PSY 845: Module 4 Problem Set SolutionsProblem 1:
You have information on one sample and want to compare it to the population to see if it is significantly different from the population mean. In each of the following situations, would it be appropriate to use the t-test OR the z-test?
You know the population mean and variance
You do not know the population mean and variance
Solution:
z-test
one sample t-test
Note: If the sample size is less than 30, even though you know the population mean and variance, it is better to use the t-test because that corrects for the extra variation you will find in smaller sample sizes.Problem 2:
Unlike the z-test that uses information about the populations variance or SD, the formula for the t-statistic uses an estimate of the populations variance or standard deviation, which is called the ______________________________ which is computed from the ____________________.
Solution:
estimated standard error of the mean (sM) which is computed from the sample variance or standard deviation.
The estimated standard error (sM) is used as an estimate of the real standard error when the value of rho is unknown. It is computed from the sample variance or sample standard deviation and provides an estimate of the standard distance between a sample mean, M, and the population mean, .
Problem 3:
a. A researcher wants to compare typing speed (the DV) in two conditions of light (IV: low, high level of light). In order to use a t-test for independent samples, the design had to have participants that were __________________________________________________.
b. To perform a t-test where we are comparing two independent samples (groups), we need three pieces of information. These are:
________________________________________________________
________________________________________________________
________________________________________________________
Solution:
The participants had to be assigned randomly to only one of the two conditions. They were either in the low light or in the high light condition of the IV.
We would need to know each samples mean, each samples SD, each samples n.
Problem 4:
The degrees of freedom (df) for each sample that is being compared using the t-test is computed as __________.
Solution:
n 1, number in sample minus 1. Thus, df is related to the sample size. The larger the sample, the larger the df.
Problem 5:
When we did the z-test, we computed the z-value and then looked at where it would fall in the distribution of all possible z-scores. Similarly, when we do a t-test, we compute the t-value and then look at where that t-value would fall in the distribution of _______________________________________. The critical values of t can be found in Table ___ in the appendix of the textbook.
Solution:
The distribution of all possible t-values that could be computed for a given df. Table B.2.
Problem 6:
a. A nondirectional hypothesis predicts there will be a significant difference between the two means that can be ___________________________________ . A directional hypothesis predicts there will be a significant difference between the two means and adds _____________________________.
A nondirectional hypothesis uses a ___-tailed test, while a directional hypothesis uses a ___-tailed test to evaluate if the difference between the means is significant.
When I set alpha at 5%, and have a nondirectional hypothesis, the 5% of outcomes that occur less frequently than my critical t-value
Fall in one tail of the distribution OR
are divided between the two tails (2 % in one tail and 2 % in the other tail)?
Solution:
Nondirectional: Mean A will be significantly larger than Mean B OR Mean B will be significantly larger than Mean A.
Directional: Limits the prediction to one type of difference: either Mean A will be significantly larger than mean B OR Mean Be will be significantly larger than Mean A.
A nondirectional hypothesis uses a two-tailed test, while a directional hypothesis uses a one-tailed test to evaluate if the difference between the means is significant.
When I set alpha at 5%, and have a nondirectional hypothesis, the 5% of outcomes that occur less frequently than my critical t-value
Fall in one tail of the distribution OR
Are divided between the two tails (2 % in one tail and 2 % in the other tail)? This is because you allowed the outcome to go in either direction, A>B or B>A
Problem 7:
Go to Table B.2 in the appendix of your textbook, and decide whether the following t-value that has been computed is significant (that is, its value is equal to or greater than the t-value beyond which only 5% or less of the proportion of all possible t-values would occur), given the df, the type of hypothesis (directional/nondirectional), the alpha level, the required (critical) t-value, and the actual t-value that was computed to compare the means of Group A with Group B.
Hypothesis: There will be a significant difference between the means of Group A and Group B; df = 25; computed t-value = 2.35.
Is this a directional or nondirectional hypothesis?____________
Given the type of hypothesis, the proportion of extreme values that I am interested in will be found in one tail or in two tails combined?
Given that the df = 25 and alpha is .05, what would be the value of t that my computed value would have to equal or exceed?
Is my computed value significant at this alpha level?
Answers to a:
Hypothesis: There will be a significant difference between the means of Group A and Group B; specifically, the mean of Group B will be significantly greater than the mean of Group A; df = 25; computed t-value = 2.35.
Is this a directional or nondirectional hypothesis?____________
Given the type of hypothesis, the proportion of extreme values that I am interested in will be found in one tail or in two tails combined?
Given that the df = 25 and alpha is .05, what would be the value of t that my computed value would have to equal or exceed?
Is my computed value significant at this alpha level?
Solution:
For Part a:
Is this a directional or nondirectional hypothesis? Nondirectional because I didnt specify if I expect A to be larger than B (A>B) or B>A.
Given the type of hypothesis, the proportion of extreme values that I am interested in will be found in one tail or in two tails combined? I didnt predict whether it would be in the right tail (B>A) or in the left tail (B
<a). so,="" i="" will="" take="" either="" outcome="" (2.5%="" in="" one="" tail="" and="" 2.5%="" the="" other="" tail).="" given="" that="" df="25" alpha="" is="" .05,="" what="" would="" be="" value="" of="" t="" my="" computed="" have="" to="" equal="" or="" exceed?="" using="" table="" b.1,first="" go="" down="" first="" column="" (df)="" find="" 25.="" gives="" me="" row="" look="" in.="" now,="" am="" looking="" for="" required (the="" critical="" t)="" has="" larger="" than.="" need="" under="" headed="" proportion="" two="" tails="" combined="" then="" 0.05="" (my="" level).="" going="" correct="" df,="" 2.06.="" significant="" at="" this="" level?="" 2.35,="" which="" than="" thus,="" a="" probability="" occurring="" less="" 5%="" time:="" (t(25)="2.35," two-tailed,="" p=""
A). So, I will only consider the 5% of all possible outcomes that fall in that one tail of the distribution.

Given that the df = 25 and alpha is .05, what would be the value of t that my computed value would have to equal or exceed? Using Table B.1, first I go down the first column (df) and find 25. That gives me the row to look in. Now, I am looking for the required value of t (the critical value of t) that my computed value of t has to equal or be larger than.

I need to look under the column headed Proportion in One Tail and then 0.05 (my alpha level). Going down that column to the correct row for the df, I find the critical value of t is 1.708.Is my computed value significant at this alpha level? My computed value of t is 2.35, which is larger than 1.708. Thus, my computed value has a probability of occurring less than 5% of the time:

(t(25)=2.35, one-tailed, p < .05). I can reject the null hypothesis and say that the means of Groups A and B are significantly different and, specifically, Group Bs mean is significantly larger than Group As mean.What if it had been Group A that was significantly larger than Group B? Well, you cant count that as a significant finding because that is not what you predicted. Problem 8: (Problem 23 on p. 349 of textbook) Research has shown that people are more likely to show dishonest and self-interested behaviors in darkness than in a well-lit environment (Zhong, Bohns, & Gino, 2010). In one experiment, participants were given a set of 20 puzzles and were paid $0.50 for each one solved in a 5-minute period. However, the participants reported their own performance and there was no obvious method for checking their honesty. Thus, the task provided a clear opportunity to cheat and receive undeserved money. One group of participants was tested in a room with dimmed lighting and a second group was tested in a well-lit room. The reported number of solved puzzles was recorded for each individual. The following data represent results similar to those obtained in the study. Group A Well-Lit Room Group B Dimly Lit Room (Is there a treatment effect of lighting?) 7 9 8 11 10 13 6 10 8 11 5 9 7 15 12 14 5 10 Is there a significant difference in reported performance between the two conditions? Use two-tailed test with alpha set at p < .01. Compute Cohens d to estimate the size of the treatment effect. After you have computed Cohens d, how would you interpret it? Solution: To compute the t-value, we first need three pieces of information for each sample: n Group A: n = 9 Group B: n = 9 Mean Group A: M = 7.55 Group B: M = 11.33 (Mean = sum of the values in the group/n in the group) SS Group A: SS = 42.22 Group B: SS = 38 To compute SS: SS = sum of the squares of the deviations Subtract the groups mean from each of the scores in the group (the difference is called a deviation) Square each of the deviations Add up the squared deviations [(X-MX)2Group A (A-MA) (A-MA)2 Group B (B-MB) (B-MB)2 7 7-7.55 = -.55 -.552= .3025 9 -2.33 5.4289 8 8-7.55 = .45 .2025 11 -.33 .1089 10 10-7.55 = 2.45 6.0025 13 1.67 2.7889 6 67.55 = -1.55 2.4025 10 -1.33 1.7689 8 87.55 = .45 .2025 11 -.33 .1089 5 5-7.55 = -2.55 6.5025 9 -2.33 5.4289 7 77.55 = -.55 .3025 15 3.67 13.4689 12 12-7.55 = 4.45 19.8025 14 2.67 7.1289 5 5-7.55 = -2.55 6.5025 10 -1.33 1.7689 Sum A = 68 SSA = 42.22 Sum B = 102 SSB = 38.00 n = 9 n = 9 MA = 68 MB = 102 9 = 7.55 9 = 11.33 SS Next, compute sample variance, s2 = df, for each sample df = n 1; 9 1 = 8 (same for both groups)42.22 38 Group A: sA2 = 8 = 5.28 ; Group B: sB2 = 8 = 4.75 Next, compute the estimated standard error of each group: sM = s2/nGroup A: sM = 5.28/9 Group B: sM = 4.79/9= .766 = .729Next, compute the combined standard error for differences between the means of the the two groups:sA2 sB2 S(MA-MB) = nA + nB = = 1.06 Then, we have all the necessary pieces to compute the t-value:MA MB 7.55 11.33 t = S(MA-MB) = 1.06 = 3.57 (Disregard whether it is a + or -)However, the minus sign tells us that the second mean (B) was larger than the first mean (A) for directionality, if we have predicted direction.So, t(16) = 3.57.Now, go to Table B.2 and look up the critical value of t when df = 16, you have a nondirectional hypothesis, and your alpha is set at .01. The critical value is 2.92.Our computed value of 3.57 is larger than the critical value and we can reject the null hypothesis (that there is no effect of room lighting on the behavior).t(16) = 3.57, two-tailed, p < .01. b. MA MB SSA + SSB 42.22+38 80.22 Cohens d = pooled variance Pooled variance, sp2 = dfA + dfB = 8 + 8 = 16 = 5.017.55 11.33 3.78 d = 5.01 = 2.24 = 1.69