Weeks 7 and 8 Flashcards
If you find p < .05 for Levene’s test it means you have “failed” the test.
a) true
b) false
a) True
Explanation … The null hypothesis of Levene’s test is that the populations from which your samples are drawn have the same variance. If Levene’s test ends with a value of p < .05 this means that the chance that this hypothesis is true is less than 5%. We should therefore conclude that the populations have different variances and that the treatment somehow affect the variance. This is bad for a t-test or z-test. We have therefore failed this auxiliary test.
For a 2-sample design, you don’t need to pass Levene’s test if both samples have n > 30.
a) true
b) false
b) false
Explanation … The central limit theorem deals with the nature of the sampling distribution of the mean. It says that if the sample sizes are large enough (n>30 for both samples), the sampling distribution of the mean will be Normal even if the populations from which the samples are taken are not Normal. In contrast, Levene’s test doesn’t deal with sampling distributions and doesn’t depend on sample size at all. Levene’s test discerns whether or not the samples in a design could have come from populations with the same variance. You always need to conduct Levene’s test, even if the sample sizes are large.
3) What is tested by Levene’s test?
a) whether 2 samples have the same variance
b) whether 2 samples have the same skew and kurtosis
c) whether 2 samples come from populations with the same variance
d) whether 2 samples come from the same population
c) whether 2 samples come from populations with the same variance
Explanation … An assumption of independent-samples t-tests is that the treatment affects the mean and not the variance. If true, this means that even if 2 samples may not have the same variance, the populations from which they are taken do have the same variance. This is what Levene’s test assesses.
What is the danger of proceeding with an independent-samples t-test if you have failed Levene’s test.
a) You could lose control over the Type I error rate
b) The difference in means could become non-zero when it is supposed to be zero
c) The sample variances might be different
d) You might lose degrees of freedom
a) You could lose control over the Type I error rate
Explanation … The problem with failing Levene’s test is that not only are the sample variances different, but that they are so different we must assume they come from populations with different variances. In other words … the treatment doesn’t just affect the population mean, it also affects the population variance. This could result in a t-test yielding a significant t-value even when the null hypothesis that M1 = M2 is true, which would be a Type I error. The hazard of proceeding after failing Levene’s test is therefore to risk an elevated Type I error rate.
In statistics you pool variances, not standard deviations.
a) true
b) false
a) True
Explanation … In statistics, it is variances that add together, not standard deviations. This is an instance of the variance sum law and is why ANOVA is called analysis of variance and not analysis of standard deviations. This all means that if you have 2 sample standard deviations, the first thing you have to do before pooling them is to square them so as to turn them into variances. Only after that is accomplished can you can pool the variances.
It makes no sense to pool variances if you have failed Levene’s test.
a) true
b) false
a) True
Explanation … You pool variances if each sample variance in a design can legitimately serve as an estimate of a single overall population variance. If you have failed Levene’s test, however, then you can assume that the treatment affects the variance and that the variances of the different samples must estimate different population variances. There is no sense in pooling the variances in these circumstances.
In a balanced design, pooling is the same as averaging.
a) true
b) false
a) True
Explanation … When you pool variances you first weight each variance and then add together all the weighted variances to find the pooled variance. In statistics, the weight associated with a variance from a particular sample is the proportion of overall observations in the design that are in that particular sample. In a 2-sample design, if half of the observations are in the first sample and the other half of the observations are in the other sample, (i.e., a balanced design) then the weights are ½ and ½. But this is exactly the same as averaging. So, for balanced designs, pooling variances is the same as averaging them. For unbalanced designs, pooling is not the same as averaging because the pooling procedure will pay more attention to the variance from the larger sample.
8) If you pool together 2 variances, the resulting pooled variance is larger than either of the 2 variances you started with.
a) true
b) false
b) False
Explanation … When you pool variances, the result is a variance that lies somewhere in value between the 2 variances you started out with. In this sense, pooling is like averaging … when you average two numbers you end up with a value somewhere between your starting points. This is a good tip if you ever have to calculate a pooled variance. The end result should end up somewhere between the 2 variances you started with. If that isn’t true then you did not do your math correctly.
Suppose that in a 2-sample design (where you have passed Levene’s test) the samples have variances s_1^2=8, and s_2^2=10. Calculate the pooled variance if the design is balanced.
a) s_pooled^2=5
b) s_pooled^2=8
c) s_pooled^2=9
d) s_pooled^2=164
c) s_pooled^2=9
Explanation … The calculation is simple when you have a balanced design because then pooling variances is the same as averaging them. For this problem, the pooled variance is the average of 8 and 10, which is 9.
Suppose that in a 2-sample design (where you have passed Levene’s test) the samples have variances s_1^2=8, and s_2^2=10. Calculate the pooled variance if the sample sizes are n1 = 11 and n2 = 31.
a) s_pooled^2=8.5
b) s_pooled^2=9
c) s_pooled^2=9.5
a) s_pooled^2=9.5
Explanation … In an unbalanced design the weights used for pooling variances come from the sample sizes. Specifically, the weight to be used for a particular sample variance is equal to the proportion of the overall degrees of freedom associated with that sample. In this problem, the independent-samples design has N = n1 + n2 = 11 + 31 = 42 observations and so there are df = N – 2 = 40 degrees of freedom overall. 10 of these degrees of freedom occur in the first sample and 30 appear in the second sample. The weights for the variances are therefore ¼ and ¾. Using these weight to pool the variances gives s_pooled^2= 1/4×8+3/4×10=38/4 = 9.5
What is the value of spooled in the previous question?
a) s_pooled=√8.5
b) s_pooled=√9
c) s_pooled=√9.5
c) s_pooled=√9.5
Explanation … This may seem like too easy a question but many students get this step wrong either in an exam or when making calculations for an assignment. The pooled standard deviation, a term that appears in the bottom of the formula for a t-test, is nothing more than the square root of the pooled variance.
Suppose that in a 2-sample design (where you have passed Levene’s test) the samples have standard deviations s_1=2, and s_2=4. Calculate the spooled if the sample sizes are n1 = 11 and n2 = 21.
a) s_pooled=4
b) s_pooled= √4
c) s_pooled= 13
d) s_pooled= √13
d) s_pooled= √13
Explanation … This question combined a number of the features of the previous questions with the extra twist that you begin and end with standard deviations, not variances. This is a very common occurrence is such calculations. Overall, the sequence of calculations here will be 1) square the sample standard deviations to get the 2 variances, 2) find the weights needed for pooling, 3) sum the weighted variances to get the pooled variance, 4) take the square root of the pooled variance to get the pooled standard deviation, spooled.
Here are the steps …
- s_1^2= (s_1 )^2= 2^2=4 , and s_2^2=(s_2 )^2= 4^2=16
- As calculated in in a previous question the weights are w1 = ¼ and w2 = ¾ (because ¼ of the overall degrees of freedom occur in the first sample and ¾ of the degrees of freedom occur in the second sample)
- Summing the weighted variances gives s_pooled^2= 1/4×4+3/4×16=52/4 = 13
- Taking the square root yields s_pooled= √13
If Cohen’s effect size for a treatment is “large” then the corresponding t-test will have p < .05.
a) true
b) false
b) False
Explanation … In general it is easier to find significance for a large effect than for a small one, but the probability of finding significance depends on how much power a test has. Even a large effect may not be significant if you don’t have enough power in your test. Thus, effect size and significance in a test are different things.
What standard deviation do you use if you fail to pass Levene’s test in an independent-samples design
a) spooled
b) sX
c) the standard deviation of the control sample
c) the standard deviation of the control sample
Explanation … If you fail Levene’s test then it makes no sense to pool variances and spooled also makes no sense. In that case a common way to construct Cohen’s effect size is to divide the difference in means by the standard deviation of the control sample. This form of Cohen’s d can be interpreted as telling you how much the treatment changes the mean against a background of baseline variability.
Suppose that researchers design a questionnaire to evaluate mood. The questionnaire is standardized so that participants in a neutral mood score with M = 50 and sX = 15. In a test of how surroundings affect mood, a researcher gives the test to N = 35 participants in a bare room with a single metal desk and no windows. The mean in the sample is M = 42.5. Calculate Cohen’s effect size for the effect of surroundings on mood.
a) d = 0.1
b) d = 0.5
c) d = 1.0
d) d = 2.0
b) d=0.5
Explanation …. In this design there is only a single sample involved and the true population mean and standard deviation is assumed to be known. If we were testing a hypothesis about the mean we would use a single-sample z-test. Cohen’s effect size in this case expresses the deviation of the sample mean from the expected population mean in standardized units. To standardize, we will divide by the population standard deviation. Thus, we get d = (50 – 42.5) / 15 = 7.5/15 = 0.5. The sample size of N = 35 does not come into the calculations – it was just thrown in there to confuse you – the whole point about effect sizes is that they don’t depend on arbitrary choices investigators make such as how many participants to recruit.
