psych exam #3 ch.8-11
Do the data provide evidence that the alcohol significantly increased (slowed) reaction time? Use a one-tailed test with α = .01.
1) IV = Alcohol level groups: sample drivers with 6 oz of wine and sober drivers in population DV = reaction time to onset of braking in an emergency situation Test: one-tail z-test 2) H0: μno alcohol ≥ μalcohol HA: μno alcohol < μalcohol*Note that greater than is used for the alcohol group because larger reaction time numbers indicate a slower reaction time. 3) Look up critical region in unit normal table α = .01 is +2.325 4) σM = σ / √n 40 / √25 40/5 = 8. z = (M-μ) / σM 420-400 / 8 = 20 / 8 = 2.5 5) I added my z obtained to the graph above and it exceeds my z critical. Therefore, I reject the null hypothesis. 6) Cohen's d = (M-μ) / σ 20 / 40 = 0.5 7) The reaction time of drivers that had 6 oz of wine (M = 420 ms) was significantly slower than the reaction time of drivers in the population (μ = 400, σ = 40), z = 2.50, p < .01, d = 0.50.
Interpret both effect sizes.
83% of the variance in the participants' estimates can be explained by the position of the vertical line at the midpoint of the horizontal line. On average, participants estimated the vertical line at 2.2 std devs higher than they would have if the lines had been side-by-side.
Construct the 90% confidence interval to estimate the mean amount of time spent on the plain side for the population of birds.
90% confidence interval means 10% in the tails, so t(14) = 1.76190% confidence interval = (M - t*sM) to (M + t*sM)= (34.5 - (1.761)(1.00)) to (34.5 + (1.761)(1.00))= 32.74 minutes to 36.261 minutes, or 34.5 ± 1.761 minutes
Construct a 95% confidence interval for this experiment.
95% CI = M1 - M2 ± t*s(M12M2) 40.8-34.0 ± (2.048* 2.097) = 6.8 ± 4.295 = 2.505 to 11.095
Under what circumstances is a t statistic used instead of a z-score for a hypothesis test?
A t statistic is used instead of a z-score for a hypothesis test when the population standard deviation and variance are not known.
Briefly explain how increasing sample size influences each of the following. Assume that all other factors are held constant. The size of the z-score in a hypothesis test.
As sample size increases, error decreases, the denominator of the z-score decreases, and the z-score increases.
Describe the similarities between an F-ratio and a t-statistic.
Both items are a ratio, have error in the denominator, detect group differences, and are inferential procedures. Mathematically, they are related/similar as well because F = t^2.
b. Compute the 80% confidence interval for the mean change in attention span for the population.
CI = 4.8 ± 1.328*2.50 = 4.8 - 3.32 = 1.48 and 4.8+3.32 = 8.12CI = 1.48 to 8.12 *Notice how zero is not contained in our estimate. If the null were true, then our CI would include zero.
b. Construct a 95% confidence interval to estimate the size of the population mean difference.
CI = 7 ± 2.306*2 = 7 ± 4.612 = 2.39 to 11.61
What does Cohen's d tell us about a significant effect (BE SPECIFIC)?
Cohen's d tells us the magnitude of groups differences. Put another way, it tells us how far apart the groups are in SD units.
Identify the critical value(s) for an alpha of .05 and make sure to include the df.
Critical t: t(13) = 1.771
If appropriate, compute effect size (include formula!). Otherwise state N/A
Estimated Cohen′s 𝑑 = 𝑀 ― 𝜇𝑠 = 116 ― 10330 = 1330 = 0.43 Variance accounted for = 𝑟2 = 𝑡2𝑡2 + 𝑑𝑓 = 2.1722.172 + 24 = 4.70928.709 = 0.16
What are the null and alternative hypotheses for Levene's test?
H0: σ2group 1 = σ2group 2 HA: σ2group 1 ≠ σ
State the null and alternative hypotheses using the appropriate symbols.
H0: ≤ 103H1: > 103
The National Retail Federation reported that in 2010 the average person spent $103.00 on Valentine's Day merchandise. Twenty-five individuals were surveyed about their 2011 Valentine's Day spending, and this sample produced a mean of M = $116 with SS =21,600. Has there been a significant increase in Valentine's Day spending this year? Use the steps below to conduct a one-sample t-test that answers this question. a. What is the IV and DV?
IV = Year spending occurred 2010 and 2011DV= Amount of money spent on Valentine's Day
Identify the critical value.
Look in t distribution table for "proportion in one tail" = 0.01 and "df" = 15Critical t value = 2.602
Identify the critical value that will serve as the decision criterion for = .05.
Look in t distribution table for "proportion in one tail" = 0.05 and "df" = 24Critical t value = 1.711
Compare parts a and b. Why is your conclusion similar or different?
My conclusions are different because moving from a two-tailed test to a one-tailed, directional test increased the size of my critical region and gave me more statistical power.
If appropriate, compute effect size (include formula!). Otherwise state N/A.
N/A
Interpret these results.
Our study did not show that students who answered questions while studying scored any higher on average on the exam (M = 78.3, s = 8.4) than students who did not answer questions while studying μ = 73.4, (t(15) = 2.33, p > .01).
State your decision regarding the null hypothesis.
Reject null hypothesis
d. State your decision regarding the null hypothesis.
Reject null hypothesis
Compute the test statistic and complete the ANOVA summary table below. Then state your decision regarding the null hypothesis. Attach a sheet of paper (or use the back of this page) with neatly organized formulas and work. To ensure accuracy, round to 2 decimal places
Source SS df MSBetweenGroups18.00 2 9.00WithinGroups6.21 9Total 24.21 110.69F = 13.04(12.96 is also ok) Decision: Reject the null hypothesis Work for 2c:dfBG = k - 1 = 3 - 1 = 2dfWG = N - k = 12 - 3 = 9dfTotal = N - 1 = 12 - 1 = 11 SSWG = MSWG * dfWGSSBG = SSTotal - SSWG MSBG = SSBG/dfBGF = MSBG/ MSWG = 12.96 or 13.04
Comparing your answers for parts a and b, explain how the size of the sample influences the outcome of a hypothesis test.
When I compare parts a and b, I can see that sample size influences the denominator or our error measurement (SE of Mean). As sample size increases, standard error decreases and the resulting z-score becomes larger.
b) What would the F-ratio be if SSBetween Groups equals zero?
Zero
c. n = 24
𝑡 =± 2.069
b. n = 15
𝑡 =± 2.145
In a study examining the effect of alcohol on reaction time, Liguori and Robinson (2001) found that even moderate alcohol consumption significantly slowed response time to an emergency situation in a driving simulation. In a similar study, researchers measured reaction time 30 minutes after participants consumed one 6-ounce glass of wine. Again, they used a standardized driving simulation task for which the regular population averages μ = 400 msec. The distribution of reaction times is approximately normal with σ = 40. Assume that the researcher obtained a sample mean of M = 420 for the n = 25 participants in the study. a. Are the data sufficient to conclude that the alcohol has a significant effect on reaction time? Use a two-tailed test with α = .01. We are going to all steps for hypothesis testing.
1) IV = Alcohol level groups: sample drivers with 6 oz of wine and sober drivers in population DV = reaction time to onset of braking in an emergency situation Test: two-tail z-test2) H0: μalcohol = 400 HA: μalcohol ≠ 4003) Look up critical region in unit normal table α = .01 is ±2.575 4) σM = σ / √n 40 / √25 40/5 = 8. z = (M-μ) / σM 420-400 / 8 = 20 / 8 = 2.5 5) I added my z obtained to the graph above and it does not exceed my z critical. Therefore, I fail to reject the null hypothesis. 6) not applicable do not compute effect size when you fail to reject the null hypothesis 7) The reaction time of drivers that had 6 oz of wine (M = 420 ms) was not significantly different than the reaction time of drivers in the population (μ = 400, σ = 40), z = 2.50, p > .01.
Researchers have noted a decline in cognitive functioning as people age (Bartus, 1990). However, the results from other research suggest that the antioxidants in foods such as blueberries can reduce and even reverse these age-related declines, at least in laboratory rats (Joseph et al., 1999). Based on these results, one might theorize that the same antioxidants might also benefit elderly humans. Suppose a researcher is interested in testing this theory. The researcher obtains a sample of n = 16 adults who are older than 65, and gives each participant a daily dose of a blueberry supplement that is very high in antioxidants. After taking the supplement for 6 months, the participants are given a standardized cognitive skills test and produce a mean score of M = 50.2. For the general population of elderly adults, scores on the test average μ = 45 and form a normal distribution with σ = 9 .a. Can the researcher conclude that the supplement has a significant effect on cognitive skill? Use a one-tailed test with α = .05.
1) IV = Antioxidant level groups: very high antioxidants elderly and elderly population DV = score on a standardized cognitive skills test Test: one-tail z-test 2) H0: μvery high antioxidant elderly ≤ 45 HA: μvery high antioxidant elderly > 45 3) Look up critical region in unit normal table α = .05 is +1.65 4) σM = σ / √n 9 / √16 9/4 = 2.25. z = (M-μ) / σM 50.2 - 45 / 2.25 = 5.2 / 2.25 = 2.31 5) I added my z obtained to the graph above and exceeds my z critical. Therefore, I reject the null hypothesis. 6) Cohen's d = (M-μ) / σ 5.2 / 9 = 0.58 7) The participants' performance on a standardized cognitive skills test was significantly better for the elderly people that consumed a supplement that is very high in antioxidants (M = 50.2) than the elderly in the population (μ = 45, σ = 9), z = 2.31, p < .05, d = 0.58.
There is some evidence suggesting that you are likely to improve your test score if you rethink and change answers on a multiple-choice exam (Johnston, 1975). To examine this phenomenon, a teacher gave the same final exam to two sections of a psychology course. The students in one section were told to turn in their exams immediately after finishing, without changing any of their answers. In the other section, students were encouraged to reconsider each question and to change answers whenever they felt it was appropriate. Before the final exam, the teacher had matched 9 students in the first section with 9 students in the second section based on their midterm grades. For example, a student in the no-change section with an 89 on the midterm exam was matched with student in the change section who also had an 89 on the midterm. The difference between the two final exam grades for each matched pair was computed and the data showed that the students who were allowed to change answers scoring higher by an average of MD = 7 points with SS = 288.a. Do the data indicate a significant difference between the two conditions? Use a two-tailed test with α = .05.
1) IV = Exam answer changes (allowed changes or allowed no changes) DV = percent of correct exam itemsType of test: related sample t-test, two tailed 2) H0: μD = 0 HA: μD ≠ 0 3) df = 8, alpha = .05 tcritical = ±2.306 4) s = √(SS/n-1) = √(288/8) = 6sMD = 6 / √9 = 2tobtained = 7 / 2 = 3.5 5) Reject the null hypothesis because the obtained exceeds the critical value. 6) Cohen's d = 7 / 6 = 1.17r2 = 12.25 / 20.25 = .60 7) The students who were allowed to change their answers scored higher than students that were not allowed to change their answers, t(8) = 3.5, p < .05, r2 = .60.*Note: I did not display descriptive statistics because we did not have this information for each group.
The stimulant Ritalin has been shown to increase attention span and improve academic performance in children with ADHD (Evans et al., 2001). To demonstrate the effectiveness of the drug, a researcher selects a sample of n = 20 children diagnosed with the disorder and measures each child's attention span before and after taking the drug. The data show an average increase of attention span of MD = 4.8 minutes with a variance of s2 = 125 for the sample of difference scores.a. Is this result sufficient to conclude that Ritalin significantly improves attention span? Use an α = .05 and one tailed test.
1) IV = Presence of Ritalin (Ritalin or no Ritalin)DV = attention spanType of test: related sample t-test, one-tailed 2) H0: μD ≤ 0 HA: μD > 0 *Hint: Think about how the difference score is created. Post minus pre so that gains are shown as positive numbers and losses are negative .3) df = 19, alpha = .05 tcritical = 1.729 4) s = √125 = 11.18sMD = 11.18 / √20 = 2.50tobtained = 4.8 / 2.50 = 1.92 5) Reject the null hypothesis because the obtained exceeds the critical value .6) Cohen's d = 4.8 / 11.18 = 0.43r2 = 3.69 / 22.68 = .16 7) Children with ADHD's attention span was longer when the children took Ritalin than when they did not take Ritalin, t(19) = 1.92, p < .05, r2 = .16.*Note: I did not display descriptive statistics because we did not have this information for each group.
Suppose an experiment is conducted with a sample of n = 12 males that look at a set of 30 photographs of women and rates the attractiveness of each woman using a 5-point scale (5 = most attractive). One photograph appears twice in the set, once with a tattoo and once with the tattoo removed. For each participant, the researcher records the difference between the two ratings of the same photograph. On average, the photograph without the tattoo is rated MD = 1.2 points higher than the photograph with the tattoo, with SS = 33 for the difference scores. Does the presence of a visible tattoo have a significant effect on the attractiveness ratings? Use an α = .05.
1) IV = Presence of tattoo (tattoo or no tattoo)DV = attractiveness ratingType of test: related sample t-test, two tailedKeep in mind that you need to use the information before the statistics to pick your tails. Other than the mean difference, the previous part of the problem did not state a direction for the hypothesis. This one was a little tricky. 2) H0: μD = 0 HA: μD ≠ 0 3) df = 11, alpha = .05 tcritical = ±2.201 4) s = √(SS/n-1) = √(33/11) = 1.73sMD = 1.73 / √12 = .50tobtained = 1.2 / .50 = 2.4 5) Reject the null hypothesis because the obtained exceeds the critical value. 6) Cohen's d = 1.2 / 1.73 = 0.69r2 = 5.76 / 16.76 = .347) Men rated the photographs of women without a tattoo as more attractive than photographs of women with tattoos, t(11) = 2.4, p < .05, r2 = .34.*Note: I did not display descriptive statistics because we did not have this information for each group.
In 1974, Loftus and Palmer conducted a classic study demonstrating how the language used to ask a question can influence eyewitness memory. In the study, college students watched a film of an automobile accident and then were asked questions about what they saw. One group was asked, "About how fast were the cars going when they smashed into each other?" Another group was asked the same question except the verb was changed to "hit" instead of "smashed into." The "smashed into" group reported significantly higher estimates of speed than the "hit" group. Suppose a researcher repeats this study with a sample of today's college students and obtains the following results. Estimated SpeedSmashed into Hitn = 15 n = 15M = 40.8 M = 34.0SS = 510 SS = 414 a. Use the 7 steps of hypothesis testing to evaluate if the results indicate a significantly higher estimated speed for the "smashed" group? Use an α = .05.
1)IV = words used to describe accident: hit and smashed intoDV = estimated speed of carsWe would use an independent samples t-test because participants were only in one group or another (i.e., hit or smashed into). If we are replicating, then we would use a directional hypothesis because past research found that smashed into had a higher reported speed than hit. 2) H0: μhit ≥ μsmashed into HA: μhit < μsmashed into 3) alpha = .05 and df = 28 so tcritical = 1.701 4) sp2 = (510+414) / (14+14) = 924/28 = 33s(m1-m2) = √(33/15)+(33/15) = √2.2+2.2 = √4.4 = 2.097tobtained = 40.8-34.0 / 2.097 = 6.8 / 2.097 = 3.24 5) the tobtained exceeds the tcritical so we would reject the null hypothesis. 6) r2 = (3.24*3.24) / (3.24*3.24) + (28) = 10.4976 / 38.4976 = .27 7) Participants estimated that the cars collided with a higher speed when "smashed into" (M=40.8, s=6.04) was used than when "hit" was used (M=34.0, s=5.44), t(28) = 3.24, p < .05, r2 = .27.
If applicable, compute the effect size and write a one-sentence PE interpretation of the effect size. Include the formula! If not applicable, write NA.
2 = SSBG/SSTOTAL = 18/24.21 = .74 Interpretation: "74% of the variability in attractiveness ratings is accounted for by facial symmetry."
b. Calculate the estimated standard error for the sample mean difference. Briefly explain what is measured by the estimated standard error.
= 4.00 / (√9) = 4 / 3 = 1.33The estimated standard error of the mean difference is the average distance that a sample mean difference score deviates from the mean of mean difference scores. Recall that we create a Sample Distribution of Means of Difference scores.
b. A researcher is examining the effect on humor on memory by presenting a group of participants with a series of humorous and not humorous sentences and them recording how many of each type of sentence is recalled by each participant.
A related samples, or repeated measures t-test is appropriate because everybody experiences both levels or groups of the independent variable (humorous and not humorous sentences).
c. A researcher is evaluating the effectiveness of a new cholesterol medication by recording the cholesterol level for each individual in a sample before they start taking the medication and again after 8 weeks with the medication.
A related samples, or repeated measures t-test is appropriate because everybody experiences both levels or groups of the independent variable (no medicine and cholesterol medicine).
For the each of the following studies determine whether a repeated-measures t test is the appropriate analysis. Explain your answers. a. A researcher is examining the effect of violent video games on behavior by comparing aggressive behaviors for one group who just finished playing a violent game with another group who played a neutral game.
A related samples, or repeated measures t-test is not appropriate because each group only experienced 1 level or 1 group of the independent variable.
Explain why ANOVA is preferred over multiple, separate t-tests when comparing 3 or more groups.
ANOVA is preferred over multiple t-tests because ANOVA will compare three or more groups while controlling for type I error rate.
Describe the basic characteristics of an independent measures, or a between-subjects, research study.
An independent samples, or between-subjects, design normally has 2 separate groups and each group experiences only 1 level/condition/group of the independent variable.
The size of Cohen's d.
As sample size increases this should decrease SD which is in the denominator of Cohen's d so Cohen's d should increase.
Comparing your answers from parts a and b, how does the variability of the scores in the sample influence the measures of effect size?
As the variability in our sample increases, our effect size decreases. Greater variability in our sample indicates that the population will likely also have greater variability (assuming we sampled our population at random) and thus makes us more conservative when we are making conclusions about the population.
Interpret the 95% confidence interval.
Based on our results, we are 95% confident that the true population mean for estimates of the length of the vertical line is between 11.79 and 12.61 inches.
Why is it important to compute effect size when there is a significant effect?
Effect size is important because it provides a standardized measure of the magnitude of group differences (Cohen's D) or tells us the amount of variance in the DV that the IV accounts for (r2). Without effect size, results across different studies could not be compared, the magnitude of groups differences is not known (only that they are different), and the amount of variation in DV scores that the IV can explain or accounts for is not known.
a. Why is it important to compute effect size when there are significant results?
Effect size is important because it provides a standardized measure of the magnitude of group differences (Cohen's d) or tells us the amount of variance in the DV that the IV accounts for (r2). Without effect size, results across different studies could not be compared, the magnitude of group differences is not known (only that they are different), and the amount of variation in DV scores that the IV can explain or accounts for.
Compute the estimated Cohen's d to measure the size of the treatment effect.
Estimated Cohen′s 𝑑 = 𝑀 ― 𝜇𝑠 = 34.5 ― 303.873 = 4.53.873 = 1.16
Determine the critical value ( = .01) and include the df.
F(2, 9) = 8.02 note: 4.26 is for = .05
Using the data in "part a" above, if you were to conduct a formal hypothesis test, then what decision would you make regarding the null hypothesis? Note: It is not necessary to actually conduct a hypothesis test to answer this question. Rather, think about what your F-ratio would be if SSBG = 0.
Fail to reject the null hypothesis because the numerator would equal zero and F = 0.
Fourteen students participated in a research study on the effect of sleep deprivation in a motor skill task. The students played the game "Operation" on two different occasions - once after a full night of sleep and once after being kept awake for 36 hours. Each task was measured two weeks apart. Half of the students were randomly assigned to the sleep deprivation condition first and the fully rested condition second, and the other half completed the two conditions in the reverse order. Below are the numbers of errors that each student committed. Does sleep deprivation increase the number of errors? Using the steps below conduct a related-samples t-test that answers this question. Pay close attention to which condition I have specified as X1 and which is X2. a. State the null and alternative hypotheses using the appropriate symbols.
H0: D ≤ 0 H1: D > 0
Friedman and Rosenman (1974) have classified people into two categories: Type A personalities and Type B personalities. Type As are hard-driving, competitive, and ambitious. Type Bs are more relaxed, easy-going people. Separate samples of Type As and Type Bs are obtained with n = 15 in each sample. The individual participants are all given a questionnaire measuring level of frustration. Higher scores indicate greater frustration. The average score for the Type As is M = 84 with SS = 641, and the Type Bs have an average of M = 71 with SS = 619. Do these data indicate a significant difference in the amount of frustration experienced by the two groups? Use the steps below to conduct an independent samples t-test that answers this question. a. State the null and alternative hypotheses using the appropriate symbols.
H0: Type A - Type B = 0 or Type A = Type BHA: Type A - Type B ≠ 0 or Type A ≠ Type B
An example of the vertical-horizontal illusion is shown in the figure at right. Although the two lines are exactly the same length, the vertical line appears to be much longer. To examine the strength of this illusion, a researcher prepared an example in which both lines were exactly 10 inches long. The example was shown to individual participants who were told that the horizontal line was 10 inches long and were then asked to estimate the length of the vertical line. For a sample of n = 25 participants, the average estimate was M = 12.2 inches with a standard deviation of s = 1.00. a. Use a one-tailed hypothesis test with α = .01 to demonstrate that the individuals in the sample significantly overestimate the true length of the line. (Note: Accurate estimation would produce a mean of μ = 10 inches.)
H0: μ ≤ 10 inchesH1: μ > 10 inchest(24) = 2.492𝑠𝑀 = 𝑠𝑛 = 1.0025 = 0.20𝑡 = 𝑀 ― 𝜇𝑠𝑀= 12.2 ― 100.20 = 2.20.20 = 11.0011.00 > 2.492, so we reject H0. On average, the participants significantly overestimated the true length of the line (M = 12.2 in., s = 1.00) vs. ruler measurements (t(24) = 11.00, μ = 10 in., p < .01).
State the null and alternative hypotheses.
H0: ≤ 73.4 H1: > 73.4
Describe the homogeneity of variance assumption and explain why it is important for the independent measures t-test.
Homogeneity of variance is the idea that both groups equally contribute to pooled variance. Or put another way, the variances among the two groups is not significantly different. This assumption is important because if one group has more variance than another group, the group with more variance will contribute more to pooled variance. Put another way, the larger variance drags the smaller variance towards it and inflates the overall pooled variance. FYI, calibrations for violating this assumption involved adjusting the degrees of freedom.
Weinstein, McDermott, and Roediger (2010) report that students who were given questions to be answered while studying new material had better scores when tested on the material compared to students who were simply given an opportunity to reread the material. In a similar study, an instructor in a large psychology class gave one group of students questions to be answered while studying for the final exam. The overall average for the exam was μ = 73.4 but the n = 16 students who answered questions had a mean of M = 78.3 with a standard deviation of s = 8.4. For this study, did answering questions while studying produce significantly higher exam scores? Use a one-tailed test with α = .01. a. What is the IV and DV?
IV = whether or not they answered questions while reading DV= exam score
A researcher is interested in the effect that type of residence has on the personal happiness of college students. She selects samples of students who live in campus residence halls, in Greek houses, in university apartments, in off-campus housing, and at home. She asks each respondent to state their gender, age, major, ethnicity, and rate their happiness on a scale of 1 (not happy) to 10 (happy). Name the following:
IV/Factor: type of residence Levels: _campus residence halls, Greek houses, university apartments,off-campus housing, at home DV: happiness Type of hypothesis test (full name): Between-subjects ANOVA
a) If the SSBetween Groups in an F-ratio equals zero, then conceptually, what does this mean?
If the SSBG = 0 then this means that there is no variability that we can explain. Put another way, there is no difference in variation among our groups due to our IV.
In 2008 Napier and Jost published an article entitled "Why Are Conservatives Happier Than Liberals?" in the journal Psychological Science. Using a nationally representative sample of American voters they found that on average conservatives had greater life satisfaction than liberals. In the paper they explain this difference in happiness by using what is called "system-justification theory." Imagine that you want to replicate a portion of this study. There are a variety of methods that could be used to examine the effect of political orientation on happiness. For example, participants could complete a survey that includes questions about political ideology and life satisfaction. Using the ideology questions you could classify people into one of two groups: conservatives and liberals. You could then compare the mean life satisfaction scores across these two groups. If you chose to use this design, which hypothesis test would you need to analyze the data? Be specific.
Independent Samples t-test
Comparing your answer for parts a and b, how does the variability of the scores in the sample affect the hypothesis test?
Larger variability in our sample means more uncertainty when we are running our hypothesis test, making it harder to know for sure whether or not the treatment has had an effect.Mathematically, this can be seen in the fact that as the variance gets larger, the denominator of the t statistic will get larger. This in turn will reduce the absolute value of the t statistic, bringing it closer to zero and less likely to fall in the critical region.
What are post-hoc tests used for and what do they do?
Post hoc tests are used to compare all group combinations while controlling for type I error rate. Post hoc tests show which groups are different from one another if the F-ratio were significant.
The power of a hypothesis test.
Power is influenced by sample size because sample size reduces error and produces a smaller denominator in the test statistic. As sample size increases, power increases.
he U.S. Department of Energy is funding 16 smart grid projects throughout the country. One of these projects is the Pacific Northwest Smart Grid Demonstration Project. As part of this, Avista Utilities is in the process of implementing smart grid technology in Pullman. One aim of this technology is to allow for real-time energy use information to be available for Avista customers via the Internet. Such information will hopefully help consumers make well-informed decisions about their energy use. Imagine that you are part of a research team that has been hired to evaluate the impact of this smart grid technology. You and your team randomly select 265 homes in Pullman, Washington. Just before the new Smart Meters are installed your team monitors the electricity meters on these homes for 2 weeks in order to establish a baseline measure of average daily energy usage for each home. After the Smart Meters are installed and customers have received instructions on how to use the new website to obtain information about the current use, a second measure of energy usage is taken. Your team monitors the energy usage for the same 265 homes for 2 weeks and computes the new average daily energy use for each home. You now want to compare energy use before and after to determine whether the Smart Meters have impacted consumers' behaviors. Which hypothesis test would be appropriate in this situation? Be specific.
Related-samples t-test, two tailed
Compute the appropriate test statistic. Show all formulas and work and circle your calculated test statistic.
SS = (D-MD)2 = 131.5 (calculation is above) s = (SS/(n-1)) = (131.5/13) = 3.18 sMD = s / (n) = 3.18/14 = 0.85 t = (MD - D)/ sMD = (2.5-0)/0.85 = 2.94
State your decision regarding the null hypothesis - Reject or Fail to Reject.
Since 2.17 > 1.711, it falls within the critical region and we would reject H0.
State your decision regarding the null hypothesis - Reject or Fail to Reject.
Since 2.33 is not greater than 2.602, we would fail to reject H0.
Interpret the results (PE statement).
Sleep deprivation increases the number of errors. Students made more errors after being deprived of sleep (M = 11.21, s = 3.07) than after a full night of sleep (M = 8.71, s = 2.20), t(13) = 2.94, p < .05, d = 0.79.
A random sample is selected from a normal population with a mean of μ = 30 and a standard deviation of σ = 8. After a treatment is administered to the individuals in the sample, the sample mean is found to be M = 33. Don't worry about using all 7 steps, please focus on computing the test statistic (i.e., the z-score for a sample mean), looking up the alpha level, and making a decision about the null hypothesis. a. If the sample consists of n = 16 scores, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with α = .05.
Start by calculating the z-score for the sample mean. You need SE to do this calculation/σM = σ / √n 8 / √16 8/4 = 2. z = (M-μ) / σM 33-30 / 2 = 3 / 2 = 1.5Using the unit normal table, the critical z-value for α = .05 is ±1.96.The obtained z-score has not exceeded the critical z-score so we fail to reject the null hypothesis and conclude that the treatment administered to individuals in the sample did not have an effect.
If the sample consists of n = 64 scores, is the sample mean sufficient to conclude that the treatment has a significant effect? Use a two-tailed test with α = .05.
Start by calculating the z-score for the sample mean. You need SE to do this calculation/σM = σ / √n 8 / √64 8/8 = 1. z = (M-μ) / σM 33-30 / 1 = 3 / 1 = 3Using the unit normal table, the critical z-value for α = .05 is ±1.96.The obtained z-score has exceeded the critical z-score so we reject the null hypothesis and conclude that the treatment administered to individuals in the sample had an effect.
Research has shown that facial symmetry impacts attractiveness. A psychologist selected a random sample of 12 people and used random assignment to place them in 1 of 3 conditions. According to condition assignment, participants were then shown a photo of a face that was perfectly symmetrical, slightly asymmetrical, or highly asymmetrical. They rated the face in terms attractiveness on a 1-7 scale. Use the steps below to conduct a between-subjects ANOVA examining whether attractiveness ratings differ across the symmetry conditions a. State the hypotheses and type of test.
Test: Between-Subjects ANOV H0: μ1 = μ2 = μ3 Note: subscripts don't have to be 1, 2, 3 & could be group name H1: Not H0 Note: μ1 ≠ μ2 ≠ μ3 is not correct for H1 or at least one group is different than the other groups
Conceptually speaking, what is the F-ratio a measure of (what is measured by the numerator and the denominator)?
The F-ratio is comprised of two variances. The variance we can explain in the numerator and the variance expected by chance, or the variance we cannot explain, is in the denominator.
Conceptually speaking, what does r2 measure?
The amount of variation in the DV that the IV accounts for.
Give a brief PE interpretation of the results at this point. Include the test results in APA format.
The attractiveness ratings differed across the facial symmetry conditions, F(2,9) = 13.04, p < .01,2 = .74. may have 12.96 Note: Must say there are differences but can't make specific comparisons
Describe what is measured by the estimated standard error in the bottom of the independent-measures t statistic.
The estimated standard error of the mean is our estimate of the distance between a sample mean and the population mean in the sample distribution of means, if the null hypothesis were true. In this case, the distribution of sample means is composed of Mean 1 minus Mean 2 scores. It is the basis for what sample mean 1 minus mean 2 score would be high probability or rather likely to be obtained if the null hypothesis was true.
What concept does the numerator of the t-ratio measure? What about the denominator?
The numerator measures the actual difference between the sample mean and population mean. If the independent variable influenced the dependent variable, this difference should be large.The denominator is the estimated distance between the population mean and sample mean that we would expect by chance if the null hypothesis was true. If we randomly selected a sample mean from the sample distribution of means, the denominator is the estimated average distance we would expect between a sample mean and the population mean. This distance is what chance would predict if the null hypothesis was true. You can also think of this distance between a sample mean and population mean as a discrepancy due to sampling error.
What is the defining characteristic of a repeated measures or within-subjects research design?
The primary feature of a repeated measures or within-subjects design is that everyone experiences both groups or levels of the independent variable. You can think of the manipulation of the IV as being a manipulation "within" participants.
Conceptually speaking, what does Cohen's d measure?
This is a measure of the magnitude of group differences in SD units.
Conceptually speaking, what makes up the tobtained , or t-ratio?
This is a ratio of difference between means over, or divided by, mean difference expected by chance, or if null hypothesis is true. Difference in means due to treatment, or IV, divided by chance differences between means.
a) Conceptually, what would it mean if the SSWithin Groups in an F-ratio equals zero?
This would mean that there is no variation among the scores within each group. Everyone has the same score in each group but not the same score between groups.
Facebook reports that the average user has 130 friends; is connected to 80 community pages, groups, and events; and creates 90 pieces of content each month. Consider our class as a representative sample of UW psychology students. Among the 84 Facebook users in our class, the average number of friends is 467.05, and this sample has a standard deviation of 296.10. Does the mean number of Facebook friends among UW psychology students differ from the mean for all Facebook users? Name the type of test that should be used to answer this question. Be specific. State why this specific test is appropriate.
Two-tailed, one-sample t-test;No population standard deviation was given and no direction of difference was specified.
Interpret the results (PE statement).
Type A personalities (M = 84, s = 6.77) have a significantly higher level of frustration than Type B personalities (M = 71, s = 6.65), t(28) = 5.31, p < .05, r2 = .50.
b) What would the F-ratio be if SSWithin Groups equals zero?
Undefined or Zero. Note: This would be a meaningless analysis because everyone would have the same score within each group and there would be no chance variation, only variation between groups.
Interpret these results. (Remember: use "plain English" and present the relevant statistics and results in APA format. Look at examples from videos/lecture!)
Valentine's Day spending in 2011 (M = 116, s = 30) was greater than Valentine's Day spending in 2010 ( = 103), t (24) = 2.17, p < .05, r2 = .16.
Calculate the estimated d and r2, the percentage of variance accounted for, to measure the size of this effect.
Variance accounted for = 𝑟2 = 𝑡2𝑡2 + 𝑑𝑓 = 11.00211.002 + 24 = 121145 = 0.8345 Estimated Cohen′s 𝑑 = 𝑀 ― 𝜇𝑠 = 12.2 ― 101 = 2.21 = 2.2
Interpret the 95% confidence intervals.
We are 95% confident that the true population mean of the difference between the smashed and hit groups is between 2.505 and 11.095
c. Write a sentence demonstrating how the results of the confidence interval would appear in a research report.
We are 95% percent confident that the true population mean difference score is between 2.39 and 11.61.
Three different drugs for treating Alzheimer's disease are compared using a between-subjects research design. Nine individuals with Alzheimer's disease are randomly assigned to receive either Drug A, Drug B, or Drug C. After several months of receiving the drug the nine participants completed a measure assessing the severity of their symptoms. Each participant received a symptom score from 0-100.a. In the table below, the mean for Condition A is given. Complete the rest of the table with the participants' scores and the remaining condition means so that SSBG = 0 and SSWG > 0. Hint: think about what makes, or how we calculate, SSBG and SSWG
You need to have some variability among the 9 scoresBut all 3 group means need to equal 50 (make sure they really are)For example, 49, 50, 51 (drug A) ; 45, 50, 55 (drug B) ; 48, 50, 52 (drug C)
Construct a 95% confidence interval for the population mean of the estimated length of the vertical line.
at p = 0.05, t(24) = ±2.06495% confidence interval = (M - t*sM) to (M + t*sM)= (12.2 - (2.064)(0.20)) to (12.2 + (2.064)(0.20))= 11.79 in. to 12.61 in., or 12.2 ± 0.4128 inches
Identify the critical value(s) for an alpha of .05 and make sure to include the df.
critical t: t(28) = +/- 2.048
If appropriate, compute Cohen's d for the effect size.
d = MD / s = 2.5/3.18 = 0.79
To evaluate the effect of a treatment, a sample of n = 8 is obtained from a population with a mean of μ = 40, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 35. Don't worry about using all 7 steps, please focus on computing the test statistic (i.e., the t-score for a sample mean), looking up the alpha level, and making a decision about the null hypothesis. a. If the sample variance is s2 = 32, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with α = .05?
df = (8 - 1) = 7, so our critical t value = ± 2.365.𝑠𝑀 = 𝑠𝑛 = 328 = 2𝑡 = 𝑀 ― 𝜇𝑠𝑀= 35 ― 402 = ―52 = ―2.500―2.500 < ― 2.365, so we reject H0 and conclude that the treatment has had a significant effect.
If the sample variance is s2 = 72, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with α = .05? Don't worry about using all 7 steps, please focus on computing the test statistic (i.e., the z-score for a sample mean), looking up the alpha level, and making a decision about the null hypothesis.
df = (8 - 1) = 7, so our critical t value = ± 2.365.𝑠𝑀 = 𝑠𝑛 = 728 = 3𝑡 = 𝑀 ― 𝜇𝑠𝑀= 35 ― 403 = ―53 = ―1.667―1.667 does not fall within the critical regions, so we fail to reject H0 and conclude that there is no evidence that the treatment has had an effect.
If appropriate, compute r2 for the effect size.
r2 = (t2)/(t2 + df) = (5.312)/(5.312 + 28) = .50
b. What does r2 tell us about the results (BE SPECIFIC)?
r2 tells us the amount of variance in the DV that the IV accounts for.
Compute the appropriate test statistic. Show all work and formulas and circle your calculated test statistic. (3 pts)
s2p = (SS1 + SS2)/(df1 + df2) = (641 + 619)/(28) = 45 s(M1-M2) = [( s2p /n1)+ ( s2p /n2)] = [(45/15)+ (45/15)] = 2.449 t = [(M1 - M2) - (1 - 2)]/ s(M1-M2) = [(84-71)-0]/2.449 = 5.31
6. A sample of n = 9 individuals participates in a repeated-measures study that produces a sample mean difference of MD = 4.25 with SS = 128 for the difference scores.a. Calculate the standard deviation for the sample of difference scores. Briefly explain what is measured by the standard deviation.
s=√(SS/n-1) = √(128/8) = 4In this case, standard deviation is for the difference scores so it is the average distance a difference score is from the mean difference score.
Many animals, including humans, tend to avoid direct eye contact and even patterns that look like eyes. Some insects, including moths, have evolved eye-spot patterns on their wings to help ward off predators. Scaife (1976) reports a study examining how eye-spot patterns affect the behavior of birds. In the study, the birds were tested in a box with two chambers and were free to move from one chamber to another. In one chamber, two large eye-spots were painted on one wall. The other chamber had plain walls. The researcher recorded the amount of time each bird spent in the plain chamber during a 60-minute session. Suppose the study produced a mean of M = 34.5 minutes on the plain chamber with SS = 210 for a sample of n = 15 birds. (Note: If the eye-spots have no effect, then the birds should spend an average of μ = 30 minutes in each chamber.) a. Is this sample sufficient to conclude that the eyespots have a significant influence on the birds' behavior? Use a two-tailed test with α = .05.
t (14) = ±2.145𝑠 = 𝑆𝑆𝑛 ― 1 = 21015 ― 1 = 21014 = 15 = 3.873𝑠𝑀 = 𝑠𝑛 = 3.87315 = 1.00𝑡 = 𝑀 ― 𝜇𝑠𝑀= 34.5 ― 301.00 = 4.504.50 > 2.145, so we reject H0. The birds spent less time in the chamber with the eye-spots than they did in the chamber without.
If other factors are held constant, explain how each of the following influences the value of the independent measures t statistic, the likelihood of rejecting the null hypothesis, and the magnitude of measures of effect size. a. Increasing the number of scores in each sample.
t statistic = sample size influences the denominator, as sample size increases the denominator becomes smaller and the t statistic becomes larger likelihood of rejecting null hypothesis = if the t statistic becomes larger and sample size increase, it would be easier to reject the null hypothesis magnitude of effect size = we have two measures of effect size Cohen's d should increase as sample size increases because variability should decrease as sample size increases. variability is in the denominator of Cohen's dr2 should decrease as sample size increases because sample size is part of denominator
b. Increasing the variance for each sample.
t statistic = variability is part of the denominator so if variance increases, the t statistic should decrease. Conceptually, you can think of this as "more statistical noise". likelihood of rejecting null hypothesis = If increasing variability decreases the t statistic, then it would be harder to reject the null hypothesis. We would need a larger group difference to detect an effect magnitude of effect size = for Cohen's d, variability is part of the denominator so as variability increases, cohen's d decreasesfor r2, variability influences the t-statistic which is in the numerator and denominator so as variability increases, t obtained decreases, and r squared also decreases
Find the estimated standard error of the mean for each of the following samples.a. n = 9 with SS = 1152
𝑠 = 𝑆𝑆𝑛 ― 1 = 11529 ― 1 = 11528 = 144 = 12𝑠𝑀 = 𝑠𝑛 = 129 = 4.00
Compute the appropriate test statistic. Show all formulas that you use as well as your work and circle your final answer.
𝑠 = 𝑆𝑆𝑛 ― 1 = 2160025 ― 1 = 2160024 = 900 = 30𝑠𝑀 = 𝑠𝑛 = 3025 = 6.00𝑡 = 𝑀 ― 𝜇𝑠𝑀= 116 ― 1036.00 = 2.17
b. n = 16 with SS = 540
𝑠 = 𝑆𝑆𝑛 ― 1 = 54016 ― 1 = 54015 = 36 = 6𝑠𝑀 = 𝑠𝑛 = 616 = 1.50
c. n = 25 with SS = 600
𝑠 = 𝑆𝑆𝑛 ― 1 = 60025 ― 1 = 60024 = 25 = 5𝑠𝑀 = 𝑠𝑛 = 525 = 1.00
Assuming that the sample standard deviation is s = 20, compute r2 and the estimated Cohen's d to measure the size of the treatment effect.
𝑠𝑀 = 𝑠𝑛 = 2016 = 5.00𝑡 = 𝑀 ― 𝜇𝑠𝑀= 49.2 ― 455.00 = 4.25.00 = 0.84Variance accounted for = 𝑟2 = 𝑡2𝑡2 + 𝑑𝑓 = 0.8420.842 + 15 = 0.705615.7056 = 0.0449Estimated Cohen′s 𝑑 = 𝑀 ― 𝜇𝑠 = 49.2 ― 4520 = 4.220 = 0.21
Compute the appropriate test statistic. Show all formulas that you use as well as your work and circle your final answer.
𝑠𝑀 = 𝑠𝑛 = 8.416 = 2.10𝑡 = 𝑀 ― 𝜇𝑠𝑀= 78.3 ― 73.42.10 = 4.92.10 = 2.33
random sample of n = 16 scores is obtained from a population with a mean of μ = 45. After a treatment is administered to the individuals in the sample, the sample mean is found to be M = 49.2.a. Assuming that the sample standard deviation is s = 8, compute r2 and the estimated Cohen's d to measure the size of the treatment effect.
𝑠𝑀 = 𝑠𝑛 = 816 = 2.00𝑡 = 𝑀 ― 𝜇𝑠𝑀= 49.2 ― 452.00 = 4.22.00 = 2.10Variance accounted for = 𝑟2 = 𝑡2𝑡2 + 𝑑𝑓 = 2.1022.102 + 15 = 4.4119.41 = 0.2272Estimated Cohen′s 𝑑 = 𝑀 ― 𝜇𝑠 = 49.2 ― 458 = 4.28 = 0.525
c. n = 30
𝑡 = 2.462
b. n = 20
𝑡 = 2.539
Find the t value that forms the boundary of the critical region in the right-hand tail for a one-tailed test with α = .01 for each of the following sample sizes .a. n = 10
𝑡 = 2.821
Find the t values that form the boundaries of the critical region for a two-tailed test with α = .05 for each of the following sample sizes: a. n = 4
𝑡 =± 3.182