9 = 1.289

Standarderror of the mean difference = (1.289/10)1/2= 0.359

Therefore

Tobt= (1.2 -0)/ 0.359

=3.34

  1. With α= .05, what is tcrit?

Fromthe t-table tcrit at 0.05 significant level and 9 df = 1.833

  1. What should the researcher conclude about this relationship?

Consideringthat 3.34 is greater than 1.833, we reject the null hypothesisconcluding that there is enough sample evidence that the childrenexhibit more aggressive acts after watching a violent television show

  1. Compute the appropriate confidence interval.

Fromthe t- table, with alpha = 0.05 and df = 9 tcrit = 2.262

Confidenceinterval = (SD)(-tcrit)+ Dbar &lt= µD&lt= (SD)(+tcrit)+ Dbar

(0.359)(-2.262)+1.2&lt= µD&lt=(0.359)(2.262)+1.2

0.3879&lt=µD&lt=2.0121

  1. How large is the effect of violence in terms of the difference it produces in aggression scores?

1.2/(1.289)1/2= 1.06 This demonstrates that the effect of changing the conditionswas to change scores by an amount that is slightly larger than onestandard deviation.

21.Anexperimenter investigated the effects of a sensitivity trainingcourse on a policeman’s effectiveness at resolving domesticdisputes (using independent samples who had or had not completed thecourse). The dependent variable was the ability to resolve a domesticdispute. These success scores were obtained:

NoCourse Course

11 13

14 16

10 14

12 17

811

15 14

12 15

13 18

912

11 11

  1. Should a one-tailed or a two-tailed test be used?

Atwo-tailed test should be used.

  1. What are H0 and Ha?

Ho: Means of the two courses are same

Ha: Means of the two courses are not the same

  1. Subtracting course from no course, compute tobt and determine whether it is significant.

Fromthe data, we can calculate the mean, Standad Vaiance and the Samplesize

Meanof no course = 11.5, standard variance = 4.72, sample size = 10

Meanof course = 14.1,standard variance = 5.88, sample size = 10

Pooledvariance = (10-1)4.72 + (10-1)5.88 / 18 = 5.29

Standarderror = (5.29*0.2)1/2= 1.03

Tobt= (11.5 – 14.1)/1.03 = – 2.53

Degreesof freedom = 18

Fromtables at 5% level the critical value = +/- 2.101.

Theresults are significant since tobt is larger than tcrit

  1. Compute the confidence interval for the difference between the µs.

(Sx1– x2)(-tcrit)+ (X1bar – X2bar) &lt= mh– mc&lt= (Sx1– x2)(+tcrit)+ (X1bar – X2bar)

(1.03)(-2.101)+(-2.6)&lt= µ1µ2&lt=(1.03)(2.101)+(-2.6)

-4.76&lt=µ1µ2&lt=-0.44

(e)Whatconclusions can the experimenter draw from these results?

Policewho undergoes through the course become more successful at solvingdisputes than police officer who do not undergo through the course

  1. Using our two approaches, compute the effect size and interpret it.

D= (X1bar – X2bar)/sqr S2p

D= (11.5 – 14.1)/5.31/2= 1.13

Effectsize = 2.532/2.532+18 = 0.26

Thisimplies that taking the course is very essential

Reference

Heiman,G. (1992).&nbspBasicstatistics for the behavioral sciences.Boston: Houghton Mifflin.

Close Menu