Psych Stats Exam #2

p ≤ α

we reject the null hypothesis and say the results are significant

Assume two independent samples (Sample A and Sample B), with Mean of Sample A = 10.81 and Mean of Sample B = 13.14. Sample A has a variance (s2) of 1.85, and Sample B has a variance of 3.48. Each sample has N = 11. Calculate tobt. Round to 2 decimal places.

((10.81 - 13.14))/√((1.85/11) + (3.48/11)) (-2.33)/√(5.33/11) (-2.33)/0.69609299 tobt = -3.3472539 tobt = -3.35

What are the TWO factors that determine the critical value of z (zcrit)?

(1) α level (2) One-tailed or two-tailed test

Eric Cartman wants to determine if hippies and non-hippies differ in the amount of showers they take per week. He randomly samples 10 hippies and 10 non-hippies from South Park and records the number of showers they take in one week. The mean for hippies is 4.5 showers, while the mean for non-hippies is 5.9 showers. s2hippies = 2.22 and s2non-hippies = 1.67. Construct the 95% confidence interval of the difference between population means. Round the final answer to 2 decimal places. What would that concluded about the H0?

(5.9-4.5) ± sqrt[(2.22/10) + (1.67/10)]*2.101 (1.4) - sqrt(3.89/10)*2.101 = 0.08960915 = 0.09 (1.4) + sqrt(3.89/10)*2.101 = 2.71039085 = 2.71 (NOTE: I reveresed the group means when substracting the mean differences in order to get positive CI values. However, if you had gotten -2.71, -.09, that would also be correct). The confidence interval does not contain 0, and so we can reject H0.

definition of p-value?

- probability of obtaining a given score or more extreme score by chance alone

Suppose you are interested if individuals who regularly watch The Daily Show differ significantly from individuals who watch regular news networks on their knowledge of current events. You randomly sample 10 Daily Show viewers and 10 regular news network viewers and administer a questionnaire assessing knowledge of events from the past week (scored 0-10). Use the following information to construct the 95% confidence interval for the difference between population means. Round the final answer to 2 decimal places. Xobt = Daily Show = 8.45 Xobt regular news = 7.60 s2Daily Show = 1.73 s2regular news = 1.65

-0.37-2.07 95% CI = (X1-X2) ± (sx1-sx2)(tcrit) sx1-x2 = sqrt ((1.73/10 + 1.65/10)) sx1-x2 = sqrt ((3.38/10) sx1-x2 = 0.58137767 tcrit = 2.101 (8.45-7.60) - (0.58137767(2.101) = (0.85) - (0.58137767)(2.101) = -0.3714745 = -0.37 (8.45-7.60) + (0.58137767)(2.101) = (0.85) + (0.58137767(2.101) = 2.07147448 =2.07

When a=0.05, the critical z-value for α1 tailed is equal to ___________, while the critical z-value for α2 tailed is equal to ________.

-1.645 or 1.645; ±1.96 If we are doing a one-tailed test, and α = 0.05, then the critical z-value corresponds to the z-score beyond which 5% of scores fall in either tail. If we are testing if the sample mean is less than the population mean, then the critical z-value is -1.645. If we are testing if the sample mean is greater than the population mean, then the critical value is +1.645. If we are doing a two-tailed test, and α = 0.05, then the critical value corresponds to the z-score beyond which 5% of scores fall in both tails (2.5% in each). Since we are testing in both tails, the critical z-value is ±1.96

A clinical psychologist conducts an experiment on 25 patients with schizophrenia to determine if a new treatment affects the amount of time these patients need to stay institutionalized. The results show that under the new treatment, the 25 patients stayed a mean duration of 78 weeks, with a standard deviation of 20 weeks. Previously collected data on a large population of patients with schizophrenia showed a mean of 85 weeks with an unknown standard deviation. Suppose α = 0.051 tail. If you were specifically testing that your sample mean was lower than the population mean, what would be the value of tcrit?

-1.711 The df are the same (25 -1 = 24) However, we now need to use the .05 significance level for one-tailed test column Since we are testing only in the left extreme side of the distribution, we just have one critical value, which is -1.711 Notice that the absolute value for tcrit for a one-tailed test is lower than tcrit for a two-tailed test. This is because we do not have to divide α by 2, as we do with a two-tailed test, and thus our critical region is larger.

A clinical psychologist conducts an experiment on 25 patients with schizophrenia to determine if a new treatment affects the amount of time these patients need to stay institutionalized. The results show that under the new treatment, the 25 patients stayed a mean duration of 78 weeks, with a standard deviation of 20 weeks. Previously collected data on a large population of patients with schizophrenia showed a mean of 85 weeks with an unknown standard deviation. Use α = 0.052 tail. Calculate tobt. Round to 2 decimal places.

-1.75 sx = 20/sqrt(25) = 20/5 = 4 tobt = (78-85)/(4) = -7/4 = -1.75

Suppose I want to determine if a certain class texts significantly more in class compared to the national average. I observe the class and record how many texts each student sends during class. With N = 75, I calculate a sample mean = 13. I look up the national average, which is μ = 11, with σ = 3.25. I will use a z-test to determine if the class texts significantly more in class than the population of college students. I will use a z-test with α = 0.052 tail. Calculate the standard error of the mean. Round to 4 decimal places.

.3753 Standard error of the mean = Population SD/sqrt(N) = 3.25/sqrt(75) = 3.25/8.66025404 = .37527767 = .3753

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. The following data are obtained: SSbetween = 26.13 SSwithin = 22.80 Estimate the size of the effect, calculating ω2. Round to 2 decimal places.

.44 ω2​ = SSB - (k-1)s2w/SST + s2w ω2​. = 26.13 - (2)(1.90)/(48.93) + 1.90 = 22.33/50.83 = 0.4393 = 0.44

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. The following data are obtained: SSbetween = 26.13 SSwithin = 22.80 Estimate the size of the effect, calculating η2. Round to 2 decimal places.

.53 η2​ = SSB/SST = 26.13/48.93 = .5340 = 0.53

You are testing how anxiety can relate to how anxious you are about taking statistics. You have separated people into 4 groups and have a sample size of 129. So far you have gathered that the SSW= 94222.296 and that the SST=94567.488 . You also know that the degree of freedom total is 128. From the information given, calculate Fobt. Round to 3.

1) Find SSB which is calculated by SST-SSW or 94567.488-94222.296= 345.192 2) Find df: You know that the k (# of groups)= 4 and that N = 129 Thus, dfB= K-1 = 3 , dfW= N-k = 129-4 = 125 3) Find s^2= SS/df SSB= 345.192/3 = 115.064 SSW=9422.296/125 = 753.778 4) Find fobt = S^2B/S^2W Thus, 115.064/74=753.778 = .153 5) Fcrit = .928

Using a one way Anova, you want to determine the effect of study patterns on exam score. You have 3 groups of students: (1) students who begin studying two weeks ahead of time (2) students who begin studying one week ahead of time (3) students who begin studying one day ahead of time n = 5 in each group. SS for Between Group: 776.133 SS for Within Group: 377.6 SS for Total: 1153.733 What can you conclude about the null, using α = 0.05?

1) Null: μ1 = μ2 = μ3 2) Between Group df: k-1 --> 3-1 = 2 Within Group df: N-k = 15-3 = 12 Total df: N-1 = 15-1 = 14 3) s2B = 776.133/2 = 388.067 4) s2W = 377.6/12 = 31.467 5) FOBT = 388.067/31.467 = 12.33 6) Find F Crit: dfnumerator = 2 dfdenominator = 12 Fcrit = 3.88 7) FOBT (12.33) > FCRIT (3.88), so reject H0 and conclude that at least one of the means (μ1, μ2, or μ3) differs from at least one of the others

If your SSr= 18.50, dfr=1, SSw = 25.33 and dfw=16, calculate Fobt for Factor A. Round to 2 decimals.

1) S^2R= SS/df = 18.50/1 = 18.50 2) S^2W = SS/df= 25.33/16 =1.58 3) Fobt (for A)= S^2R/S^2W =18.50/1.58 = 11.69 11.69

t crit depends on 3 factors

1) α level 2) One-tailed or two-tailed test 3) Degrees of freedom

Suppose you are interested if individuals who regularly watch The Daily Show differ significantly from individuals who watch regular news networks on their knowledge of current events. You randomly sample 10 Daily Show viewers and 10 regular news network viewers and administer a questionnaire assessing knowledge of events from the past week (scored 0-10). α = 0.052 tail. Use the following information to calculate tobt. Round the final answer to 2 decimal places. Xobt Daily Show = 8.45 Xobt regular news = 7.60 s2Daily Show = 1.73 s2regular news = 1.65

1.46 tobt = (8.45 - 7.60)/sqrt((1.73/10 + 1.65/10)) tobt = (0.85)/sqrt(3.38/10) tobt = (0.85)/0.58137767 tobt = 1.46204446 tobt = 1.46

If we are conducting a z-test to determine if a sample mean is significantly higher than a population mean, what would be the value of zcrit, if α = 0.051 tail? 1.645 -1.645 1.96 -1.96 ±1.645 ±1.96

1.645 If we are conducting a one-tailed test, and we are specifically testing if the sample mean is significantly higher than the population mean, then we are only looking at extreme scores in the right tail of the z-distribution. Thus, our critical value of z must be positive. This rules out answer options B, D, E, and F. The last piece of information we need to consider is the alpha level. Since it is one-tailed, the critical value will be lower than a two-tailed test. zcrit would be equal to 1.645, because this is the value for which 5% of scores fall at or above this z-score.

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. The following data are obtained: SSbetween = 26.13 SSwithin = 22.80 Calculate s2w. (Note: s2W = MSW).

1.90 s2w = SSwithin/dfwithin = 22.80/12 = 1.90

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. If the mean of group 1 is 6.00, the mean of group 2 is 8.00, and the mean of group 3 is 9.20, what would be Qobt if we did a Tukey HSD test comparing Group 2 to Group 3? Round to 2 decimal places. s2w = 1.90

1.95 Qobt = (larger mean - smaller mean)/sqrt(s2w/n) Qobt = (9.20 - 8)/sqrt(1.90/5) = (1.2)/sqrt(1.90/5) = 1.94665705 = 1.95

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. The following data are obtained: SSbetween = 26.13 SSwithin = 22.80 Calculate dfwithin

12 dfwithin = N - k N = 15 dfwithin = 15 - 3 = 12

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. The following data are obtained: SSbetween = 26.13 SSwithin = 22.80 Calculate s2B. Do not round. (Note: s2B = MSB).

13.065 s2B = SSbetween/dfbetween = 26.13/2 = 13.065

Suppose you want to determine the average amount of time first-year students spend studying. You randomly sample 61 first-year students and ask them how many hours per week they study. The mean of the sample is 20 hours, with a standard deviation of 6.5 hours. The population standard deviation is unknown. Construct the 95% confidence interval for the population mean. Round the final answer to 2 decimal places.

18.34-21.66 sX = (6.5)/sqrt(61) sX = 6.5/7.81024968 sx = 0.83223972 tcrit = 2.00 lower limit = 20 - 0.83223972(2.00) = 18.3355206 upper limit = 20 + 0.83223972(2.00) = 21.6644794 95% confidence interval = 18.34-21.66

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. The following data are obtained: SSbetween = 26.13 SSwithin = 22.80 Calculate dfbetween

2 dfbetween = k -1 k = 3 dfbetween = 3 - 1 = 2

f β = .20, there is a _______ chance of making a ______ error. 20%; Type I 80%; Type I 20%; Type II 80%; Type II

20%; Type II

A clinical psychologist is interested in the effect that anxiety level has on the ability of individuals to learn new material. She also wants to know if the effect of anxiety level depends on the difficulty of the new material. She conducts a 3 x 3 factorial ANOVA with 45 individuals (15 low anxious, 15 medium anxious, and 15 high anxious) and assigns 5 each to three material difficulty levels (low, medium, and high). She sets α = 0.05. What kind of design is this?

3 x 3 design

24 depressed patients are divided into four treatment groups: (1) CBT; (2) fluoxetine; (3) CBT + fluoxetine; and (4) placebo. SSbetween = 185.78 and SSwithin = 48.21. You set α = 0.05. Determine Fcrit.

3.10 df1 = 3 df2 = 20 Fcrit = 3.10

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. If the mean of group 1 is 6.00, the mean of group 2 is 8.00, and the mean of group 3 is 9.20, what would be Qobt if we did a Tukey HSD test comparing Group 1 to Group 2? Round to 2 decimal places. s2w = 1.90

3.24 Qobt = (larger mean - smaller mean)/sqrt(s2w/n) Qobt = (8 -6)/sqrt(1.90/5) = 3.24442842 = 3.24

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. Using α = 0.05, determine Qcrit.

3.77 Qcrit (k = 3, df = 12) = 3.77

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. The following data are obtained: SSbetween = 26.13 SSwithin = 22.80 Using α = 0.05, determine Fcrit. (Note: Degrees of freedom Numerator = dfbetween; Degrees of freedom Denominator = dfwithin).

3.88 Fcrit (df = 2, 12) = 3.88

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. If the mean of group 1 is 6.00, the mean of group 2 is 8.00, and the mean of group 3 is 9.20, what would be Qobt if we did a Tukey HSD test comparing Group 1 to Group 3? Round to 2 decimal places. s2w = 1.90

5.19 Qobt = (larger mean - smaller mean)/sqrt(s2w/n) Qobt = (9.20 - 6)/sqrt(1.90/5) = (3.2)/sqrt(1.90/5) = 5.19108548 = 5.19

Suppose I want to determine if a certain class texts significantly more in class compared to the national average. I observe the class and record how many texts each student sends during class. With N = 75, I calculate a sample mean = 13. I look up the national average, which is μ = 11, with σ = 3.25. I will use a z-test to determine if the class texts significantly more in class than the population of college students. I will use a z-test with α = 0.052 tail. Next, calculate zobt. Round to 2 decimal places. Use the rounded standard error you just calculated (.3753).

5.33 zobt = (sample mean - μ)/(std. error) = (13 - 11)/(.3753) = (2)/(.3753) = 5.32907008 = 5.33

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. The following data are obtained: SSbetween = 26.13 SSwithin = 22.80 Calculate Fobt. Round to 2 decimal places.

6.88 Fobt = s2B/s2w = 13.065/1.90 = 6.87631579 = 6.88

If p < α, |zobt| must be greater than zcrit. True False

A This is true. If the p-value of zobt is lower than our significance level (α), then by default the absolute value of zobt must be greater than zcrit.

Assuming the same α level, the critical z-value for a two-tailed test will be higher than the critical z-value for a one-tailed test. True False

A This is true. Since in a two-tailed test, we are testing the effect in both tails, we require a larger critical value to bound the region of rejection.

What is a robust test?

A test that can that can still be used even if assumptions are violated.

Raising α from 0.01 to 0.05 would _______ our chance of a ________ error, and would ________ our chance of a __________ error. A) increase; Type I; increase; Type II B) increase; Type I; decrease; Type II C)decrease; Type I; decrease; Type II D) decrease; Type I; increase; Type II

B

In an independent samples t-test, if the null is false, we would expect the group _____ to be ______, and the group _______ to be ______. means; different; variances; different means; different; variances; equal means; equal; variances; equal means; equal; variances; different

B If the null is false, we would expect the group means to be different, but we assume homogeneity of variance.

If the z-score for our sample mean (zobt) is greater than the critical value of z (zcrit), then _____. A) p > α B) p < α C) p = α D) cannot be determined

B The correct answer is B. If zobt is greater than zcrit, then we know the p-value corresponding to zobt must be lower than our significance level (α).

If we retain the null hypothesis, we say the results are significant. True False

B This is false. If we retain the null hypothesis, we conclude that it is reasonable to assume our results were due to chance alone. Or, more specific to the z-test, that our sample mean could have come from a random sample drawn from a population with a given mean. This would mean we conclude our results are not significant.

Which of the following would be an example of a Type I error? A. concluding that a guilty defendant is guilty B. concluding that an innocent defendant is guilty C. concluding that an innocent defendant is innocent D. concluding that a guilty defendant is innocent

B. concluding that an innocent defendant is guilty A Type I error is when the researcher decides to reject the null and conclude there is an effect (or, there is a significant difference), when in reality the null is actually true. This would be analogous to a jury concluding that a defendant is guilty, when in reality the defendant is innocent.

If in reality the null is true, the probability of correctly retaining the null is equal to ______. 1) α 2) β 3) 1 - α 4) 1 - β

C

If Cohen's d = .65, this means that one group has a mean that is _________ higher than the other group mean. This would be considered a _________ effect size. A) .65 standard deviations; large B) .65 standard errors; large C) .65 standard deviations; medium D) .65 standard errors; medium

C Cohen's describes the difference between the two sample means in standard deviation units. 0.20 - 0.49 = small effect 0.50 - 0.79 = medium effect 0.80+ = large effect

Power is to β as _______ . N is to power σ is to σX α is to confidence level effect size is to power

C Confidence level refers to the probability of failing to reject the null when the null is true (e.g., correctly retaining the null). It is equal to 1-α. If we increase power, we lower the chance of a Type II error (β). Similarly, if we raise α, we lower our confidence level.

In a t-test, if we lowered the α level, this would _____ power, and if the sample standard deviation was increased, this would _____ power. increase; increase increase; decrease decrease; decrease decrease; increase

C If we lowered the alpha level, this would reduce our probability of making a Type I error, but would increase our probability of making a Type II error, which would decrease power. If we increased the sample standard deviation, this would lower the value of tobt, which would also decrease power.

If the results of an experiment are statistically significant, which of the following statements describes the most accurate conclusion? A) The results are important B) The difference between group means must be large C) The results are reliable D) The results are likely due to chance

C The correct answer is C. A good synonym for significant is reliable. Answer choices A and B could be true, but they are not necessarily true. That will depend upon the effect size. In very large samples, it is possible for there to be a significant difference between our group means, but the difference to be quite small and not practically important. Answer choice D is incorrect because this would be the case if we failed to reject the null. Answer choice C is the only accurate description. To say the results are reliable means that if we repeated the experiment, we would expect to consistently get results that would allow us to reject the null hypothesis.

How to interpret Cohen's d?

Cohen's d describes the difference between two sample means in SD units. In terms of power/effect: 0.20-0.49 = small effect 0.50-0.79 = medium effect 0.80+ = large effect In terms of ETAsquared/Omega: 0.01-0.05= small effect 0.06-0.13= medium effect 0.14-1.00= large effect

When doing an experiment with many groups, what is the main problem with doing t-tests between all possible groups? How does ANOVA avoid that problem?

Conducting multiple t-tests without correction increases the overall chance of making a Type I error. Using the F-test (ANOVA) allows us to make one overall comparison among groups while keeping the chance of a Type I error at α.

If Factor B has a significant effect, we would expect _______ to be large relative to _________. A) s2rows; s2columns B) s2columns; s2rows C) s2rows; s2within-cells D) s2columns; s2within-cells

D

If power = .80, this means: A) if the null is true, there is an 80% chance of committing a Type I error B) if the null is false, there is an 80% chance of committing a Type II error C) if the null is true, there is an 80% chance of not committing a Type I error D) if the null is false, there is an 80% chance of not committing a Type II error

D Power is the probability of rejecting the null when the null is actually false. In other words, correctly rejecting the null. A Type II error is failing to reject the null when the null is false. Since power + β (chance of a Type II error) = 1, then if power = .80, β = .20. Another way to think about power is that it is the probability of NOT committing a Type II error, if the null is false. Answer choice A is incorrect because if power = .80, there is only a 20% probability of committing a Type II error, if the null is false. Answer choice B is incorrect because this would be the case if α = .20 (and we are not given any information about α). Answer choice D is incorrect because this would be the confidence level if α = .20 (and we are not given any information about α).

All of the following would be instances in which we would reject H0 EXCEPT: A) |tobt| > tcrit B) p < α C) the confidence interval of the difference between means does not contain 0 D) all of the above would be instances in which we would reject H0

D The correct answer is D. All of the statments are referring to situations in which we would reject H0. If |tobt| > tcrit, then the p-value for our statistic must be lower than the α level. In the case of comparing two group means, if the confidence interval of the difference between means does not contain 0, then we would conclude there is a significant difference between the population means and thus reject H0.

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. The following data are obtained: SSbetween = 26.13 SSwithin = 22.80 Calculate dfbetween

H0: μ1 = μ2 = μ3

Assume two independent samples (Sample A and Sample B), with Mean of Sample A = 10.81 and Mean of Sample B = 13.14. Sample A has a variance (s2) of 1.85, and Sample B has a variance of 3.48. Each sample has N = 11. Determine tcrit for α = 0.052 tail.

Degrees of freedom = (N1 - 1) + (N2 - 1) = 10 + 10 = 20 tcrit = ±2.086

There is a ________ relationship between α and the chance of a Type I error. Direct Inverse

Direct α is the probability of making a Type I error. If α = 0.05, there is a 5% chance of making a Type I error. If α = 0.01, there is only a 1% chance of making a Type I error. As we raise α, we raise our chance of a Type I error. It is a direct relationship.

If the assumption of homogeneity is violated, you cannot use the t-test. True False

False The t-test is a robust test, meaning that it can still be used even if assumptions are violated. For the independent samples t-test, we can often still use the t-test, even if homogeneity of variance is violated, especially if our sample sizes are equal and n ≥30.

A clinical psychologist is interested in comparing the effectiveness of short-term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive-behavioral therapy group, and an attention placebo group. The following data are obtained: SSbetween = 26.13 SSwithin = 22.80 Using α = 0.05, what do you conclude?

Fobt > Fcrit, so we reject H0 and can say that at least one of the conditions is significantly different from at least one of the other conditions

That clinical psychologist finishes her experiment and her data shows to be that SSr (difficulty level) = 506.844 , SSc (anxiety level) = 36.844 , SSrxc (interaction) = 40.756 , SSw (within-cells) = 98.80 , and SST= 683.244 with a df(t) =44. Remember, this is calculated in a 3x3 design. Calculate Fobt for Factor A (difficulty level), Factor B, and for Interaction . Round to 2 decimal places. Find F crit for all 3 variables.

Fobt A : 506.844/2 = 253.422 s2w = 98.80/36 = 2.744444 253.422/2.74444 = 92.34 Fobt B: df= c-1 = 2 S^2B= SS/df = 36.844/2 = 18.442 S2w= 2.744 18.442/2.744 = 6.71 Fobt Interaction: df= (c-1)(r-1) = (3-1)(3-1) = 2x2= 4 s^2rxc= SS/df = 40.756/4 = 10.189 FcritA= dfr/dfw = 2/36= 3.26 FcritB= dfc/dfw= 2/36 = 3.26 Fcrit interaction = dfRxC/dfw = 4/36= 2.63

Difference between planned comparisons and post hoc tests? When do you use them?

In an ANOVA, a significant F value (effect) tells us that 1 condition differs from one of the other conditions. We use both planned comparisons and post hoc tests when we are are determining what those conditions are. Planned Comparisons is when the comparisons were planned in advance of the experiment and arise from predictions based on theory and prior research. They are better than post hoc, but they should be few in numbers. Post Hoc Comparisons is when the comparisons arise after the experimenter sees data and picks groups with mean scores that are far apart. It also arises from doing all the possible comparisons with no theoretical prior basis. Because these comparisions were not planned before, must correct for higher probability of Type 1 error that increases due to multiple comparisions. MAIN DIFFERENCE: planned comparisons do not correct for an increases probability of making a Type 1 error, thus, they are more powerful than post hoc tests.

Two Sample Means?

Independent Groups t-test

How to interpret Levene's Test?

Levene's Test tests the assumption that the two population variances are equal to each other. If p>.05, assume homogeneity of variance if p <.05, violation of assumption of homogeneity of variance

difference between ω2 (omega squared) and η2 (eta squared)?

Omega squared is an unbiased estimate of the total variance (effect) in the dependent variable that is accounted for by the independent variable. Eta squared, on the other hand, does not provide an unbiased estimate of the effect of the independent variable. Rather, it provides a biased estimate that usually overestimates the true size of effect.

What is power? What 3 things increases power?

Power is the probability that the results of an experiment will allow the rejection of the null if the null is false. (Beta - 1) 1) increase sample size (standard error of the mean is inversely related to the mean- thus, as sample size increases, the variability decreases) 2) Raise alpha levels (increase alpha, area size for power increases. This also increases type 1 error, but decreases type 2 error) 3) Use a 1 tail test (more powerful than the other 2 because it focuses alpha in one direction) 4) having a large effect size

The police department of a major city reports that the mean number of auto thefts per neighborhood per year is μ = 6.88 with σ = 1.19. You are curious how the auto theft rate in your community compares. Based on a random sample of 15 neighborhoods in your community, you calculate a sample mean of 8.13. Assuming α = 0.052 tailed, what do you conclude?

Reject H0

A clinical psychologist conducts an experiment on 25 patients with schizophrenia to determine if a new treatment affects the amount of time these patients need to stay institutionalized. The results show that under the new treatment, the 25 patients stayed a mean duration of 78 weeks, with a standard deviation of 20 weeks. Previously collected data on a large population of patients with schizophrenia showed a mean of 85 weeks with an unknown standard deviation. Suppose α = 0.051 tail. Recall that tobt = -1.75. Given the tcrit value you just determined, what would be your decision about H0 in this situation?

Reject H0, because tobt < -tcrit, or stated otherwise, |tobt| > tcrit Note that we were able to reject H0 with a one-tailed test, but we were not able to with a two-tailed test, due to the higher critical value of a two-tailed test. However, note that if you had gotten the same result but in the opposite direction (tobt = +1.75), we would have failed to reject H0, since we are only testing in one direction.

Eric Cartman wants to determine if hippies and non-hippies differ in the amount of showers they take per week. He randomly samples 10 hippies and 10 non-hippies from South Park and records the number of showers they take in one week. Is there a significant difference between these two groups? Use α = 0.052 tail. Make a decision about H0 and justify your answer.

Reject H0, because |tobt| > tcrit

Single Sample Mean and σ is unknown?

Single sample t test

Difference between statistical significance and Practical significance (importance)?

Statistical significance means the differences in group means are not likely due to error. The problem is that statistically significant differences can be found even with small difference. Practical significance determines if the difference between samples are big enough to have real meaning/importance enough.

If you know the population standard deviaton, you should use a ______, whereas if you do not know the population standard deviation, you should use a _________. z-test; t-test t-test; z-test z-test; z-test t-test; t-test

The answer is A. A z-test is used when we know the value of the population standard deviation, whereas a t-test is used when we are estimating the population standard deviation with the sample standard deviaton.

Homogeneity of Variance definition?

The assumption that if the null was false, the means of the populations are expected to be different, but their variances are equal. Levene's Test tests this assumption.

All of the following are true about a 2-way ANOVA except: if you have a significant interaction effect, you must have at least 1 significant main effect the within-cell variance captures the inherent variability in the data the main effect of Factor A is captured by s2rows in a 3x3 design, the two main effects will have the same Fcrit value

The correct answer is A. It is possible to have a significant interaction effect but no significant main effects. There is referred to as a cross-over interaction. All the remaining statements are true.

If α = 0.05, and the probability level (p-value) of the observed statistic is 0.04, you would _________. A) reject H0 B) retain H0 C) accept H0 D) redo the experiment

The correct answer is A. Since p< α, we reject the H0 and conclude there is an effect of the independent variable.

Compared to the t-test, the z-test is a ______________ test. more powerful less powerful equally powerful cannot be determined

The correct answer is A. The z-test is used when we know the value of σ, so it will have less variability than a t-test, which must estimate σ using s. As a result, the t distribution has more elevated tails and higher critical values, than the z distribution. Therefore, the t-test has less power to detect an effect, since the obtained statistic has to meet a higher threshold than in the z-test in order to reject the null hypothesis.

Which of the following is true about the F distribution? it is normally distributed the mean of the F distribution equals 0 it has no negative values it is negatively skewed

The correct answer is C. Since the F ratio is based upon variances, which is a squared value, it has no negative values. As a result, it is positively skewed, with a median of 1.

The power of the t-test increases with _________. increases in N increases in the effect of the IV decreases in sample variance(s) all of the above

The correct answer is D. All of the above would increase the power of the t-test.

A clinical psychologist conducts an experiment on 25 patients with schizophrenia to determine if a new treatment affects the amount of time these patients need to stay institutionalized. The results show that under the new treatment, the 25 patients stayed a mean duration of 78 weeks, with a standard deviation of 20 weeks. Previously collected data on a large population of patients with schizophrenia showed a mean of 85 weeks with an unknown standard deviation. Use α = 0.052 tail. Explain why you should use a single-sample t-test instead of the z-test.

The population standard deviation is unknown

There are _______ variance estimates in a one-way ANOVA, and _______ variance estimates in a 2-way ANOVA.

There are 2 variance estimates (between-group, within-group) in a one-way ANOVA, and 4 variance estimates (rows, columns, rows x columns, within-cells) in a 2-way ANOVA.

Suppose I want to determine if a certain class texts significantly more in class compared to the national average. I observe the class and record how many texts each student sends during class. With N = 75, I calculate a sample mean = 13. I look up the national average, which is μ = 11, with σ = 3.25. I will use a z-test to determine if the class texts significantly more in class than the population of college students. I will use a z-test with α = 0.052 tail. If we reject H0, what does that mean with regard to the difference between our sample mean and the population mean?

There is a significant difference between our sample mean and the population mean. In other words, it is unlikely our sample mean came from a population of scores with μ = 11.

Increasing the within-group variance would increase the power of a one-way ANOVA. True False

This is false. Increasing the within group variance would reduce the value of Fobt, making it less likely it will be greater than Fcrit, and that you would be able to reject the null hypothesis. Thus, increasing within-group variance would decrease power in a one-way ANOVA.

If p < .05 for the Levene's test, we can assume homogeneity of variances. True False

This is false. The null hypothesis for the Levene's Test is that the group variances are equal to each other. If p <.05, we reject the null, which would indicate that homogeneity of variance has been violated.

Changing the α level affects the p-value of an obtained statistic. True False

This is false. The p-value refers to the probability of obtaining a result or more extreme results by chance alone (or, in other words, if the null is true). α is the decision level for whether or not to reject the null and is set prior to the experiment. Changing α affects the probability of committing a Type I or Type II error, but not the p-value of a statistic.

If we fail to reject the null hypothesis, when in reality the null is actually false, then we have committed a Type I error. True False

This is false. This statement describes a Type II error, which is when we fail to reject the null, but in reality the null is actually false, and we should have rejected it. In other words, there really is a difference, but we failed to "catch" it. A Type I error is when we decide to reject the null, but in reality we should have failed to reject the null since the null is actually true. In other words, there really is not a difference, but we concluded there is.

Results could be statistically significant without being practically significant. True False

This is true. Statistical significance means we were able to reject the null hypothesis and that our results are reliable. In other words, if we repeated the experiment, we would consistently get results that would allow us to reject the null. It says nothing about practical significance, or how large the difference is, or how important the results are.

Given the same α level, there is a lower chance of committing a Type II error with a planned comparison than with a Tukey HSD post hoc test. True False

This is true. Planned comparisons are more powerful than post hoc tests. Thus, there is a lower probability of committing a Type II error.

A significant Fobt indicates there is a significant difference among the group means, but does not indicate which means are significantly different from each other. True False

This is true. The F-test is an omnibus test, meaning that it tells us that at least one of the group means is significantly different from at least one of the other group means, but it does not tell us where those differences are. We need to follow-up with planned comparisons or post hoc tests to find which specific group means are different from each other.

Since planned comparisons are more powerful than post hoc tests, it is preferable to conduct planned comparisons when we wish to compare all possible pairs of means. True False

This statement is false. Although planned comparisons are more powerful than post hoc tests, they should be limited in number and only done when planned in advance of data analysis based on prior research, as they do not control for the higher risk of a Type I error.

If the Levene's Test is significant (in other words, if p < α), then the assumption of homogeneity of variance has been met. True False

This statement is false. The null hypothesis for the Levene's Test is that the groups have equal variance. If the test is significant, then we would reject this null and conclude that the groups have unequal variances. This is one occassion in which we are hoping to fail to reject the null.

The p-value is the probability that the null hypothesis is true. True False

This statement is false. The p-value refers to the probability of obtaining the results we got, or more extreme results, if the null hypothesis were true.

Assuming the same α level, and that both involve a two-tailed test, the critical value of z is the same for N = 50 and N = 100. True False

This statement is true. Since there is only one z distribution, the critical value of z does not depend upon sample size.

The absolute values of the critical values of z for α = 0.051 tailed and α = 0.102 tailed are equal to each other. True or False?

True

In a 3 x 2 design, Fcrit for the interaction effect and Fcrit for the main effect of Factor A will be equal values. True False

True ex. df-rows = (r-1) = (3-1) = 2 df-columns = (c-1) = (2-1) = 1 df-interaction = (r-1)(c-1) = (3-1)(2-1) = 2

Suppose you are interested if individuals who regularly watch The Daily Show differ significantly from individuals who watch regular news networks on their knowledge of current events. You randomly sample 10 Daily Show viewers and 10 regular news network viewers and administer a questionnaire assessing knowledge of events from the past week (scored 0-10). Since we failed to reject the null, we are at risk of making a _________ error, which could be due to low _________.

Type II; Power

Two differences between t test and z test

We have to use the estimated standard error of the mean in calculating tobt We need to use the t-table when determining tcrit

Why do we use ANOVA for 3+ groups?

We use ANOVA for comparing 3+ group because ANOVA allows for one overall comparison that says whether or not there is a significant difference between group means. (he may compare it to the t test, so we don't use the t test for +2 groups because it increases type 1 error and every comparison has 5% chance of making an error. (adds up))

Type II Error

When we fail to reject the null when the null is false. Remember: a type II error would be if a guilty person was arrested but was given a not guilty verdict.

Type 1 Error

When we reject the null when the null is true. Remember: a Type 1 error would be if an innocent person was arrested but given a guilty verdict.

How do we interpret ω2 and η2?

With either omega or eta, we're calculating the difference between groups due to variance. With our results from the formula, we can create a percentage and state, "___% of the variability in the dependent variable that is NOT explained by the main effect of Factor __( not being tested A/B) and the interaction effect is explained by the main effect of Factor (being tested A/B)."

It is a riveting Friday night. You and your friend are having a discussion about hypothesis testing. Your friend says that the p-value is the probability that the null hypothesis (H0) is true. Is your friend correct? If not, what is the more correct interpretation?

Your friend is incorrect. The null hypothesis is either true, or it isn't - but we never know for sure. The p-value is the probability we would have obtained the results we got, or even more extreme results, if the null hypothesis were true and chance alone was responsible for the results.

Single Sample Mean and σ is known?

Z test

Whats the difference between a one tailed test and a two-tailed test? (besides obvious differences)

a one tailed test is used when the experimenter is looking for an extreme difference in one direction. In other words, whether or not the sample mean is significantly higher or significantly lower than the population mean. 1) provides more power to an effect a two tailed test is used when the researcher is looking at extreme differences in either directions.

Eric Cartman wants to determine if hippies and non-hippies differ in the amount of showers they take per week. He randomly samples 10 hippies and 10 non-hippies from South Park and records the number of showers they take in one week. Is there a significant difference between these two groups? Use α = 0.052 tail. Determine tcrit.

df = (10-1) + (10-1) = 18 tcrit = ±2.101 Since it is two-tailed, don't forget the ± sign!

Power is ____________ related to α and _____________ related to β. directly; directly inversely; inversely inversely; directly directly; inversely

directly; inversely Power is directly related to the alpha level - increasing α increases power. Power is inversely related to β (the chance of making a Type II error). Remember that power + β = 1. If we increase power, we lower β. Another way to think about power is the probability of not making a Type II error, if the null is actually false.

A significant interaction effect indicates that one _______ has a different effect on the ________ at different levels of the ____________.

factor; dependent variable; other factor A significant interaction effect indicates that one factor has a different effect on the dependent variable at different levels of the other factor.

What is the p-value defined as in terms of the null?

p-value is the probability we would have gotten our result or a more extreme result if the null hypothesis were true. the p-value is the probability of obtaining a given score or a more extreme score.

24 depressed patients are divided into four treatment groups: (1) CBT; (2) fluoxetine; (3) CBT + fluoxetine; and (4) placebo. SSbetween = 185.78 and SSwithin = 48.21. You set α = 0.05. Calculate Fobt. Round to 2 decimal places.

s2B = 185.78/k -1 = 185.78/4-1 = 185.78/3 = 61.9267 s2w = 48.21/N - k = 48.21/24-4 = 48.21/20 = 2.4105 Fobt = 61.9267/2.4105 = 25.69

In a two sampled test, the experimenters were testing the difference between Selective Serotonin Reuptake Inhibitor Medicine and the Placebo. They found for the SSRI Medication, where their sample size was 15, that the sample mean was 18.1429 and that the squared standard deviation was 14.1271. They also found in the Placebo, where their sample size was also 15, that the sample mean was 28.1429 and that the squared standard deviation was 24.7138. Calculate the 95% Confidence Interval of Difference of Means. Use the formula given.

sx1-x2 = √((14.1271/15 + 24.7138/15)) = √(38.9309/15) = 1.61102245 (18.1429 - 28.1429) - (1.61102245)(2.048) = -10 - (1.61102245)(2.048) = -13.299374 (18.1429-28.1429) + (1.61102245)(2.048) = -10 + (1.61102245)(2.048) = -6.700626 -13.30, -6.70

A researcher examines courtship (how long couples date) and marital success. Based on a longitudinal study of 30 couples, the researcher finds that the mean courtship length for happily married couples (n = 15) was 755 days, while the mean courtship length for those who divorced within the first 7 years (n = 15) had a mean courtship of 1025 days before marrying. s2 happily married = 242.85 and s2early divorced = 143.93. Calculate the 95% confidence interval for µ1 - µ2. Round the final answer to 2 decimal places.

sx1-x2 =√((242.85/15+143.93/15))= √(386.78/15)= 5.07792609 tcrit= (N+N)-2 = 15+15 -2= df=28 = 2.048 (755-1025) - (5.07992609)(2.048) = -280.39959 (755 - 1025) + (5.07792609)(2.048) = 259.60041 -280.40, -259.60

In a two sampled test, the experimenters were testing the difference between Selective Serotonin Reuptake Inhibitor Medicine and the Placebo. They found for the SSRI Medication, where their sample size was 15, that the sample mean was 18.1429 and that the squared standard deviation was 14.1271. They also found in the Placebo, where their sample size was also 15, that the sample mean was 28.1429 and that the squared standard deviation was 24.7138. Calculate tobt.

t = (18.1429 - 28.1429)/√((14.1271/15) + (24.7138/15)) t = -10/√(38.8409/15) t = -10/√2.58939333 t = -10/1.6091592 t = -6.2144255 ≈ -6.21

steps of null hypothesis testing?

the null states that there is NO difference between sample means and population mean. Thus, the null hypothesis would state that the sample mean and the population mean are the same. It also implies that any difference is due to sampling error.

You want to test if Tufts students spend, on average, more hours per week preparing for class than the national average of college students. The national average is 15.55 hours/week, with an unknown standard deviation. Using a sample of 16 Tufts students, you calculate a sample mean of 19.75 hours/week and a standard deviation of 8.35 hours/week. Use α = 0.052 tail. Calculate tobt. Round to 2 decimal places. Find t crit. Compare t-obt and t-crit, what can you conclude about the null?

tobt = (19.75-15.55)/(8.35/√16) tobt = (4.2)/(8.35/4) tobt = (4.2)/(2.0875) tobt = 2.01197605 tobt = 2.01

Eric Cartman wants to determine if hippies and non-hippies differ in the amount of showers they take per week. He randomly samples 10 hippies and 10 non-hippies from South Park and records the number of showers they take in one week. The mean for hippies is 4.5 showers, while the mean for non-hippies is 5.9 showers. s2hippies = 2.22 and s2non-hippies = 1.67. Is there a significant difference between these two groups? Use α = 0.052 tail. Perform an independent samples t-test and calculate tobt. Round the final answer to 2 decimal places.

tobt = (4.5 - 5.9)/sqrt[(2.22/10) + (1.67/10)] tobt = (-1.4)/sqrt(3.89/10) tobt = -2.2446738 = -2.24

|zobt| < |zcrit|

we fail to reject the null

p > α

we fail to reject the null hypothesis and say the results are not significant

|zobt| ≥ |zcrit|

we reject the null

If we are hoping to reject the null hypothesis, we would want the ___________ variance to be as small as possible, and the __________ variance to be as large as possible.

within-group; between-group If we are hoping to reject the null, we want Fobt to be as large as possible. Thus, we would want to minimize within-group variance (the denominator) and maximize between-group variance (the numerator).

You want to test if Tufts students spend, on average, more hours per week preparing for class than the national average of college students. The national average is 15.55 hours/week, with a standard deviation of 8.33 hours. Using a sample of 16 Tufts students, you calculate a sample mean of 19.75 hours/week. Calculate zobt. Round to 2 decimal places.

z = (19.75 - 15.55)/[8.33/√16] z = (4.2)/(8.33/4) z = (4.2)/(2.0825) z = 2.02

Suppose I want to determine if a certain class texts significantly more in class compared to the national average. I observe the class and record how many texts each student sends during class. With N = 75, I calculate a sample mean = 13. I look up the national average, which is μ = 11, with σ = 3.25. I will use a z-test to determine if the class texts significantly more in class than the population of college students. I will use a z-test with α = 0.052 tail. Determine zcrit for α = 0.052 tail.

±1.96 Since we are using a two-tailed test, we are testing in both directions (if the sample mean is significanty higher than the population mean, and if the sample mean is significantly lower than the population mean). Thus, we need to find the positive z-value in which 2.5% of scores fall above it, and the negative z-value for which 2.5% of scores fall below it. Thus, we need the z-value that corresponds to a Column C value of .025. This value is ±1.96. Since we are testing in both directions, zcrit can be either -1.96 or +1.96. If zobt is negative and falls below -1.96, we reject the H0. If zobt is positive and falls above +1.96, we reject the H0. zcrit with α = 0.052 tail = ±1.96

A clinical psychologist conducts an experiment on 25 patients with schizophrenia to determine if a new treatment affects the amount of time these patients need to stay institutionalized. The results show that under the new treatment, the 25 patients stayed a mean duration of 78 weeks, with a standard deviation of 20 weeks. Previously collected data on a large population of patients with schizophrenia showed a mean of 85 weeks with an unknown standard deviation. Use α = 0.052 tail. What is the value of tcrit?

±2.064 df = N-1 = 25 - 1= 24

Suppose you are interested if individuals who regularly watch The Daily Show differ significantly from individuals who watch regular news networks on their knowledge of current events. You randomly sample 10 Daily Show viewers and 10 regular news network viewers and administer a questionnaire assessing knowledge of events from the past week (scored 0-10). α = 0.052 tail. Determine tcrit.

±2.101 df = 10 + 10 - 2 = 20 - 2 = 18 tcrit (df = 18) = ±2.101

If we conduct a one-way ANOVA with two groups, and Fobt = 49, what would be the value of tobt if instead we did an independent groups t-test?

±7 If k = 2, we can do either an independent groups t-test or a one-way ANOVA, and t² = F Fobt = 49 tobt = ±7

α is always .05 or .01. True False

α is whatever the researcher sets at the beginning of the experiment, depending on the consequences for a Type I or Type II error. .05 is commonly used, as it keeps the chance of a Type I error at a low rate (5%), but not so low that there is a higher risk of a Type II error. But, other values could be used (.10, .20). However, the higher the α, the greater the chance of a Type I error.

what does α level mean?

α level , also known as the level of significance, is the probability of making the wrong decision when the null hypothesis is true.

Suppose I want to determine if a certain class texts significantly more in class compared to the national average. I observe the class and record how many texts each student sends during class. With N = 75, I calculate a sample mean = 13. I look up the national average, which is μ = 11, with σ = 3.25. I will use a z-test to determine if the class texts significantly more in class than the population of college students. I will use a z-test with α = 0.052 tail. State H0.

μ = 11 The null hypothesis (H0) states that the sample mean is not significantly different from the population mean. Stated another way, in this example, the population from which our sample was drawn has a mean equal to 11 (the national average).

Suppose you are interested if individuals who regularly watch The Daily Show differ significantly from individuals who watch regular news networks on their knowledge of current events. You randomly sample 10 Daily Show viewers and 10 regular news network viewers and administer a questionnaire assessing knowledge of events from the past week (scored 0-10). α = 0.052 tail. State H0.

μ1 = μ2 or μ1 - μ2 = 0

24 depressed patients are divided into four treatment groups: (1) CBT; (2) fluoxetine; (3) CBT + fluoxetine; and (4) placebo. SSbetween = 185.78 and SSwithin = 48.21. You set α = 0.05. Calculate omega squared. Round to 4 decimal places.

ω2 = (185.78) - (3)(2.4105)/(233.99) + (2.4105) = (185.78) - (7.2315)/(236.4005) 178.5485/236.4005 0.7553