Stats

Explain how the size of the sample standard deviation influences the likelihood of finding a significant mean difference.

A larger standard deviation increases variability, which decreases the likelihood of finding a significant difference.

What is the defining characteristic of a repeated-measures or within-subjects research design?

A repeated-measures design uses the same group of participants in all of the treatment conditions.

What are the basic characteristics of an independent-measures, or a between-subjects, research study?

An independent-measures study requires a separate sample for each of the treatments or populations being compared. It is sometimes called a between-subjects design. The design of the study involves separate and independent samples and makes a comparison between two groups of individuals.

Data similar to the results obtained show an average of M = 786 calories per meal with s = 85 for n = 100 children and adolescents before the labeling, compared to an average of M = 772 calories with s = 91 for a similar sample of n = 100 after the mandatory posting. Use a two-tailed test with α = .05 to determine whether the mean number of calories after the posting is significantly different than before calorie content was posted. (Use a t Distributions table to find the critical values of t.)

Because n is the same for both samples, we can add the sum of squares for both samples and divide by 2. 852+9122=7,753852+9122=7,753. The estimated standard error is 12.45, and t = 14/12.45 = 1.12. With df = 198 the critical value is 1.97 (if you are using a distribution calculator or program), otherwise it is 1.98, as this is the larger value between 1.98 and 1.96 on the t-table. Fail to reject the null hypothesis and conclude that there was no significant change in calorie consumption after the mandatory posting. r² = t²/(t² + df) = 1.12²/(1.12² + 198) = 1.25/199.25 = 0.0063 or 0.63%

Construct a 90% confidence interval to estimate the size of the mean difference.

For 90% confidence with df = 14, the t values are ±1.761 and the interval extends from 4.34 - 1.761(2) = 0.818 to 4.34 + 1.761(2) = 7.862.

Are the neurological test scores significantly lower for the contact athletes than for the noncontact athletes in the control group? Use a one-tailed test with α = .05.

For a one-tailed test, the critical boundary is t = 1.761. For the noncontact athletes, M = 9 and SS = 44. For the contact athletes, M = 6 and SS = 56. The pooled variance is SS1 + SS2df1 + df2SS1 + SS2df1 + df2 = 44 + 567 + 744 + 567 + 7 = 1001410014 = 7.14, the estimated error is sM1 − M2M1 − M2 = √(s2pn1sp2n1 + s2pn2sp2n2) = √(7.1487.148 + 7.1487.148) = 1.34, and t(14) = M1 − M2sM1 − M2M1 − M2sM1 − M2 = 9 − 61.349 − 61.34 = 3.01.343.01.34 = 2.24. The data show that the contact athletes have significantly lower scores.

Explain the difference between a matched-subjects design and a repeated-measures design.

For a repeated-measures design, the same subjects are used in both treatment conditions. In a matched-subjects design, two different sets of subjects are used. However, in a matched-subjects design, each subject in one condition is matched with respect to a specific variable with a subject in the second condition so that the two separate samples are equivalent with respect to the variable the researcher would like to control.

Use a one-tailed test with α = .05 to determine whether these data are sufficient to conclude that high-creativity people are more likely to cheat than people with lower levels of creativity.

For these data, the pooled variance is s2pp2 = (SS₁ + SS₂) / (df₁ + df₂) = (749.5 + 830) / (26 + 26) = 1579.5/52 = 30.38, the estimated standard error is sM1 - M2M1 - M2 = √(s2pp2/n₁ + s2pp2/n₂) = √(30.38/27 + 30.38/27) = √(1.125 + 1.125) = √2.25 = 1.50, and t(52) = (M₁ - M₂)/sM1 - M2M1 - M2 = (7.41 - 4.78)/1.50 = 2.63/1.50 = 1.75. With df = 52 and α = .05, the critical value is t = 1.675. The sample mean difference is in the right direction and is large enough to be significant. Reject H₀.

Participants enter a research study with unique characteristics that produce different scores from one person to another. For an independent-measures study, these individual differences can cause problems. How can these problems be eliminated or reduced with a repeated-measures study?

In a repeated-measures study, individual differences do not add to the variance of the scores because the same participants are used in all treatment conditions. In a repeated-measures study, both samples consist of the same individuals, eliminating systematic differences between the two samples, which bias the results.

How does larger sample size affect the magnitude of the standard error for the sample mean difference?

Larger sample size produces a smaller standard error.

Is there a significant difference in reported pain between the two conditions? Use a two-tailed test with α = .01.

M₁ = ∑Xn₁∑Xn₁ = 689689 = 7.56 M₂ = ∑Xn₂∑Xn₂ = 10291029 = 11.33 SS₁ = ∑(X - M₁)² = 42.23 SS₂ = ∑(X - M₂)² = 38.00 s2pp2 = 42.23 + 388 + 842.23 + 388 + 8 = 80.231680.2316 = 5.01 s(M1 - M2)(M1 - M2) = √(5.01/9 + 5.01/9) = √(0.557 + 0.557) = √1.114 = 1.055 t = 7.56 - 11.33 1.0557.56 - 11.33 1.055 = −3.771.055−3.771.055 = -3.573 With critical boundaries of ±2.921, reject H₀; there is a significant difference in reported pain between the two conditions.

Estimated Cohen's d

The estimated Cohen's d = (M₁ - M₂) / √s2pp2 = (7.41 - 4.78) / √30.38 = 2.63/5.51 = 0.48

Compute Cohen's d to measure the size of the treatment effect.

The estimated Cohen's d is MDsMDs = 2.632.63 = 0.867.

A repeated-measures study with a sample of n = 16 participants produces a mean difference of MDD = 4 with a standard deviation of s = 8. Use a two-tailed hypothesis test with α = .05 to determine whether it is likely that this sample came from a population with μDD = 0.

The estimated standard error is sMDMD = s√ns√n = 8484 = 2.00 points, and t(15) = MD - μDsMDMD - μDsMD = 42.0042.00 = 2.000. With a critical boundary of ±2.131, the results indicate failure to reject the null hypothesis.

One sample has SS = 60 and a second sample has SS = 48. If n = 7 for both samples, find each of the sample variances and compute the pooled variance. Because the samples are the same size, you should find that the pooled variance is exactly halfway between the two sample variances.

The first sample has s² = SS1df1SS1df1 = 606606 = 10.00, and the second has s² = SS2df2SS2df2 = 486486 = 8.00. The pooled variance is s2pp2 = SS1 + SS2df1 + df2SS1 + SS2df1 + df2 = 1081210812 = 9.00 (halfway between).

Describe the homogeneity of variance assumption. Why is homogeneity of variance important for the independent-measures t test?

The homogeneity of variance assumption states that the variances are equal for both populations being tested. If homogeneity of variance is violated, the sample variances cannot be meaningfully pooled

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 a 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 scored higher by an average of MDD = 7 points with SS = 288. Do the data indicate a significant difference between the two conditions? Use a two-tailed test with α = .05.

The null hypothesis says that changing answers has no effect, H₀: μDD = 0. With df = 8 and α = .05, the critical values are t = ±2.306. For these data, the standard error is sMDMD = s/√n = 6/3 = 2, and t(8) = (MDD - μDD)/ sMDMD = (7 - 0)/2 = 3.50. Reject H₀ and conclude that changing answers has a significant effect on exam performance.

For n = 25 participants, the mean difference between the two types of advertisements is MDD = 1.32 points (neutral ads rated higher) with SS = 150 for the difference scores. Is this result sufficient to conclude that there is a significant difference in the ratings for two types of advertisements? Use a two-tailed test with α = .05.

The null hypothesis states that there is no difference in the ratings for two types of advertisements; as such, μDμD = 0. For these data the sample variance is s² = SS / (n - 1) = 150/24 = 6.25, the estimated standard error is sMDMD = s/√n = 2.5/5 = 0.50 and t(24) = MD−μDsMDMD−μDsMD = (1.32 - 0)/0.50 = 2.64. With df = 24, the critical value is 2.064. Reject the null hypothesis and conclude that there is a significant difference in the ratings for two types of advertisements.

For each of the following, assume that the two samples are obtained from populations with the same mean, and calculate how much difference should be expected, on average, between the two sample means. Each sample has n = 4 scores with s² = 68 for the first sample and s² = 76 for the second. (Note: Because the two samples are the same size, the pooled variance is equal to the average of the two sample variances.)

The pooled variance is s2pp2 = (s2112 + s2222) / 2 = (68 + 76) / 2 = 144/2 = 72. The estimated standard error for the sample mean difference is sM1 - M2M1 - M2 = √(s2pp2/n1 + s2pp2/n2) = √36 = 6 points.

For each of the following, calculate the pooled variance and the estimated standard error for the sample mean difference. The first sample has n = 4 scores and a variance of s² = 17, and the second sample has n = 8 scores and a variance of s² = 27.

The pooled variance is s2pp2 = df1s21 + df2s22df1 + df2df1s12 + df2s22df1 + df2 = (3)(17)+(7)(27)3 + 7(3)(17)+(7)(27)3 + 7 = 51 + 1891051 + 18910 = 2401024010 = 24.00. Estimated standard error is s(M1 - M2)(M1 - M2) = √(s2pn1sp2n1 + s2pn2sp2n2) = √(24.00424.004 + 24.00824.008) = √(6.00 + 3.00) = √9.00 = 3.00.

An increase in the number of scores in each sample.

The size of the two samples influences the magnitude of the estimated standard error in the denominator of the t statistic. As sample size increases, the value of t also increases (moves farther from zero), and the likelihood of rejecting H₀ also increases; however, there is little or no effect on measures of effect size.

Describe what is measured by the estimated standard error in the bottom of the independent-measures t statistic.

The standard error for the independent-measures t provides an estimate of the standard distance between a sample mean difference (M₁ - M₂) and the population mean difference (μ₁ - μ₂). When the two samples come from populations with the same mean (in other words, when H₀ is true), the standard error indicates the standard amount of error (distance) between two sample means (M₁ - M₂ - [μ₁ - μ₂] = M₁ - M₂ - 0 = M₁ - M₂).

An increase in the variance for each sample.

The variability of the scores influences the estimated standard error in the denominator. As the variability of the scores increases, the value of t decreases (becomes closer to zero), and the likelihood of rejecting H₀ decreases; also, measures of effect size decrease.

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 eight weeks with the medication.

Yes, a repeated-measures t test is the appropriate analysis. There are two scores (before and after) for each individual.

Compute the value of r² (percentage of variance accounted for) for these data.

r² = t2t2 + dft2t2 + df = 5.01819.0185.01819.018 = 0.264 (26.4).

The lower economic class participants were significantly more generous with their points compared with the upper-class individuals. Results similar to those found in the study show that n = 12 lower-class participants shared an average of M = 5.2 points with SS = 11.91, compared to an average of M = 4.3 with SS = 9.21 for the n = 12 upper-class participants.

s2pp2 = (11.91 + 9.21) / (11 + 11) = 21.12/22 = 0.96 s(M1 - M2)(M1 - M2) = √(0.96/12 + 0.96/12) = √(0.08 + 0.08) = √0.16 = 0.4 t = (5.2 - 4.3) / 0.4 = 2.25 With critical boundaries of ±2.074, reject H₀; there is a significant mean difference between the two economic populations.

Is there a significant difference in performance on new problems for these two groups? Use a two-tailed test with α = .05.

s2pp2 = 108 + 1167 + 7108 + 1167 + 7 = 2241422414 = 16 s(M1 - M2)(M1 - M2) = √(16/8 + 16/8) = √(2 + 2) = √4 = 2 t = 10.50 - 6.16210.50 - 6.162 = 4.3424.342 = 2.17 With critical boundaries of ±2.145, reject H₀; there is a significant mean difference between the results of the two methods of instruction

Suppose that a researcher intends to examine this phenomenon using a sample of 20 high school students. The researcher first surveys the students' parents to obtain information on the family's TV viewing habits during the time that the students were 5 years old. Based on the survey results, the researcher selects a sample of n = 10 students with a history of watching Sesame Street and a sample of n = 10 students who did not watch the program. The average high school grade is recorded for each student, and the data are as follows:

s2pp2 = SS1+SS2df1+df2SS1+SS2df1+df2= 200 + 1609 + 9200 + 1609 + 9 = 3601836018 = 20 sM1 − M2M1 − M2 = √(s2pn+s2pnsp2n+sp2n) = √(20102010 + 20102010) = √(2 + 2) = √4 = 2 t = 93 − 85293 − 852 = 8282 = 4.00 With critical boundaries of ±2.878, reject H₀; there is a significant difference in reported performance between the two groups of students.

Two samples are selected from the same population. For each of the following, calculate how much difference is expected, on average, between the two sample means. One sample has n = 4, the second has n = 6, and the pooled variance is 60.

sM1 − M2M1 − M2 = √(s2pp2/n₁ + s2pp2/n₂) = √(60/4 + 60/6) = √(15 + 10) = √25 = 5

Explain how the size of the sample mean difference influences the likelihood of finding a significant mean difference.

the null hypothesis states that there is no difference between the means of the two groups, the larger the difference between the sample means, the more likely the difference is statistically significant.

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 MDD = 4.8 minutes with a variance of s² = 125 for the sample of difference scores. Is this result sufficient to conclude that Ritalin significantly improves attention span? Use a one-tailed test with α = .05.

μDμD = 0 because the null hypothesis states that the difference between the means is 0. The estimated standard error is sMDMD = s/√n = 11.18/4.47 = 2.5 and t(19) = (MD−μDD−μD)/sDD = (4.8 - 0)/2.5 = 1.920. With a critical value of 1.729, reject the null hypothesis and conclude that attention span is significantly longer with the medication.

n = 16 male athletes during the first three weeks of the semester and recorded the number of minutes that each student was late to class. The athletes were then asked to begin texting their arrival at the classroom, and the researcher continued to monitor attendance for another three weeks. For each athlete, the average lateness for the first three weeks and for the second three weeks were calculated, and the difference score was recorded. The data showed that lateness to class decreased by an average of MDD = 21 minutes with SS = 2940 when the students were texting. Use a two-tailed test with α = .01 to determine whether texting produced a significant change in attendance.

μDμD = 0 because the null hypothesis states that the difference between the means is 0. The sample variance is s² = SSn - 1SSn - 1 = 294015294015 = 196, the estimated standard error is sMDMD = √s2ns2n = √1961619616 = 3.5, and t(15) = MD−μDsMDMD−μDsMD = (21 - 0)3.5(21 - 0)3.5 = 6.000. With a critical value of ±2.947, reject the null hypothesis and conclude that texting produced a significant change in attendance.

A sample of n = 12 males looks at a set of 30 photographs of women and rates the attractiveness of each woman using a 5-point scale (5 = most positive). 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 MDD = 1.2 points higher than the photograph with the tattoo, with SS = 33 for the difference scores.

μDμD = 0 because the null hypothesis states that the difference between the means is 0. The sample variance is s² = SSn - 1SSn - 1 = 33113311 = 3, the estimated standard error is sMDMD = √s2ns2n = √312312 = √1414 = 0.50, t(11) = MD−μDsMDMD−μDsMD = 1.2 - 00.501.2 - 00.50= 2.40. With critical boundaries of ±2.201, reject H₀. The tattoo has a significant effect on attractiveness ratings.

In a similar study, n = 16 participants were asked to rate the quality of low-priced items under two scenarios: purchased by a friend or purchased yourself. The results produced a mean difference of MDD = 2.6 and SS = 135, with self-purchases rated higher. Is the judged quality of objects significantly different for self-purchases than for purchases made by others? Use a two-tailed test with α = .05 and the Distributions tool.

μDμD = 0 because the null hypothesis states that the difference between the means is 0. The sample variance is s² = SSn - 1SSn - 1 = 1351513515 = 9, the estimated standard error is sMDMD = √s2ns2n = √916916 = 3434 = 0.750, and t(15) = MD−μDsMDMD−μDsMD = 2.6 - 00.752.6 - 00.75 = 3.467. With critical boundaries of ±2.131, reject H₀.