PSYC 60 Quiz 3

Suppose you want to know if the population of stressed children has more behavior problems than the population of typical children. The Youth-Self Report form measures behavior problems in which μ = 50 and σ = 10. Suppose we sample 5 high stressed children (x̅ = 56, s = 7.65) and we use α .01 for a two-tailed test. Calculate the Confidence Interval (CI) and provide a conclusion.

The critical values for a two-tailed test using α .01 and N-1 (4) df are ±4.60. CI = x̅ ± t subscript df(s /√N) CI(.99) = 56 ± 4.60(7.65 /√5) CI(.99) = 40.26 ≤ μ ≤ 71.74 Conclusion: There is a 99% probability that the interval 40.26 to 71.74 includes the true μ for behavior scores for stressed children.

Suppose you want to know if the population of stressed children has more behavior problems than the population of typical children. The Youth-Self Report form measures behavior problems in which μ = 50 and σ = 10. Suppose we sample 5 high stressed children (x̅ = 56, s = 7.65) and we use α .05 for a two-tailed test. Calculate the Confidence Interval (CI) and provide a conclusion.

The critical values for a two-tailed test using α .05 and N-1 (4) df are ±2.78. CI = x̅ ± t subscript df(s /√N) CI(.95) = 56 ± 2.78(7.65 /√5) CI(.95) = 46.49 ≤ μ ≤ 65.51 Conclusion: There is a 95% probability that the interval 49.49 to 65.51 includes the true μ for behavior scores for stressed children.

The notation for the null and alternative hypotheses are same for ________________________________________________.

independent samples and related samples

t tests assume...

normality in the population from which the sample was drawn (regular bell-shaped curve).

Independent-samples t tests use... (df)

n₁+n₂-2 or N-2

Disadvantages of related samples:

order effect and carry-over effect

In a related-samples t test, N is...

the number of pairs, not the number of scores.

Sample size is __, not the __.

the number of scores in each sample, not the number of samples

Related samples, repeated measures, and matched samples are...

the same thing.

Confidence Interval (CI) formula for related samples:

CI = d-bar ± t subscript df(s subscript D ÷ √N) (t subscript df = the critical values)

What do one-sample t tests ask?

"Is there a difference between a sample mean and a population mean?"

Guidelines for interpreting d^:

< .2 = trivial effect size ≥ .2 and < .5 = small effect size ≥ .5 and < .8 = medium effect size ≥ .8 = large effect size

Two assumptions for an independent-samples t test:

1. Homogeneity of variance (the situation in which two or more populations have equal variances) 2. The samples come from populations with normal distributions

The first advantage of related samples:

1. It avoids the problem of person-to-person variability (Suppose therapy improved everyone's weight by 10 pounds. If many individuals' pre-therapy weight was extremely low, we may not notice the gain of 10 pounds because the post-therapy weight would still be relatively low. Unless we have a pre-therapy score to compare to the post-therapy score, we wouldn't know the therapy is working. We get around this problem with related samples because we're using difference scores and not just post-treatment scores.)

The first factor that affects whether the t value will be statistically significant:

1. The difference between x̅ and μ (One-sample t-tests ask, "Is there a difference between the sample mean and the population mean?" Well, the further they are apart, i.e., the bigger the difference, the more likely you'll have statistical significance. You can also think about it this way: The bigger the numerator in the t-test formula, the bigger the overall fraction, and the bigger the t, the more likely it is to beat the critical value.)

What makes a CI smaller?

1. larger α (α .05 has a smaller CI than α .01 because a 95% chance that the CI has captured μ is less than 99%) 2. smaller s (the less spread out the scores are, the easier it is to know exactly where μ is) 3. larger N (A larger sample size or lower variability will result in a tighter confidence interval with a smaller margin of error. A smaller sample size or a higher variability will result in a wider confidence interval with a larger margin of error.)

What makes a CI wider?

1. smaller α (α .01 has a wider CI than α .05 because a 99% chance that the CI has captured μ is bigger than 95%) 2. larger s (the more spread out the scores are, the harder it is to know exactly where μ is) 3. smaller N (A smaller sample size or a higher variability will result in a wider confidence interval with a larger margin of error. A larger sample size or lower variability will result in a tighter confidence interval with a smaller margin of error.)

Seventeen anorexic girls have undergone family therapy, and the girls' weight before and after treatment were recorded. mean weight before: x̅ = 83.23 s = 5.02 mean weight after: x̅ = 90.49 s = 8.48 s subcript D = the standard deviation of the 'difference scores' = 7.16 What is N?

17 (the number of PAIRS of scores, not the total number of scores)

The second advantage of related samples:

2. It controls for extraneous variables (If we measured one group of anorexic girls before therapy and a different group of girls after therapy, there could be several reasons why we may or may not see a weight difference between the two groups that are not a function of the therapy. Perhaps the therapy is effective but the second group didn't gain weight because of a stressful event, or perhaps the therapy is ineffective but the second group gained weight anyway for some other reason. In either case, we may draw the wrong conclusion about the effectiveness of therapy. Again, related samples get around this because they use difference scores.)

The second factor that affects whether the t value will be statistically significant:

2. N, sample size (The bigger the sample size, the more likely you'll have statistical significance. That's why we trust larger sample sizes more. You can also think about it this way: On the table of critical values, as N increases, the critical values decrease, meaning they get easier to beat which increases your chance of having statistical significance.)

The third advantage of related samples:

3. It requires fewer participants than independent samples, which gives us more power (Every additional participant increases variability, so it's better to use just one person. Every additional participant requires extra time, effort, and money.)

The third factor that affects whether the t value will be statistically significant:

3. s, sample standard deviation (The bigger the s, the LESS likely you'll have statistical significance. One-sample t-tests ask, "Is there a difference between the sample mean and the population mean?" Well, the greater the standard deviation, the more variability in both the sample and population scores, which makes us less certain if there really is a statistically significant difference. Low variability is the most likely to be statistically significant because there is very little overlap and we are more sure that the sample and population are different. You can also think about it this way: If you increase s in the t-test formula, the denominator gets bigger. A big denominator makes the whole fraction smaller. A smaller t is less likely to beat the critical value.)

The fourth factor that affects whether the t value will be statistically significant:

4. α, alpha (The greater the alpha, i.e., the larger the rejection region, the more likely you'll have statistical significance. Using α .05 means you're more likely to reject H0 than α .01.)

The fifth factor that affects whether the t value will be statistically significant:

5. One-tailed vs two-tailed tests (A one-tailed test has a larger rejection region on one side of the distribution compared to a two-tailed test and no rejection region on the other side. For a one-sample t-test, if you predict x̅ to be larger than μ, i.e., to the right of μ in the distribution, and x̅ does indeed turn out to be larger, you'll have double the chance of statistical significance compared to a two-tailed test because you have twice the rejection region on the right side of the distribution compared to a two-tailed test. If you're wrong about the direction, you'll have zero chance of statistical significance because there is no rejection region on the left side of the distribution.)

Confidence Interval (CI) formula for independent-samples t tests:

CI = (x̅1 - x̅2) ± t subscript df • s subscript x̅1 - x̅2 (t subscript df = critical values)

The Central Limit Theorem states that:

Given a population with a mean μ and a variance σ^2, the sampling distribution of the mean will have a mean equal to μ and a variance equal to σ^2/N. The distribution will approach the normal distribution as N, the sample size, increases.

Suppose you want to know if the population of stressed children has more behavior problems than the population of typical children. The Youth-Self Report form measures behavior problems in which μ = 50 and σ = 10. What is the null hypothesis? The alternative hypothesis?

H0: μ = 50 H1: μ ≠ 50 Null hypothesis: There is no statistically significant difference in the population means of behavior problems between typical children and stressed children.

Seventeen anorexic girls have undergone family therapy, and the girls' weight before and after treatment were recorded. mean weight before: x̅ = 83.23 s = 5.02 mean weight after: x̅ = 90.49 s = 8.48 s subcript D = the standard deviation of the 'difference scores' = 7.16 What is the null hypothesis? Alternative hypothesis?

H0: μ1 = μ2 H1: μ1 ≠ μ2 Null hypothesis: There is no statistically significant difference in the population means of anorexic girls' weight before versus after family therapy.

Adams et al. (1996) predicted that homophobia may be related to anxiety about becoming homosexual. They administered the Index of Homophobia to 64 self-proclaimed heterosexual males, classed them as homophobic or non-homophobic, and then had them watch a video of heterosexual and homosexual erotic stimuli and recorded their level of sexual arousal. What is the null hypothesis? Alternative hypothesis?

H0: μ1 = μ2 H1: μ1 ≠ μ2 Null hypothesis: There is no statistically significant difference in the population means of sexual arousal to homosexual stimuli between homophobic and non-homophobic men.

When you don't have homogeneity of variance, you have...

Heterogeneity of variance (a situation in which samples are drawn from populations having different variances) ex. one leptokurtic distribution, one platokurtic distribution

What do related-samples t tests ask?

Is there a difference between two samples that are related to each other?

What do independent-samples t tests ask?

Is there a difference between two samples whose scores are not related to each other?

Related-samples t tests use... (df)

N - 1 df

One-sample t tests use... (df)

N- 1 df

Can effect sizes be negative?

No.

Understand why we calculate effect size.

Not all statistically significant differences are actually meaningful. Effect size helps us understand the actual magnitude, or strength, of the difference that we found. You wouldn't calculate effect size if you didn't have statistical significance on your t-test. There are different ways of calculating effect size, but for Ch. 12- 14, we only use Cohen's d.

Seventeen anorexic girls have undergone family therapy, and the girls' weight before and after treatment were recorded. mean weight before: x̅ = 83.23 s = 5.02 mean weight after: x̅ = 90.49 s = 8.48 s subcript D = the standard deviation of the 'difference scores' = 7.16 Calculate the CI for related samples using α .05 for a two-tailed test and provide a conclusion.

The critical values for a two-tailed test using α .05 and N-1 df are ±2.12. CI(.95) = 7.26 ± 2.12(7.16 ÷ √17) CI(.95) = 7.26 ± 3.68 CI(.95) = 3.58 ≤ μ ≤ 10.94 Conclusion: There is a 95% probability that the interval 3.58 to 10.94 lbs includes the population mean gain.

Adams et al. (1996) predicted that homophobia may be related to anxiety about becoming homosexual. They administered the Index of Homophobia to 64 self-proclaimed heterosexual males, classed them as homophobic or non-homophobic, and then had them watch a video of heterosexual and homosexual erotic stimuli and recorded their level of sexual arousal. Homophobic men: Non-homophobic men: x̅1 = 24 x̅2 = 16.5 s2 = 148.87 s2 = 139.16 n1 = 35 n2 = 29 Calculate the CI for independent samples using alpha .05 and a two-tailed test and provide a conclusion.

The critical values for alpha .05 and a two-tailed test with N-2 df are ±2.00. CI = (24 - 16.5) ± 2.00 • 3.018 CI = 7.5 ± 6.04 = 1.46, 13.54 CI(.95) = 1.46 ≤ μ ≤ 13.54 Conclusion: There is a 95% probability that the interval 1.46 to 13.54 includes the population mean difference in arousal to homosexual stimuli between homophobic and non-homophobic men.

Seventeen anorexic girls have undergone family therapy, and the girls' weight before and after treatment were recorded. mean weight before: x̅ = 83.23 s = 5.02 mean weight after: x̅ = 90.49 s = 8.48 s subcript D = the standard deviation of the 'difference scores' = 7.16 What is d-bar?

The difference of the means. 90.49 - 83.23 = 7.26

Suppose you draw 1,000 samples of 5 people from a population, calculate the mean of each sample, and then plot these means in a frequency distribution (i.e., create a sampling distribution of the mean). Now, suppose you draw 1,000 samples of 50 people from a population and create a sampling distribution of the mean. Which sampling distribution is wider?

The one where N = 5

Difference scores (gain scores):

The set of scores representing the difference between the subjects' performance on two occasions

Understand the effect of N, sample size.

There is a 50/50 chance of getting heads or tails when you toss a coin, but let's say you don't know that. You toss a coin 4 times--an unacceptably small sample size. It comes up as heads 3 times, so you conclude that there is a 75% chance of getting heads, but this is wrong. You need to increase your sample size. This is why we trust large sample sizes.

Pooled variance (formula):

a weighted average of separate sample variances

Matched samples:

an experimental design in which the same subject is observed under more than one treatment

Related samples:

an experimental design in which the same subject is observed under more than one treatment (ex. comparing students' two sets of quiz scores (DV) after they used two different studying strategies (IV))

Confidence Interval (CI):

an interval, with limits at either end, having a specified probability of including the parameter being estimated. CI = x̅ ± t subscript df(s /√N) (t subscript df = the critical values)

Calculate Cohen's d given the following: μ = 7,200 σ = 1,200 x = 6,000 s = 1,200 N = 16 alpha = .01, two tailed

d^ = -1 But we would interpret the absolute value of d^, which is 1, indicating a strong effect.

Suppose you want to know if the population of stressed children has more behavior problems than the population of typical children. The Youth-Self Report form measures behavior problems in which μ = 50 and σ = 10. Suppose we sample 5 high stressed children (x̅ = 56, s = 7.65) and we use α .05 for a two-tailed test. Calculate Cohen's d and provide a conclusion.

d^ = .78 Conclsuion: The sample of stressed children scored .78 standard deviations above the population mean, indicating a medium effect.

Seventeen anorexic girls have undergone family therapy, and the girls' weight before and after treatment were recorded. mean weight before: x̅ = 83.23 s = 5.02 mean weight after: x̅ = 90.49 s = 8.48 s subcript D = the standard deviation of the 'difference scores' = 7.16 Calculate the effect size (Cohen's d) for related samples and provide a conclusion.

d^ = 1.45 Conclusion: Girls gained an average of 1.45 standard deviations of pretreatment weight over the course of therapy, indicating a large effect.

Calculate related samples Cohen's d given the following: x̅1 = 7.6 x̅2 = 5.6 s1 = 1.14 s2 = 1.14 d-bar = 2 s subscript D = .71 N = 5

d^ = 1.75

Effect size (Cohen's d) for independent-samples t tests formula:

d^ = x̅1 - x̅2 / s subscript p

Related samples effect size (Cohen's d) formula:

d^ = x̅1 - x̅2 / s1 (The order of x̅1 and x̅2 doesn't matter--just know that you can't have a negative effect size.)

Repeated measures:

data in a study in which you have multiple measurements for the same participant (ex. measuring a person's cortisol level before and after a competition)

Confidence Intervals (CI) for related-samples t tests capture the _________________________.

population mean change (increase/decrease)

Confidence intervals for independent-samples t tests capture the _________________________________.

population mean difference

The homogeneity of variance assumption for independent-samples t tests is for ___________________________, not _______________________.

population variances; sample variances

Adams et al. (1996) predicted that homophobia may be related to anxiety about becoming homosexual. They administered the Index of Homophobia to 64 self-proclaimed heterosexual males, classed them as homophobic or non-homophobic, and then had them watch a video of heterosexual and homosexual erotic stimuli and recorded their level of sexual arousal. Homophobic men: Non-homophobic men: x̅1 = 24 x̅2 = 16.5 s2 = 148.87 s2 = 139.16 n1 = 35 n2 = 29 Calculate the effect size (Cohen's d) for an independent-samples t test and provide a conclusion. *Hint: Find the pooled variance, then in the denominator of the formula do the square root of the pooled variance.

s2p (pooled variance) = 144.48 d^ = 24 - 16.5 / √144.48 = .62 Conclusion: Homophobic men showed .62 standard deviations more arousal to homosexual stimuli than non-homophobic men, indicating a moderate effect.

Adams et al. (1996) predicted that homophobia may be related to anxiety about becoming homosexual. They administered the Index of Homophobia to 64 self-proclaimed heterosexual males, classed them as homophobic or non-homophobic, and then had them watch a video of heterosexual and homosexual erotic stimuli and recorded their level of sexual arousal. Homophobic men: Non-homophobic men: x̅1 = 24 x̅2 = 16.5 s1 = 148.87 s2 = 139.16 n1 = 35 n2 = 29 Calculate an independent samples t-test using alpha .05 for a two-tailed test, provide the result in APA format, and provide a conclusion.

s2p (pooled variance) = 144.48 t = 2.48, 62 df The critical values for a two-tailed test using alpha .05 and 62 df are about ± 2.00, so we would reject H0. Result in APA format: t(62) = 2.48, p < .05 Conclusion: The population mean sexual arousal of homophobic men to homosexual stimuli is a statistically significant higher value than for non-homophobic men.

Calculate an independent-samples t test using an alpha .05 two-tailed test given the following and provide the result in APA format: x̅1 = 53.60 x̅2 = 42.20 s1 = 11 s2 = 16 n1 = 5 n2 = 5

s2p (pooled variance) = 188.50 t = 1.31, 8 df The critical values are ± 2.31, so we do not reject H0. Result in APA format: t(8) = 1.31, p > .05

Calculate the effect size (Cohen's d) for independent-samples given the following: x̅1 = 53.60 x̅2 = 42.20 s2 = 5.50 s2 = 6.30 n1 = 5 n2 = 5

s2p (pooled variance) = 34.97 d^ = 53.6 - 42.2 / 5.914 = 1.93 Interpret the absolute value of d^, which is 1.93, indicating a strong effect.

Calculate an independent-samples t test using an alpha .05 two-tailed test given the following and provide the result in APA format: x̅1 = 53.60 x̅2 = 42.20 s1 = 5.50 s2 = 6.30 n1 = 5 n2 = 5

s2p (pooled variance) = 34.97 t = 3.05, 8 df The critical values are ± 2.31, so we reject H0. Result in APA format: t(8) = 3.05, p < .05

Suppose you want to know if the population of stressed children has more behavior problems than the population of typical children. The Youth-Self Report form measures behavior problems in which μ = 50 and σ = 10. Here, we are getting a ____________________________________ and seeing if it's different from the __________________________________________.

sample of stressed children; population of typical children

If your sample sizes are the same for an independent-samples t-test, then you can...

simply take the average of both variances, since it will be the same as your weighted average if you used the pooled variance formula.

As you increase N, σ^2 gets ___________ but μ ____________________.

smaller; stays the same

Conduct a one-sample t-test given the following and provide the result in APA format: μ = 7,200 σ = 1,200 x = 6,000 s = 1,200 N = 16 alpha = .01, two tailed

t = -4, 15 df The critical values for a two-tailed test using α .01 and 15 df are ± 2.947, so we can reject H0. Result in APA format: t(15) = -4, p < .01

Conduct a one-sample t-test given the following and provide the result in APA format: μ = 100 σ = 15 x = 108 s = 13.53 N = 5 alpha = .05, two tailed

t = 1.32, 4 df The critical values for a two-tailed test using α .05 and 4 df are ± 2.78, so we would not reject H0. Result in APA format: t(4) = 1.32, p > .05

Suppose you want to know if the population of stressed children has more behavior problems than the population of typical children. The Youth-Self Report form measures behavior problems in which μ = 50 and σ = 10. Suppose we sample 5 high stressed children (x̅ = 56, s = 7.65) and we use α .05 for a two-tailed test. Conduct a one-sample t-test, providing the result in APA format and the conclusion.

t = 1.754, 4 df The critical values for a two-tailed test using α .05 and 4 df are ±2.78. Our test stat does not fall in the rejection region, so we don't reject H0. Result in APA format: t(4) = 1.754, p > .05 Conclusion: There is no statistically significant difference in the population means of behavior problems between typical children and stressed children.

Calculate a related-samples t test using an α .01 two-tailed test given the following and provide the result in APA format: x̅1 = 7.6 x̅2 = 6.9 s1 = 1.14 s2 = 1.14 d-bar = .7 s subscript D = .71 N = 5

t = 2.20, 4 df The critical values = ±4.60, so we would not reject H0. Result in APA format: t(4) = 2.20, p > .01

Seventeen anorexic girls have undergone family therapy, and the girls' weight before and after treatment were recorded. mean weight before: x̅ = 83.23 s = 5.02 mean weight after: x̅ = 90.49 s = 8.48 s subcript D = the standard deviation of the 'difference scores' = 7.16 Calculate a related-samples t-test using α .05 for a two-tailed test, provide the result in APA format, and provide a conclusion.

t = 4.18, 16 df The critical values for a two-tailed test using α .05 and 16 df are ± 2.12, so we reject H0. Result in APA format: t(16) = 4.18, p < .05 Conclusion: There is a statistically significant difference in the population means of anorexic girls' weight before vs. after family therapy, such that girls gained weight during the course of therapy.

Calculate a related-samples t test using an α .01 two-tailed test given the following and provide the result in APA format: x̅1 = 7.6 x̅2 = 5.6 s1 = 1.14 s2 = 1.14 d-bar = 2 s subscript D = .71 N = 5

t = 6.29, 4 df The critical values = ±4.60, so we would reject H0. Result in APA format: t(4) = 6.29, p < .01

Related-samples t test formula:

t = d-bar / s subscript D ÷ √N

One-sample t test formula:

t = x̅ - μ / s ÷√N

As you increase N, the distribution becomes ___________ and ___________ until it passes normality (becomes leptokurtic).

taller; skinnier

Effect size (Cohen's d; effect size in terms of standard deviations)

the difference between two populations divided by the standard deviation of either population d^ = x̅ - μ / s

d-bar:

the difference of the sample means

Sampling distribution of difference between means:

the distribution of the differences between means over repeated sampling from the same populations (Two independent populations. Take a sample from each population, calculate the mean of those samples, and the difference between those means = one difference of mean score. Do this over and over again, plot all of the difference of mean scores, and you get a sampling distribution of the difference between means. The standard deviation of that sampling distribution is s subscript x̅1 - x̅2.)

Carry-over effect:

the effect of previous trials (conditions) on a subject's performance on subsequent trials (ex. You want to know if Drug A or Drug B improves depression more. You give Drug A for a time, then measure symptoms and see no improvement. Then you give Drug B and see much improvement. We don't know if there is a carry-over effect--does Drug A have a slow-acting effect? Or is it Drug B that is the effective one? You solve this problem by altering the order of treatment, i.e., counterbalancing.)

Order effect:

the effect on performance attributable to the order in which treatments were administered (ex. You want to know if Stimulant A or B improves reaction time more. You administer A to one group and see lots of improvement in reaction time. Then you administer B to them and see no improvement. You administer B to a different group and see lots of improvement in reaction time. Then you administer A to them and see no improvement. Both stimulants work, but the first one works better in each case. If you didn't alter the order, you may falsely conclude that whichever stimulant you use first was better.)

The distribution will approach the normal distribution as N, the sample size, increases, regardless of...

the shape of the the population distribution (whether it's skewed, bimodal, negative, etc.)

s subscript D:

the standard deviation of the 'difference scores'

Standard error of difference between means:

the standard deviation of the sampling distribution of the differences between means (Two independent populations. Take a sample from each population, calculate the mean of those samples, and the difference between those means = one difference of mean score. Do this over and over again, plot all of the difference of mean scores, and you get a sampling distribution of the difference between means. The standard deviation of that sampling distribution is s subscript x̅1 - x̅2.)

s subscript x̅1 - x̅2:

the standard error of difference between means. is also the same as the denominator of the independent-samples t-test formula. (Two independent populations. Take a sample from each population, calculate the mean of those samples, and the difference between those means = one difference of mean score. Do this over and over again, plot all of the difference of mean scores, and you get a sampling distribution of the difference between means. The standard deviation of that sampling distribution is s subscript x̅1 - x̅2.)

Independent-samples t tests are used to compare __________________ whose scores are ________________.

two samples; not related (ex. comparing male and female grades on a language verbal test, comparing testosterone levels of men in their 20s vs. men in their 60s)

If your sample sizes are not the same for an independent-samples t test, then you would...

use the pooled variance formula to calculate the weighted average of both variances.

Independent-samples t test formula:

x̅1 - x̅2 / √[(pooled variance/n1) + (pooled variance/n2)]

Which CI is wider: the one calculated using α .01 or α .05?

α .01 (a 99% chance that the CI has captured μ is bigger than 95%)

Confidence Intervals (CI) for one-sample t tests capture ___.

μ.