Stats 1073 chap 8-13

Test statistic (step 2)

(Z-score) forms a ratio comparing the obtained diff b/w the sample means & the hypothesized population mean vs the amount of diff we would expect w/o any treatment effect (the standard error)

Steps of hypothesis testing

1. State the hypotheses 2. Set the criteria for a decision 3. Collect data and compute sample statistic 4. Decide whether to reject or fail to reject the null hypothesis (make a decision)

Which of the following is an advantage that an independent-measures study has over a repeated-measures study? a. An independent-measures design can eliminate time-related factors. b. An independent-measures design can use fewer participants. c. An independent-measures design can be used on populations with large variances. d. There is no advantage to using an independent-measures study.

A. An independent-measures design can eliminate time-related factors.

Factors that affect statistical power

As the effect size increases, the probability of rejecting H0 also increases, which means that the power of the test increases. One factor that has a huge influence on power is the size of the sample. Reducing the alpha level for a hypothesis test also reduces the power of the test. If the treatment effect is in the predicted direction, changing from a two-tailed test to a one-tailed test increases power.

Alternative hypothesis (step 1)

H1, states that there's a change in the general population following an intervention. In the context of an experiment, predicts that the independent variable DID HAVE AN EFFECT on the dependent variable.

Which of the following is a correct interpretation of the F-ratio in a repeated-measures analysis of variance? I : F= variance bw treatments (wo individual differences)/ variance w no treatment effect (w individual differences) II: F= bw treatment variance/ error variance III: treatment effects+ random, unsystematic differences/ random, unsystematic differences

I, II and III

Assumptions for hypothesis tests w z-score (2)

Standard deviation for unknown population (after treatment) is assumed to be same as for the population before treatment To evaluate hypotheses w z-scores, we have used the unit normal table to identify the critical region. - table can be used only if distribution for sample means is normal.

Hypothesis test

Statistical method that uses sample data to evaluate a hypothesis about a population. The general goal is to rule out chance (sampling error) as a plausible explanation for the results from a research

Concerns about hypothesis testing: measuring effect size

There are two serious limitations with using a hypothesis test to establish the significance of a treatment effect. When the null hypothesis is rejected, we are actually making a strong probability statement about the sample data, not about the null hypothesis. Demonstrating a significant treatment effect does not necessarily indicate a substantial treatment effect.

Z-score statistic

Used in hypothesis test in the first specific ex of what's called a test statistic Test statistic indicates the sample data are converted into a single, specific statistic that's used to test the hypotheses.

Z-score formula

Z=M-m/sM= Sample mean- hypothesized population mean/ standard error between M and m.

Which of the following is not an assumption for using the independent-measures t formula? a. homogeneity of variance b. independent observations within each sample c. two normal populations d. homogeneity of population means

d. homogeneity of population means

For a repeated-measures t test, the cutoff value for α = _____ using a one-tailed test is the same as the cutoff value for α = _____ using a two-tailed test. a. 0.05, 0.025 b. 0.05, 0.01 c. 0.025, 0.05 d. 0.05, 0.05

C. 0.025, 0.05 Because the two-tailed test splits the critical area into the left and right tails, the α value for a one-tailed test is half that of a two-tailed test for the same cutoff value.

A scientist is studying the impact of certain vitamins on a person's ability to remember. The sample size for the experimental group was 25. When a two-tailed t test was calculated, the t statistic came out to be 1.54. What are the percent of variance (r2) and the size of the effect? a. 0.06, which indicates a medium effect b. 0.06, which indicates a large effect c. 0.09, which indicates a medium effect d. 0.09, which indicates a small effect

C. 0.09, which indicates a medium effect r2= t2/t2+24=1.54^2/1.54^2+24=0.09

What is the repeated-measures t statistic for a two-tailed test using the following? I. II 3. 7 2. 6 8. 6 7. 5 A. -1.732 B. -0.577 C. 0.577 D. 1.732

C. 0.577 The difference scores are 4, 4, -2, -2 so the sum of the difference scores is 4, and the mean difference score is mD=4/4=1. The sum of the squared deviations is SS=9+9+9+9=36. With n=4, df=3. So the sample variance is s2= SS/(n-1)= 36/3=12. The estimated standard error is sMD= sqrt s2/n= sqrt 12/3= 1.732. The t statistic is t=MD-uD/sMD= 1-0/1.732= 0.577

The results from a repeated—measures ANOVA are presented in the following summary table. Compute η2 to measure the size of the treatment effect. Source. SS. Df. MS. F Bw treatments. 46. 2. 20. F(2,14)=10 Wt treatments 90. 22. Between subjects. 60. 8 Error. 30. 14. 3 Total. 136. 24 a. 0.388 b. 0.395 c. 0.605 d. 0.667

C. 0.605 η2 =SSbetween treatments/ SSbetween treatments + SSerror= 46/46+30= 0.605

An independent-measures study with n = 8 in each treatment produces M = 75 for the first treatment and M = 71 for the second treatment with a pooled variance of 9. Construct a 95% confidence interval for the population mean difference. a. 0< u1-u2<7.2175 b. -2.145<u1-u2<2.145 c. 0.7825< u1-u2 < 7.2175 d. -0.7825< u1-u2< 7.2175

C. 0.7825<u1-u2< 7.2175

For a repeated-measures study comparing two treatments with 12 scores in each treatment, what is the df value for the t statistic? a. 23 b. 22 c. 11 d. 1

C. 11 For a repeated-measures study comparing two treatments with 12 scores in each treatment, the degrees of freedom is n-1= 12-1= 11

If exactly 5% of the t distribution is located in the tail beyond t = 2.353, how many degrees of freedom are there? a. 1 b. 2 c. 3 d. 4

C. 3 (Use t-distribution table, 9.1)

Steps for directional tests

1.) (most critical) state the statistical hypotheses 2.) null hypothesis states that there is no treatment effect & that the alternative hypothesis says that there is an effect 3.) 2 hypotheses are mutually exclusive and cover all of the possibilities

What are the steps in a hypothesis test where we have a formula for z-scores but we do not know the value for the population mean, u?

1.) make a hypothesis about the value of u. This is the null hypothesis. 2.) plug the hypothesized value in the formula along w the other values 3.) if formula produces a z-score near zero (which is where z-scores are supposed to be), we conclude the hypothesis as correct. 4.) on the other hand, if the formula produces an extreme value (a very unlikely result), we conclude the hypothesis as wrong

A researcher is evaluating the influence of a treatment using a sample selected from a normally distributed population with a mean of μ = 30 and a standard deviation of σ = 3. The researcher expects a 1-point treatment effect and plans to use a two-tailed hypothesis test with α = 0.05. Compute the power of the test if the researcher uses n = 9 individuals. a. 17% b. 50% c. 83% d. Nearly 100%

A. 17% Explanation: For σ = 3 and n = 9, the standard error is 1. The critical points on the non-treatment distribution for α = .05 are 28.04 and 31.96. For an expected 1-point effect, the mean on the treatment distribution is 31, and the area of the critical region in the two-tailed test is very close to 17%.

A researcher expects a treatment to produce a decrease in the population mean. The treatment is evaluated using a one-tailed hypothesis test. Which z-scores would lead us to reject the null hypothesis with α = .05? I. z = -1.75 II. z = 1.75 III. z = -1.6 IV. z = 1.6 a. I only b. I and II only c. II, III, and IV only d. None of the four

A. I only Explanation: Since the researcher is interested in a decrease in the population mean, he will need a one-tailed test, and the cutoff for α = .05 is z = -1.65. So anything below -1.65 leads us to reject the null hypothesis. This happens only for I.

Three different treatments (A, B, and C) are being studied using ANOVA. The means are computed: MA = 9, MB = 4, MC = 6. If MSwithin = 6.21, n = 7, and q = 3.67, which of the following is true using Tukey's HSD test? I. Treatment A is significantly different from Treatment B. II. Treatment A is significantly different from Treatment C. III. Treatment C is significantly different from Treatment B. a. I only b. II only c. I and II only d. I, II, and III

A. I only HSD= (3.67)sqrt of 6.21/7= 3.457. Since Ma- Mb= 5>3.457, treatment A is significantly different from Treatment B. Since Ma-Mc= 3<3.457, treatment A is not significantly different from Treatment C. Since Mc- Mb=2< 3.457, Treatment C is not significantly different from Treatment B.

Which of the following is a problem with using the z-score statistic? A. It requires knowing the population variance, which is often difficult to obtain. B. It requires very large samples in order to be effective. C. It is too cumbersome to calculate. D. It requires very small standard deviations in order to be effective.

A. It requires knowing the population variance, which is often difficult to obtain.

A treatment is administered to a sample selected from a population with a mean of μ = 40 and a standard deviation of σ = 6.25. After treatment, the sample mean is M = 45. Based on this information, the effect size as measured by Cohen's d can be classified as which of the following? a. Large effect b. Medium effect c. Small effect d. No effect

A. Large effect. Explanation: Cohen's d is (45 - 40)/6.25 = 0.8, which is a "large effect" according to Cohen's criteria.

To evaluate the effect of a treatment, a sample is obtained from a population with a mean of μ = 25, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 27.4 with SS = 64. If the sample consists of 9 individuals, what is the t statistic, and are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with α = .05? a. t = 2.55, yes b. t = 2.38, no c. t = 2.28, yes d. t = 2.28, no

A. T=2.55, yes With n=9, df=8, so s2=64/8=8 and sM= square root of 8/9=0.9428. Therefore, t=27.4-25/0.9428=2.55. With a=.05 and df=8, the boundaries of the critical region are -2.306 and 2.306. Therefore, 2.55 is inside the critical region. We reject the null hypothesis and conclude that the treatment has a significant effect.

Which of the following most accurately describes the F-ratio in ANOVA testing? a. The ratio of variances b. The ratio of sample mean difference to standard error c. The ratio of sample mean difference to sample variance d. The ratio of variance to sample mean difference

A. The ratio of variances

In a normal sample distribution with n = 16, the null hypothesis is rejected. If the sample size is changed to 64 with all other factors staying the same, what happens to the z-score and the decision about the null hypothesis? a. The z-score is doubled, and the null hypothesis is still rejected. b. The z-score is multiplied by 6, and the null hypothesis is still rejected. c. The z-score is doubled, and we fail to reject the null hypothesis. d. The z-score is multiplied by 6, and we fail to reject the null hypothesis.

A. The z-score is doubled, and the null hypothesis is still rejected. Explanation: Because the sample is increased by a factor of 4, the z-score is increased by a factor of . Since the original z-score is in the critical region, the double z-score will be even more in the critical region, so the null hypothesis is still rejected.

Which of the following correctly describes the effect that decreasing sample size and decreasing the standard deviation have on the power of a hypothesis test? a. A decrease in sample size will decrease the power, but a decrease in standard deviation will increase the power. b. Both will increase the power. c. A decrease in sample size will increase the power, but a decrease in standard deviation will decrease the power d. Both will decrease the power.

A. decrease in sample size will decrease the power, but a decrease in standard deviation will increase the power. Explanation: as n decreases, the standard error increases, and therefore the power decreases. However, as the standard deviation increases, the power decreases.

For a repeated-measures study comparing two treatments with a sample of n = 16 participants, a researcher obtains a sample mean difference of MD = 3.3 with SS = 315 for the difference scores. Calculate the repeated-measures t statistic for these data using a two-tailed test with α = .01, and determine if it is enough to reject the null hypothesis. a. t = 2.88, the null hypothesis is not rejected b. t = 2.88, the null hypothesis is rejected c. t = 3.11, the null hypothesis is not rejected d. t = 3.11, the null hypothesis is rejected

A. t= 2.88, the null hypothesis is not rejected S2= SS/(n-1)= 315/15= 21 sMD= sqrt s2/n= sqr 21/16= 1.146 t= mD-uD/sMD= 3.3-0/1.146= 2.88

Assumptions for hypothesis tests w z-score

Assumed that participants used in study were selected randomly The values in sample must consist of independent observations - 2 events (or observations) are independent if the occurrence of the first event has no probability of 2nd event.

A repeated-measures study is done to measure the change in IQ test scores taken on a Monday versus those taken on a Friday. There were n = 9 participants in the study. The mean difference was MD = -5 points, and the standard error for the mean difference was = 0.51. Construct a 95% confidence interval to estimate the size of the population mean difference. a. -6.18 < μ < 0 b. -6.18 < μ < -3.82 c. -2.306 < μ < 2.306 d. 0 < μ< 2.306

B. -6.18<u<-3.82 For a 95% confidence interval on a two-tailed test, use a t table with df = 9 - 1 = 8 to get the t values of 2.306 and -2.306. The confidence interval is between -5 + (2.306)(0.51) = -3.82 and -5 - (2.306)(0.51) = -6.18.

A researcher is conducting an ANOVA test to measure the influence of the time of day on reaction time. Participants are given a reaction test at three different periods throughout the day: 7 a.m., noon, and 5 p.m. In this design, there are _______ factor(s) and ______ level(s). a. 2, 3 b. 1, 3 c. 2, 6 d. 3, 1

B. 1, 3 There is only one factor: the time of day The three times represents the three levels of this factor

An analysis of variance produced an F-ratio with df values 14, 1. If the same data had been evaluated with an independent-measures t test, what would be df? a. 2 b. 14 c. 15 d. 16

B. 14 With df = 14, 1, there were two treatment groups of 8 scores each. The same data in a t test would have df = (8 - 1) + (8 - 1) = 14.

A random sample is selected from a normal population with a mean of μ = 200 and a standard deviation of σ = 12. After a treatment is administered to the individuals in the sample, the sample mean is found to be M = 196. How large a sample is necessary for this sample mean to be statistically significant using a two-tailed test with α = .05? a. 30 b. 35 c. 40 d. 70

B. 35 Explanation: To be statistically significant with α = .05, 196 needs to be 1.96 standard errors away from the norm. Some 196 is 4 away from 200, this means the standard error must be 2.04. This means that . Solving, we get n = 34.6. So we need a population of n = 35

For a posttest following ANOVA, there are four different treatment groups. How many pairwise comparisons must be made to gain a complete understanding of which treatment effects differ significantly from others? a. 4 b. 6 c. 12 d. 24

B. 6 With four treatment groups (A, B, C, and D), there are 6 pairwise comparisons that must be made in posttests: (1) A with B, (2) A with C, (3) A with D, (4) B with C, (5) B with D, and (6) C with D.

For the following table from a repeated-measure analysis of variance, calculate SSbt subjects Subject. T1. T2. T3 A. 2. 3. 5 B. 1. 1. 4 C. 5. 5. 2 a. 3.11 b. 6.22 c. 12.44 d. 18.66

B. 6.22 Since there are 3 treatments, k= 3. The sum of scores, G , is 28. The total # of scores, N is 9. So, SSbt subjects= (E)P^2/k-G^2/N= 100/3+36/3+144/3-784/9= 6.22

A research report concludes that there are significant differences among treatments, with "F(3,28) = 5.62, p < .01, η2 = 0.28." If the same number of participants was used in all of the treatment conditions, then how many individuals were in each treatment? a. 6 b. 8 c. 9 d. cannot determine without additional information

B. 8 The citation states that df within = 28 and df between = 3. Therefore, the total degrees of freedom is 28 + 3 = 31. The total number of scores must be N = 31 + 1 = 32. Since df between = 3, there must be 4 treatment conditions. If the number of participants in each treatment group is the same, then each group must have 32/4 = 8 participants.

If other factors are held constant, then how does the sample size affect the likelihood of rejecting the null hypothesis and the value for Cohen's d? a. A larger sample size increases the likelihood of rejecting the null hypothesis and increases the value of Cohen's d. b. A larger sample size increases the likelihood of rejecting the null hypothesis but does not change the value of Cohen's d. c. A larger standard deviation decreases the likelihood of rejecting the null hypothesis but increases the value of Cohen's d. d. A larger standard deviation decreases the likelihood of rejecting the null hypothesis and does not change the value of Cohen's d.

B. A larger sample size increases the likelihood of rejecting the null hypothesis but does not change the value of Cohen's d. Explanation: A larger sample size decrease the standard error, which in turn increases the z-score, which then increase the likelihood of rejecting the null hypothesis. However, since sample size is not part of the calculation of Cohen's d, this value remains unchanged.

Four individuals are given an experimental treatment for cancer. White blood count will be used as one measure of the effectiveness of the treatment. The white blood count of the four individuals is measured before the treatment, one week after the treatment, six months after the treatment, and one year after the treatment. What kind of study design is most appropriate? a. An individual-measures analysis of variance b. A repeated-measures analysis of variance c. An individual-measures t test d. A repeated-measures t test

B. A repeated-measures analysis of variance

A study is conducted to see if teenagers drive at faster average speeds than the general population of drivers. The average speed of the driving population is 35 mph. The null hypothesis is H0: μaverage driving speed of teenagers = 35 mph. What is the alternative hypothesis? a. H1: μdriving speed of teenagers = 35 mph b. H1: μdriving speed of teenagers ≠ 35 mph c. H1: μdriving speed of teenagers > 35 mph d. H1: μdriving speed of teenagers < 35 mph

B. H1: μdriving speed of teenagers ≠ 35 mph Explanation: the alternative hypothesis is the hypothesis that the sample mean is NOT EQUAL TO the population mean.

Which of the following would require a repeated-measures analysis of variance? I. An experimental study in which the researcher manipulates an independent variable to create two or more treatment conditions, with the same group of individuals tested in all of the conditions II. A nonexperimental study in which the same group of individuals is simply observed at two or more different times III. An experimental study in which the researcher applies the same treatment to two different groups of individuals a. I only b. I and II only c. III only D. I , II and III

B. I and II only

What is the main advantage that ANOVA testing has compared with t testing? a. It can be used with populations that have very high variances. b. It can be used to compare two or more treatments. c. It requires a smaller number of subjects. d. There is no advantage. They are simply different tests for different situations.

B. It can be used to compare two or more treatments.

For a sample of n = 16 scores with SS = 375, compute the sample variance and the estimated standard error for the sample mean. a. s2 = 23.44, sM = 1.21 b. s2 = 25, sM = 1.25 c. s2 = 25, sM = 1.5625 d. s2 = 25, sM = 1.29

B. S2=25, sM=1.25 S2= 375/16-2=25 and sM= square root of 25/16=1.25

A random sample of n = 30 individuals is selected from a population with μ = 15, and a treatment is administered to each individual in the sample. After treatment, the sample mean is found to be M = 23.1 with SS = 400. In order to determine if the treatment had a significant effect, which of the following can we use? a. A z-score only. There is not enough information to use a t statistic. b. A t statistic. There is not enough information to use a z-score. c. Either a z-score or a t statistic. There is enough information for both. d. Neither a z-score nor a t statistic. There is not enough information to use either.

B. T-statistic. There isn't enough info to use a z-score. Because the sample variance (or standard deviation) is unknown, a t statistic must be used.

If the variance between treatments increases and the variance within treatments decreases, what will happen to the F-ratios and the likelihood of rejecting the null hypothesis in an ANOVA test? a. The F-ratio will increase, but the likelihood of rejecting the null hypothesis will decrease. b. The F-ratio and the likelihood of rejecting the null hypothesis will increase. c. The F-ratio will decrease, but the likelihood of rejecting the null hypothesis will increase. d. The F-ratio and the likelihood of rejecting the null hypothesis will decrease.

B. The F-ratio and the likelihood of rejecting the null hypothesis will increase

For an independent-measures experiment, the F-ratio is 3.1. If dfbetween = 5, and dfwithin = 14, what will the researcher conclude? a. The null hypothesis will be rejected with α = 0.01, but not with α = 0.05. b. The null hypothesis will be rejected with α = 0.05, but not with α = 0.01. c. The null hypothesis will be rejected with both α = 0.01 and α = 0.05. d. The null hypothesis will not be rejected with α = 0.01 or α = 0.05.

B. The null hypothesis will be rejected with a= 0.05, but not with a= 0.01 Using an F table, the critical value for dfbetween = 5 and dfwithin = 14 is 2.96 for α = 0.05 and 4.69 for α = 0.01. Therefore, the null hypothesis will be rejected for 0.05, but not for 0.01.

In measuring the effect of hours of sleep on performance on a mathematics test using a repeated-measures study, which of the following is an example of an order effect? a. The participants' moods may be different during the first assessment than during the second, thereby affecting performance. b. The participants may have gained experience in taking the test the first time, thereby making it difficult to determine whether the change in hours of sleep or the experience causes the difference in performance. c. Because the participants were ordered to take the assessments, this may negatively impact their performance. d. The order in which participants finish the test may be the factor that impacts their performance rather than the hours of sleep.

B. The participants may have gained experience in taking the test for the first time, thereby making it difficult to determine whether the change in hours of sleep or the experience causes the difference in performance.

What is the main reason why you would want to remove individual differences by using a repeated-measures ANOVA? a. Removing individual difference lowers the sample variances. b. The presence of a treatment effect may be masked otherwise. c. Removing individual differences allows for more than two treatment conditions. d. Removing individual differences allows for a smaller number of participants

B. The presence of a treatment effect may be masked otherwise When individual differences are large, the presence of a treatment effect may be masked if an independent-measures study is performed. In this case, a repeated-measures design would be more sensitive in detecting a treatment effect because individual differences do not influence the value of the F-ratio

Assuming all other factors stay the same, what happens to the proportion of the data in both tails as the degrees of freedom increases with a t statistic? a. The proportion in the two tails combined increases. b. The proportion in the two tails combined decreases. c. The proportion in the two tails stays the same. d. It is impossible to tell without knowing the estimated standard error.

B. The proportion in the two tails combined decreases.

The results of a hypothesis test with a repeated-measures t statistic are reported as follows: t(9) = 2.28, p < .05. Which of the following is consistent with the report? a. The study used a total of 10 participants, and the mean difference was not significant. b. The study used a total of 10 participants, and the mean difference was significant. c. The study used a total of 11 participants, and the mean difference was not significant. d. The study used a total of 11 participants, and the mean difference was significant.

B. The study used a total of 10 participants, and the mean difference was significant. The research reports the degrees of freedom as 9. This means, for a repeated-measures t statistic, that n = 10. With α = .05, the boundary of the critical region is t = 2.262. 2.28 is in the critical region, so the mean difference was significant.

The results of a repeated-measures ANOVA are reported as follows: F(3, 20) = 1.05, p > .05. Which of the following is true? a. There are 4 participants, each receiving 11 treatments. b. There are 11 participants, each receiving 4 treatments. c. There are 9 participants, each receiving 3 treatments. d. There are 9 participants, each receiving 4 treatments.

B. There are 11 participants, each receiving 4 treatments The literature always reports F(dfbetween treatments, dferror). So dfbetween treatments = 3, which means that 3 = k - 1. Therefore, the number of treatments is 4. Since dferror = 20, it follows that 20 = 2(n - 1), which means that the number of participants is n = 11

In an independent-measures ANOVA, individual differences contribute to the variance in the numerator and in the denominator of the F-ratio. For a repeated- measures ANOVA, what happens to the individual differences in the numerator and denominator of the F-ratio? a. They do not exist in either the numerator or the denominator. b. They do not exist in the numerator and are measured and subtracted out from the denominator. c. They are measured and subtracted out during the analysis in both the numerator and the denominator. d. The do not exist in the denominator and are measured and subtracted out from the numerator.

B. They don't exist in the numerator and are measured and subtracted out from the denominator

What is the primary concern when selecting an alpha value? a. To make the null hypothesis easy to test b. To minimize Type I errors c. To minimize Type II errors d. To make z-scores easy to calculate

B. To minimize Type I errors.

What is the purpose of matching subjects on a variety of variables in a matched-subjects design? a. To create a smaller sample variance b. To reduce or eliminate the effect of these variables on the variable that is being studied c. To allow for a smaller sample size d. To increase the likelihood of rejecting the null hypothesis

B. To reduce or eliminate the effect of these variables on the variable that is being studied

When might removing individual differences in ANOVA not be advantageous in a repeated-measures design? a. When the treatment effects are reasonably consistent for all of the participants b. When the treatment effects are not consistent across participants c. When the sample sizes are less than 10 d. When the population variances are not equivalent across treatment conditions

B. When the treatment effects are not consistent across participants

Under what conditions might a post hoc test be performed following ANOVA? a. When there are two treatments and the null hypothesis was rejected. b. When there are three treatments and the null hypothesis was rejected. c. When the null hypothesis is not rejected. d. All of the above

B. When there are three treatments and the null hypothesis was rejected.

Which value is not included in the calculation of an estimated Cohen's d? a. μ b. n c. M d. s

B. n The formula for an estimated Cohen's d is M-u/s. n is not used.

Which of the following explains why it is easier to reject the null hypothesis with a one-tailed test than with a two-tailed test with all the same parameters? a. Because the standard deviation in a one-tailed test is larger than that for a two-tailed test b. Because z-scores are calculated differently in a one-tailed test c. Because the critical region is all on one side in a one-tailed test and needs to be split between the two tails in a two-tailed test d. Because a two-tailed test uses a bimodal distribution

C. Because the critical region is all on one side in a one-tailed test and needs to be split between the two tails in a two-tailed test. Explanation: In a two-tailed test, the critical region for a given α needs to be split among both tails, which pushes the cutoff values further away from the mean. In a one-tailed test, the critical region is entirely on one side of the mean, allowing the single cutoff value to be closer to the mean.

Which of the following is an assumption for a related-samples t statistic? a. The observations within each treatment condition must be independent. b. The population distribution of the difference scores must be normal. c. Both a and b. d. Neither a nor b.

C. Both A and B In order to use a related-samples t statistic, the observations within each treatment condition must be independent, and the population distribution of the difference scores must be normal.

In a repeated-measures ANOVA, what is another name for the F-ratio? a. residual variance b. error variance c. Both a and b d. None of the above

C. Both a and b

Which of the following will increase the likelihood of rejecting the null hypothesis using ANOVA? a. A decrease of SSwithin b. An increase in the sample sizes c. Both a and b d. None of the above

C. Both a and b.

Which of the following best explains why individual differences are not measured and removed from the numerator in the F-ratio for a repeated-measures analysis of variance? a. The individual difference are so small because the two samples are matched in such a way as to nearly eliminate them. b. The sample sizes are large enough so that it is not a problem. c. Exactly the same individuals participate in all of the treatment conditions. d. None of the above

C. Exactly the same individuals participate in all of the treatment conditions.

For a repeated-measures ANOVA, if SSwithin treatments = 86, SSbetween treatments = 80, SSbetween subjects = 62, dfwithin treatments = 12, dfbetween treatments = 1, dfbetween subjects = 4, find the F-ratio. a. F = 2.389 b. F = 11.16 c. F = 26.667 d. F = 72.833

C. F= 26.667 SSerror = SSwithin treatments - SSbetween subjects = 86 - 62 = 24. dferror = dfwithin treatments - dfbetween subjects = 12 - 4 = 8. MSerror = SSerror/dferror = 24/8 = 3. MSbetween treatments = SSbetween treatments/dfbetween treatments = 80/1 = 80. So F = MSbetween treatments/MSerror = 80/3 = 26.667.

An analysis of variance is used to evaluate the mean differences for a research study comparing five treatment conditions with a separate sample of n = 6 in each treatment. If SSbetween treaments = 24 and SStotal = 74, find the F-ratio. a. F = 1 b. F = 2 c. F = 3 d. F = 4

C. F= 3 If SSbetween treatments=24 and SStotal=74, then SSwithin treatments=74-24= 50. In this example, there are 30 scores. With five treatment conditions, df within= N-k= 30-5= 25. Df between= k-1= 5-1=4. Therefore MSbetween= SSbetween/ dfbetween= 24/4= 6, and MSwithin= SSwithin/ df within= 50/25= 2. Finally, F= MSbetween/ MSwithin= 6/2= 3

Which of the following are correct ways of defining the power of a statistical test? I. The probability that the test will correctly reject a false null hypothesis II. The probability that the test will result in a Type II error III. The probability that the test will not result in a Type II error a. I only b. I and II c. I and III d. II only

C. I and III Explanation: The power of a statistical test is defined as the probability that the test will correctly reject a false null hypothesis. Because failing to reject a false null hypothesis is also known as a Type II error, the power can also be defined as the probability that the test will not result in a Type II error.

For a repeated-measures study comparing two treatments with n = 26 scores in each treatment, the data produce t = 2.13. If the mean difference is in the direction that is predicted by the researcher, then which of the following is the correct decision for a hypothesis test with α = .05? a. Reject H0 for a one-tailed test but not for a two-tailed test. b. Reject H0 for a two-tailed test but not for a one-tailed test. c. Reject H0 for either a one-tailed test or a two-tailed test. d. Fail to reject H0 for both a one-tailed test and a two-tailed test.

C. Reject H0 for either a one-tailed test or a two-tailed test For a repeated-measures study with n = 26, it follows that df = 25. Using a t table, the cutoff for α = .05 is 1.708 for a one-tailed test and 2.060 for a two-tailed test. Either way, t = 2.13 is in the critical region, so the null hypothesis is rejected in both cases.

In a repeated-measures analysis of variance, there are 6 subjects who each underwent 3 treatments. What are dfbetween subjects and dferror? a. dfbetween subjects = 15, dferror = 10 b. dfbetween subjects = 10, dferror = 15 c. dfbetween subjects = 5, dferror = 10 d. dfbetween subjects = 10, dferror = 5

C. df between subjects= 5, dferror= 10 dfbetween subjects = n - 1 = 6 - 1 = 5. dfwithin = N - k = 18 - 3 = 15. dferror = dfwithin - dfbetween subjects = 15 - 5 = 10.

Critical region (step 2)

Consists of outcomes that are very unlikely to occur if the null hypothesis is true. It's defined by sample means that are almost impossible to obtain if the treatment has no effect.

In an ANOVA study on the impact that various forms of cellphone use have on driving speed, a researcher concludes that there are no systematic treatment effects. What was the F-ratio closest to? a. 0 b. c. d. 1

D. 1 An F-ratio near 1 indicates that the differences bw treatments (numerator) are random and unsystematic, just like the differences in the denominator. With an F-ratio near 1, we conclude that there's no evidence to suggest that the treatment has any effect.

A repeated-measures ANOVA was calculated with df = 1, 14. If the same data were analyzed with a repeated-measures t test, how many degrees of freedom would be used with the t statistic? a. 2 b. 3 c. 13 d. 14

D. 14 The df value for the t statistic is identical to the df value for the denominator of the F-ratio. The deonominator is dferror, and the ANOVA report (df = 1, 14) indicates that dferror = 14. So df for the t test is also 14.

An analysis of variance is used to evaluate the mean differences for a research study comparing 4 treatment conditions and 7 scores in each sample. How many total degrees of freedom are there? a. 3 b. 7 c. 28 d. 27

D. 27 If there are 4 treatment conditions with 7 scores in each sample, then there are 4 × 7 = 28 scores. The total degrees of freedom is 28 - 1 = 27

ANOVA is to be used in a research study using two therapy groups. For each group, scores will be taken before the therapy, right after the therapy, and one year after the therapy. How many different sample means will there be? a. 2 b. 3 c. 5 d. 6

D. 6 With two therapies and three measurement times, there will be 2 × 3 = 6 different sample means.

If SSbetween = 125 and SSwithin = 65, what is the effect size, η2, for the corresponding ANOVA? a. 34% b. 48% c. 52% d. 66%

D. 66% If SSbetween = 125 and SSwithin = 65, then SStotal = 125 + 65 = 190. So η2 = SSbetween/ SStotal = 125/190 = 66%.

Purpose of hypothesis test

Decide b/w 2 explanations: 1.) the diff b/w the sample & the population can be explained by sampling error (there does not appear to be a treatment effect) 2.) the diff b/w the sample & the population is too large to be explained by sampling error (there does appear to be a treatment effect)

An analysis of variance is used to evaluate the mean differences for a research study comparing three treatment conditions and the same number of scores in each sample. If SSbetween treaments = 24 and SSwithin = 72, and F = 4, how many scores are in each sample? a. 30 b. 27 c. 10 d. 9

D. 9 Dfbetween= k-1= 3-1= 2. Therefore, MSbetween/df between= 24/2. Since F=4, it follows that 12/ MSwithin=4. Therefore, MSwithin= 3. Since MSwithin= SSwithin/df within, it follows that 3=72/df within. Therefore df within= 24. Finally, since df within= N-k, it follows that 24= N-3. So N=27. If all three treatment groups have the same number of participants, there must be 9 participants in each group.

Which of the following is a common limitation of hypothesis testing? a. Conclusions are made about the data set rather than about the hypothesis itself. b. Demonstrating a significant treatment effect does not necessarily indicate a substantial treatment effect. c. Hypothesis testing loses its effectiveness for small samples. d. Both a and b

D. Both a and b. Explanation: Hypothesis testing demonstrates that a specific sample mean is very unlikely (p < .05) if the null hypothesis is true but does not make a claim on the probability of the hypothesis actually being true. Further, demonstrating a significant treatment effect does not necessarily indicate a substantial treatment effect. While small samples make it difficult to form conclusions, this is not a limitation of the hypothesis testing process, but rather a limitation of the sample size itself.

Which of the following is not a correct interpretation of the F-ratio in ANOVA testing? A. F= variance bw treatments/ variance wt treatments B. F= difference including any treatment effects/ differences w no treatment effects C. F= systematic treatment effects+random, unsystematic differences/ random, unsystematic differences D. F= variance bw treatments/ total standard error

D. F= variance bw treatments/ total standard error

When n is especially small, the t distribution is __________ and _______________. a. taller, less spread out b. taller, more spread out c. flatter, less spread out d. flatter, more spread out

D. Flatter, more spread out.

In a repeated-measures ANOVA, when treatment effects are consistent from one individual to another, which of the following is true? I. The individual differences also tend to be consistent and relatively large. II. A larger value of F is produced. III. The likelihood of rejecting the null hypothesis is increased. a. I only b. II only c. II and III only d. I, II, and III

D. I, II and III

A repeated-measures t test is performed, and the 95% confidence interval due to the treatment is 2.2 < μ < 6.4. Which of the following is true? I. For the general population, the treatment will result in an increase of score between 2.2 and 6.4. II. We are 95% confident that the true mean difference is in this interval. III. The sample mean difference was 4.4. a. I only b. I and II only c. II and III only d. I, II, and III

D. I, II and III Given a 95% confidence interval, the general population will result in a change of score reflected by that interval, and we are 95% confident that the true mean difference is in the confidence interval. Further, the confidence interval is always centered about sample mean difference. Therefore, the sample mean difference is (2.2 + 6.4)/2 = 4.4.

Which of the following is not an assumption for using a repeated-measures ANOVA? a. The observations within each treatment condition must be independent. b. The population distribution within each treatment must be normal. c. The variances of the population distributions for each treatment should be equivalent. d. The number of treatments must be less than three.

D. The number of treatments must be less than three

A repeated-measures study with n = 26 participants produces a mean difference of MD = 3 points, SS = 500 for the difference scores, and t = 2.50. Calculate Cohen's d and r2 to measure the effect size for this study. a. d = 0.15, r2 = 0.2 b. d = 0.15, r2 = 0.45 c. d = 0.67, r2 = 0.45 d. d = 0.67, r2 = 0.2

D. d= 0.67, r2= 0.2

A population of trees has a mean leaf length of 6.2 inches. A sample of 17 of these trees in a particular neighborhood has a mean length of 3.2 inches. If SS = 144 for this sample, what is Cohen's d for this example, and what is the strength of the treatment effect, which in this case is growing in a particular neighborhood? a. d = 0.5, which demonstrates a small effect. b. d = 0.5, which demonstrates a medium effect. c. d = 1, which demonstrates a medium effect. d. d = 1, which demonstrates a large effect.

D. d=1, which demonstrates a large effect Since n=17, s= square root of 144/17=3. So the estimated d is 6.2-3.2/3=1, which demonstrates a large treatment effect.

A sample is selected from a population and a treatment is administered to the sample. If there is a 3-point difference between the sample mean and the original population mean, which set of sample characteristics has the greatest likelihood of rejecting the null hypothesis? a. s 2 = 10 for a sample with n = 50 b. s 2 = 4 for a sample with n = 10 c. s 2 = 10 for a sample with n = 10 d. s 2 = 4 for a sample with n =50

D. s2= 4 for a sample with n=50 The likelihood of rejecting the null hypothesis increases when the sample size increases and the variance decreases. Therefore, s2=4 for a sample with n=10 would be the most convincing.

Measuring effect size

Intended to provide a measurement of absolute magnitude of a treatment used, independent of size of samples being used

Diff of Type I and II error

It's impossible to determine a single exact probability for type II error—> depends on variety of factors & therefore is a function, rather than a specific #, represented by symbol B (beta)

One-tailed vs two-tailed

Major distinction is the criteria they use for rejecting H0 One tailed allows to reject the null hypothesis when diff b/w the sample & pop is relatively small, provided the diff is in specified direction Two-tailed requires a relatively large diff independent of direction

Cohen's d

Measures size of mean diff in terms of standard deviation Cohen's = mean difference/ standard deviation

Type II error

Occurs when researcher fails reject a null hypothesis that is really false. Occurs when sample mean not in the critical region even tho the treatment has no effect on sample. (Consequence not so serious)

Statistical power

Power of statistical test is the probability that the test will correctly reject a false null hypothesis. - power is the probability that the test will identify a treatment effect if one really exists

Alpha level

Probability that the test will lead to a type I error. It determines the probability of obtaining sample data in the critical region even tho the null hypothesis is true.

The alpha level a researcher sets at the beginning of the experiment is the level to which he wishes to limit the probability of making the error of

Rejecting the null hypothesis when it is true

Factors that influence a hypothesis test

The final decision in a hypothesis test is determined by the value obtained for the z-score statistic. Two factors help determine whether the z-score will be large enough to reject H0. In a hypothesis test, higher variability can reduce the chances of finding a significant treatment effect. Increasing the number of scores in the sample produces a smaller standard error and a larger value for the z-score.

Selecting an alpha level

The primary concern when selecting an alpha level is to minimize the risk of a Type I error. Thus, alpha levels tend to be very small probability values. By convention, the largest permissible value is α = .05. However, as the alpha level is lowered, the hypothesis test demands more evidence from the research results.

A sample of size n1 = 16 is taken from a group of bees in one hive, and the sample variance in weight is 1.2 grams. A sample of size n2 = 16 is taken from another hive, and the sample variance is 0.8 grams. What is the estimated standard error for the difference between the two means? a. 0.35 b. 0.8 c. 0.89 d. 2.08

a. 0.35 Bc the two sample sizes are the same, we can use the simple formula. Therefore, S(m1-m2)=sqrt of (s1^2/n1)+(s2^2/n2)= sqrt (1.2/16)+(0.8/16)=0.35

For an independent-measures research study, the two sample means are found to be M1 = 15.5 and M2 = 17. If the pooled variance is 1.23, what would be the reported value of Cohen's d, and how would the effect size be described? a. 1.35, which demonstrates a large effect size b. -1.35, which demonstrates a large effect size c. -1.22, which demonstrates a medium effect size d. 1.22, which demonstrates a small effect size

a. 1.35, which demonstrates a large effect size The estimated Cohen's d is d= M1-M2/sqrt of Sp^2=15.5-17/sqrt 1.23= -1.35

The results of a hypothesis test are reported as follows: "t(35) = 1.65, p < .05." Based on this report, how many individuals were in the sample? a. 36 b. 35 c. 34 d. It cannot be determined from the information provided.

a. 36 The degrees of freedom are found in parenthesis. Bc df=35, the sample size was 36.

Which of the following is the null hypothesis for a repeated-measures analysis of variance using three different treatment groups? a. H0: μ1 = μ2 = μ3 b. H0: μ1 = μ2 or μ2 = μ3 c. H0: μ1 > μ2 > μ3 d. H0: μ1 - μ2 = μ3

a. H0: μ1 = μ2 = μ3

Which of the following will cause a researcher the most problems when trying the demonstrate statistical significance using a two-tailed independent-measures t test? a. High variance b. Low variance c. High sample means d. Low sample means

a. High variance High variance will lead to a low likelihood of rejecting a null hypothesis in an independent-measures t test.

How is an independent-measures design different from a study that makes inferences about the population mean from a sample mean? a. In an independent-measures design, there are two independent samples that are compared to one another. b. In an independent-measures design, two different measures are being taken. c. In an independent-measures design, means do not play a role. d. In an independent-measures design, studies are never two-tailed.

a. In an independent-measures design, there are two independent samples that are compared to one another.

Why might a repeated-measures study require half the number of subjects compared to a similar matched-subjects study with the same number of scores? a. In the repeated-measures study, each subject could be measured twice. b. It is easier to reject the null hypothesis in a repeated-measures study. c. The variance in a repeated-measures study is smaller. d. The sample means in a matched-subjects study tend to be higher.

a. In the repeated-measures study, each subject could be measured twice. In a matched-subjects study, each individual in one sample is matched with an individual in the other sample. In a repeated-measures study, one individual could be measured twice.

For a repeated-measures ANOVA, Tukey's HSD and the Scheffé test can be used in the exact same manner as is done for the independent-measures ANOVA, provided that you substitute _________ in place of MSwithin treatments in the formulas and use ________ in place of dfwithin treatments when locating the critical value in a statistical table. a. MSerror, dferror b. SSwithin, dferror c. SSbetween subjects, dfbetween subjects d. MSbetween subjects, dfbetween subject

a. MSerror, dferror

A researcher obtains an independent-measures t statistic of t = 2.12 for a study comparing two treatments with a sample of n1 = 14 in one treatment and n2 = 10 in the other treatment. What is the correct decision for a regular one-tailed hypothesis test? a. Reject the null hypothesis with α = .05 but not with α = .01. b. Reject the null hypothesis with α = .05 or with α = .01. c. Reject the null hypothesis with α = .01 but not with α = .05. d. Fail to reject the null hypothesis with either alpha level.

a. Reject the null hypothesis with a=.05 but not with a=.01 With n1 = 14 and n2 = 10, df = 14 + 10 - 2 = 22. Using a t table, the cutoff value for α = .05 is 1.717 and for α = .01 is 2.508. 2.12 is in the critical region for α = .05, but not for α = .01. Therefore, we reject the null hypothesis with α = .05 but not with α = .01.

A researcher administers a treatment to a sample of n = 100 participants and uses a hypothesis test to evaluate the effect of the treatment. The hypothesis test produces a z-score of z = 2.1. Assuming that the researcher is using a two-tailed test, what should the researcher do? a. The researcher should reject the null hypothesis with α = .05, but not with α = .01. b. The researcher should reject the null hypothesis with either α = .05 or α = .01. c. The researcher should fail to reject the null hypothesis with either α = .05 or α = .01. d. Cannot answer without additional information.

a. The researcher should reject the null hypothesis with α = .05, but not with α = .01. Explanation: bc 2.1 is beyond 1.96, we would reject the null hypothesis for a=0.5. However, because 2.1 is not beyond 2.58, we would fail to reject the null hypothesis for a= .01

What is a Type I error? a. When a researcher rejects a null hypothesis that is actually true b. When a researcher fails to rejects a null hypothesis that is actually false c. When a researcher does not set an alpha level high enough d. When a researcher sets an alpha level too high

a. When a researcher rejects a null hypothesis that is actually true

A researcher runs a repeated-measures ANOVA on two treatment conditions. The null hypothesis is rejected with F = 9. If the researcher were to run a repeated-measures t test on the same data, what would she find? a. t = 3, and the null hypothesis would be rejected. b. t = 3, and the null hypothesis would fail to be rejected. c. t = 9, and the null hypothesis would be rejected. d. t = 9, but there is not enough information to determine whether the null hypothesis is rejected.

a. t = 3, and the null hypothesis would be rejected. In making the transition from a repeated-measures ANOVA to a repeated-measures t test, the two tests always reach the same conclusion about the null hypothesis, and F = t2. Since

A researcher selects a sample of n = 25 individuals from a population with a mean of μ = 103 and administers a treatment to the sample. If the research predicts that the treatment will decrease scores, then what is the correct statement of the null hypothesis for a directional (one-tailed) test? a. μ ≥ 103 b. μ > 103 c. μ ≤ 103 d. μ < 103

a. μ ≥ 103 If the prediction is that the treatment will decrease scores, then the null hypothesis is that scores will not decrease, hence μ ≥ 103

Two samples are taken from separate populations. The sample and population mean from one group are M1 = 16 and μ1 = 14. The sample and population mean from the second groups are M2 = 21 and μ2 = 13. If the estimated standard error is S(M1 - M2) = 2, what is the t statistic? a. -6 b. -3 c. 3 d. 6

b. -3 T= (m1-m2)-(u1-u2)/s(m1-m2)=(16-21)-(14-13)/2=-3

One sample from an independent-measures study has n = 16 with a variance of s2 = 65. The other sample has n = 25 and s2 = 70. What is the df value for the t statistic? a. 38 b. 39 c. 40 d. 41

b. 39 For two independent samples, df=df1+df2= (16-1)+(25-1)=39.

An analysis of variance produces SS between = 64 and MS between = 8. In this analysis, how many treatment conditions are being compared? a. 8 b. 9 c. 16 d. 32

b. 9 Since MS between = SS between/df between, it follows that 8 = 64/df between, so df between = 8. Further, since df between = k - 1, 8 = k - 1, so k = 9. There are 9 treatment conditions being compared.

If all other factors are held constant, which of the following is true if the sample sizes and the sample variance increase in a one-tailed independent-measures t test? a. Because the increase in sample sizes causes an increase in effect size and the increase in sample variance causes a decrease in effect size, it is impossible to tell what the combined effect will be on the effect size. b. Because the increase in sample sizes has little or no effect on effect size and the increase in sample variance causes a decrease in effect size, the combined effect will be a decrease in effect size. c. Because the increase in sample sizes causes an increase in effect size and the increase in sample variance has little or no effect on effect size, the combined effect will be an increase in effect size. d. Because the increase in sample sizes causes an increase in effect size and the increase in sample variance also causes an increase in effect size, the combined effect will be an increase in effect size.

b. Bc the increase in sample sizes has little or no effect on effect size and the increase in sample variance causes a decrease in effect size, the combined effect will be a decrease in effect size.

An independent-measures design is also known as a _______________ design. a. dependent-measures b. between-subjects c. repeated-measure d. within-subject

b. Between-subjects

Assuming a normal distribution, which of the following would call for a one-tailed hypothesis test rather than a two-tailed test? a. Determining if attending Harvard influences IQ b. Determining if driving a red car increases the number of speeding tickets per year c. Determining if being male influences height d. Determining if being a teenager influences the number of hours of sleep per night

b. Determining if driving a red car increases the number of speeding tickets per year Explanation: A one-tailed test is used when there is a specific interest in one direction on of influence: either increase or decrease. The only situation is the interest in determining if driving a red car increases the number of speeding tickets per year.

In an independent measures test, a researcher is looking at the average height of Americans versus the average height of Australians. If μ1 and μ2 are the two population mean heights, what is the null hypothesis? a. H0: μ1 − μ2 ≠ 0 b. H0: μ1 − μ2 = 0 c. H0: μ1 − μ2 > 0 d. H0: μ1 > μ2

b. H0: μ1 − μ2 = 0

What is the null hypothesis for a related-samples test? a. H0: μ1 - μ2 = 0 b. H0: μD = 0 c. H0: μD > 0 d. H0: μ1 - μ2 > 0

b. H0: μD = 0 The null hypothesis for a related-samples test is based on the mean difference, μD. . The null hypothesis is that the treatment had no effect, or H0: μD = 0

Which of the following is a correct calculation for MSerror in a repeated-measures ANOVA? a. MSerror = MSwithin treatments - MSbetween subjects b. MSerror = SSerror/dferror c. Both a and b d. None of the above

b. MSerror= SSerror/ dferror

A researcher obtains an independent-measures t statistic of t = 3.01 for a study comparing two treatments with a sample of n = 16 in each treatment. What is the correct decision for a regular two-tailed hypothesis test? a. Reject the null hypothesis with α = .05 but not with α = .01. b. Reject the null hypothesis with α = .05 or with α = .01. c. Reject the null hypothesis with α = .01 but not with α = .05. d. Fail to reject the null hypothesis with either alpha level.

b. Reject the null hypothesis with a=.05 or with a= .01

What is another name for a repeated-measures design? a. Matched-subjects design b. Within-subjects design c. Matched samples design d. Between-subjects design

b. Within-subjects design

Which of the following is not an assumption for hypothesis tests with z-scores? a. random sampling b. small standard deviations c. normal sampling distribution d. independent observations

b. small standard deviations Explanation: The four assumptions of hypothesis tests with z-scores are (1) random sampling, (2) independent observations, (3) the standard deviation is unchanged by the treatment, and (4) normal sampling distributions. The size of the standard deviation is not relevant.

To evaluate the effect of a treatment, a sample is obtained from a population with a mean of μ = 31, and the treatment is administered to the individuals in the sample. After a treatment, the sample mean is found to be M = 32.7 with a sample variance of s2 = 4. If the sample size is n = 9, what is the t statistic, and is the data sufficient to conclude that the treatment increased the scores significantly? Use a one-tailed test and α = .01. a. t = 2.55, which is sufficient to reject the null hypothesis b. t = 2.55, which is not sufficient to reject the null hypothesis c. t = 3.355, which is not sufficient to reject the null hypothesis d. t = 3.355, which is sufficient to reject the null hypothesis

b. t = 2.55, which is not sufficient to reject the null hypothesis With a sample variance of s2=4, sM= sqrt of 4/9=0.667. So t=32.7-31/0.667=2.55. For df=8 and a=.01, the critical value on a one-tailed t test is 2.896. This isn't enough to reject the null hypothesis.

In an analysis of variance, the primary effect of large mean differences within each sample is to increase the value for ______. a. the variance between treatments b. the variance within treatments c. the total variance d. Large mean differences will not directly affect any of the three variances.

b. variance within treatments

Which of the following is not an advantage of a repeated-measures study over an independent-measures study? a. A repeated-measures design typically requires fewer subjects than an independent-measures design. b. A repeated-measures design is especially well suited for studying changes that take place over time. c. A repeated-measures design can be used on populations with high variances. d. A repeated-measures design reduces or eliminates problems caused by individual differences.

c. A repeated-measures design can be used on populations with high variances. The primary advantages of a repeated-measures design are that it requires fewer subjects, is well suited for studying changes over time, and reduces or eliminates the problem of individual differences.

Why does a change in sample sizes have little or no effect on Cohen's d in an independent-measures t statistic? a. Because sample sizes do not appear in the formula for Cohen's d. b. Because the sample sizes are too small to have an effect. c. Because the sample size occurs in the numerator as part of the difference in sample means and in the denominator as part of the pooled variance, and the effect of one is virtually cancelled out by the effect of the other. d. Because the difference in the sample sizes is too small to have an effect.

c. Because the sample size occurs in the numerator as part of the difference in sample means and in the denominator as part of the pooled variance, and the effect of one is virtually cancelled out by the effect of the other.

Which of the following is NOT a step in a hypothesis test? a. State the null hypothesis about a population b. Set the alpha level. c. If the sample data is not located in the critical region, we accept the null hypothesis. d. If the sample data is located in the critical region, we reject the null hypothesis.

c. If the sample data is not located in the critical region, we accept the null hypothesis.

A researcher is looking at the impact that television has on children. Children are placed in a room with a variety of toys and a television playing a cartoon. The researcher predicts that the children will spend more than half of their 30 minutes looking at the television. The researcher tested 15 children and found a sample mean of M = 17 minutes spent watching the television with SS = 79. In order to test this hypothesis, what does the researcher need? a. A one-tailed z-score b. A two-tailed z-score c. A one-tailed t statistic d. A two-tailed t statistic

c. One-tailed t statistic Because the researcher is predicting a value with a lower bound ("more than half"), he wants a one-tailed test. Because the population variance (or standard deviation) is unknown, he will need a t statistic.

A research report summarizes the results of the hypothesis test by stating, "z = 3.11, p < .01." Which of the following is a correct interpretation of this report? a. The null hypothesis was not rejected, and the probability of a Type I error is less than .01. b. The null hypothesis was not rejected, and the probability of a Type II error is less than .01. c. The null hypothesis was rejected, and the probability of a Type I error is less than .01. d. The null hypothesis was rejected, and the probability of a Type II error is less than .01.

c. The null hypothesis was rejected, and the probability of a Type I error is less than .01. Explanation: With a z-score of 3.11, this point falls in the critical region for p < .01, so we reject the null hypothesis, and the probability of a Type I error is less than .01

Which of the following is not an assumption for hypothesis testing using the t statistic? a. The population sampled must be normal. b. The values in the sample must consist of independent observations. c. The sample size must be greater than 30. d. The population standard deviation is unknown.

c. The sample size must be greater than 30.

A researcher is conducting a directional (one-tailed) test with a sample of n = 10 to evaluate the effect of a treatment that is predicted to increase scores. If the researcher obtains t = 2.770, then what decision should be made? a. The treatment has a significant effect with either α = .05 or α = .01. b. The treatment does not have a significant effect with either α = .05 or α = .01. c. The treatment has a significant effect with α = .05 but not with α = .01. d. The treatment has a significant effect with α = .01 but not with α = .05.

c. The treatment has a significant effect with α = .05 but not with α = .01. For a sample size of n=10, the cut off t value for alpha=.05 is 1.8331. For alpha=.01, it is 2.8214. When alpha=.05, t=2.770 is in the critical region, but for alpha=.01, it is not. The treatment has a significant effect at alpha=.05, but not with alpha=.01.

A research study measures the effect of age on the average driving speed by selecting a sample of teenage drivers and a sample of drivers between the ages of 20 and 25. Why is this an example of an independent-measures design? a. The speed of one driver in the samples has no impact on the speed of another driver. b. The two sample sizes may not be the same. c. There is one measure being taken for two non-overlapping samples. d. Each driver is self-reporting their speeds.

c. There is one measure being taken for two non-overlapping samples.

A repeated-measures study with a sample of n = 10 participants produces a mean difference of MD = 4.1 points with SS = 810 for the difference scores. For these data, find the variance for the difference scores and the estimated standard error for the sample mean. a. s2 = 9.49, sMD= 3 b. s2 = 9.49, sMD= 9 c. s2 = 90, sMD= 3 d. s2 = 90, sMD= 9

c. s^2=90, sMD=3 If n=10, then df=9. Therefore s2= SS/df= 810/9= 90, and sMD= sqrt of s^2/n= sqrt of 90/10=3

A random sample is normally distributed. If all values in the sample and all values in the population are multiplied by 2, what is the impact on Cohen's d? a. Decreases b. Increases c. Stays the same d. It is impossible to tell

c. stays the same Explanation: If all values in the sample and all values in the population are multiplied by 2, then the sample mean and the population mean are multiplied by 2, which means the difference is also multiplied by 2. Further, the standard deviation is multiplied by 2. Since Cohen's d is the ratio of the mean difference to the standard deviation, there is no change in the value of Cohen's d.

For a sample size of 9 with SS = 100 and sample mean M = 18.2, what is the probability that the true population mean will be between 16.56 and 19.84? a. 20% b. 50% c. 70% d. 80%

d. 80% 18.2 is in the middle of 16.56 and 19.84, with the difference being 1.64. The confidence interval for the population mean is given by M+/-t(sM). So t(sM)= 1.64. Since n = 9, df = 8, which means that s2 = 100/8 = 12.5, and sM=sqrt of 12.5/9=1.18. So, if t(1.18) = 1.64, then t = 1.39 (and -1.39), which are the t-scores marking off the middle 80%.

What is the advantage of a repeated-subject research study? a. It uses exactly the same individuals in all treatment conditions. b. There is no risk that the participants in one treatment are substantially different from the participants in another c. A smaller number of subjects is required. d. All a, b, and c.

d. All a, b and c. The main advantage of a repeated-measures study is that it uses exactly the same individuals in all treatment conditions. Thus, there is no risk that the participants in one treatment are substantially different from the participants in another.

In a matched-subjects design, how many variables can subjects be matched on? a. 1 b. 2 c. 3 d. All of the above

d. All of the above In a matched-subjects design, the subjects can be matched on as many variables as the researcher needs and can measure.

What would be the result of setting an alpha level extremely small? a. There would be almost no risk of a Type I error. b. It would be very difficult to reject the null hypothesis. c. Neither a nor b d. Both a and b

d. Both a and b Explanation: An extremely small alpha level, such as .000001 (one in a million), would mean almost no risk of a Type I error but would push the critical region so far out that it would become essentially impossible to ever reject the null hypothesis.

Which of the following would represent independent measures? a. The average number of traffic tickets among men and women is measured to see if there is a statistically significant difference. b. A researcher measures whether coffee intake affects hunger by recording the number of calories eaten by a sample of people in one month, then having the same group of people drink coffee for a month and record the number of calories eaten. c. Measuring whether age has an effect on the amount of time an individual can hold his breath under water. A sample of 10-year-olds is compared with a sample of 18-year-olds. d. Both a and c

d. Both a and c An independent measure is one that is applied to two completely different samples. Choice b uses the same group and measures the effect of a treatment. Both choice a and choice c use two different sample groups.

A researcher failed to reject the null hypothesis with a two-tailed test using α = .05. If the researcher had used the same data with a one-tailed test, what can we conclude? a. The researcher would definitely reject the null hypothesis using a one-tailed test. b. The researcher would definitely not reject the null hypothesis using a one-tailed test. c. The researcher would probably reject the null hypothesis using a one-tailed test. d. It is impossible to tell whether or not the researcher would reject the null hypothesis using a one-tailed test.

d. It is impossible to tell whether or not the researcher would reject the null hypothesis using a one-tailed test. Because the cutoff values for a one-tailed test are less than those of a two-tailed test, just because the t statistic does not fall in the critical region for a two-tailed test does not mean that it will not fall in the region for a one-tailed test. It may be less than both cutoff values, or it may be between the two cutoff values. There is no way of knowing without seeing the actual t statistic

A sample is selected from a population with μ = 50, and a treatment is administered to the sample. If the sample variance is s2 = 121, which set of sample characteristics has the greatest likelihood of rejecting the null hypothesis? a. M = 49 for a sample size of n = 15 b. M = 49 for a sample size of n = 75 c. M = 45 for a sample size of n = 15 d. M = 45 for a sample size of n = 75

d. M = 45 for a sample size of n = 75 The likelihood of rejecting the null hypothesis increases when the sample size increases and the difference between the sample mean and population mean increases. Therefore M=45 for a sample size of n=75 would be the most convincing.

For a two-tailed independent-measures t test with sample sizes 7 and 5, the 99% confidence interval for the population mean difference is -1<u1-u2<9. If M 1 = 31, find M 2 and S(m1-m2) a. M 2 = 36, = 3.169 b. M 2 = 27, = 3.169 c. M 2 = 36, = 1.5778 d. M 2 = 26, = 1.57

d. M2= 26, S(m1-m2)= 1.5778 Since the 99% confidence interval is centered at 4, the difference in the sample means M 1 - M 2 = 9 - 4 = 5. With M 1 = 31, M 2 must be 27. For sample sizes 7 and 5, df = 10. Using a t table, with a confidence interval of 99% and df = 10, the cutoff value of t is 3.169 and -3.169. Since the right side of the confidence interval is 9, it must be that 9=4+(3.169)(sm1-m2) . Solving, -1<u1-u2<9 = 1.5778. This confirms the left boundary of the confidence interval since 4 - (3.169)(1.5778) = -1. (10.4)

For a research study measuring the effect of two treatments, the sample sizes are n 1 = 8 in one treatment and n 2 = 10 in the other treatment, and the pooled variance is 1.23. The sample means are M 1 = 21.2 and M 2 = 24.1, with corresponding population means of μ1 = 21.2 and μ2 = 23. Using a two-tailed t test, determine the t value for α = .05 and make a conclusion about the null hypothesis. a. t = -3.01, therefore we reject the null hypothesis. b. t = -3.01, therefore we do not reject the null hypothesis. c. t = -2.09, therefore we reject the null hypothesis. d. t = -2.09, therefore we do not reject the null hypothesis.

d. T=-2.09, therefore we do not reject the null hypothesis

Which of the following sets of data is least likely to reject the null hypothesis in a test with the independent-measures t statistic. Assume that other factors are held constant. a. n = 30 and SS = 190 for both samples b. n = 15 and SS = 190 for both samples c. n = 30 and SS = 375 for both samples d. n = 15 and SS = 375 for both samples

d. n=15 and SS=375 for both samples The likelihood of rejecting the null hypothesis increase as the sample size increases and as the SS decreases. The least likely situation is the one with the lowest sample size and the highest SS, hence n = 15 and SS = 375.

Type I errors

occurs when a researcher rejects a null hypothesis that is actually true. Also occurs when researcher unknowingly obtains an extreme, non representative sample.

Directional tests

one-tailed tests, the statistical hypotheses (H0 and H1) specify either an increase or decrease in population mean. They make a statement about the direction of the effect.