2. Describe what is measured by the estimated standard error in the bottom of the independentmeasures t statistic. 4. One sample has SS= 60 and a second sample has SS=48. a. If n=7 for both samples, find each of the sample variances and calculate the pooled variance. b. Assume n=7 for the first sample and n=5 for the second. Again, calculate the sample variances and the pooled variance. 6. Two separate samples receive different treatments. After treatment, the first sample has n=9 with SS=462 and the second has n=7 and the SS=420. a. Compute pooled variances for the two samples. b. Calculate the estimated standard error for the sample mean difference. c. If the sample mean difference is 10 points, is this enough to reject the null hypothesis using a two-tailed test with an alpha = .05? 8. In an attempt to examine a phenomenon, a researcher surveys a group of college students to determine the negative life events they experienced in the past 5 years and their current feelings. For n=18 participants with 2 or fewer negative experiences, the average well being score M=42 with SS=398 and for n=16 participants with 5 to 10 negative experiences, average score is M=48.6 with SS=370. a. Is there a significant difference between the two populations? Use a two tailed test with an alpha= .01 b. Compute Cohen's D to measure the size of the effect. c. Write a sentence demonstrating how the outcome of the hypothesis test and the measure of effect sixe would appear in a research report. 10. Estimated Speed Smashed Into Hit N=15 n=15 M=40.8 M=34.0 SS=510 SS=414 a. Do the results indicate a signifigantly higher estimated speed for the \"smashed into\" group? Use a one tailed test with alpha =.01 b. Compute the estimated value for Cohen's D to measure the size of the effect. c. Write a sentence demonstrating how the outcome of the hypothesis test and the measure of effect size would appear in a research report. 12. Recent research has demonstrated that music-based physical training for elderly people can improve balance and walking efficiency and reduce the risk of falls (Trombetti et al., 2011). As part of the training, participants walked in time to music and responded to changes in the music's rhythm during a 1-hour per week exercise program. After 6 months, participants in the training group increased their walking speed and their stride length compared to individuals in the control group. The following data are similar to the results obtained in the study. Exercise Group Control Group Stride length Stride length 24, 25, 22, 24 26, 23, 20, 23 26, 17, 21, 22 20, 16, 21, 17 22, 19, 24, 23 18, 23, 16, 20 23, 28, 25, 23 25, 19, 17, 16 a. Do the results indicate a significant difference in the stride length for the two groups? Use a two-tailed test with =.05 14. In the Chapter Preview we presented a study showing that handling money reduces the perception pain (Zhou,Vohs, &Baumeisyer,2009).In the experiment, a group of college students was told that they were participating in a manual dexterity study. Half of the students were given a stack of money to count and the other half got a stack of blank pieces of paper. After the counting task, the participants were asked to dip their hands into bowls of very hot water (122 F) and rate how uncomfortable it was. The following data show rating of pain similar to the results obtained in the study. Counting Money 7 8 10 6 8 5 7 12 5Counting Paper 9 11 13 10 11 9 15 14 10 a) Is there a significant differences in reported pain between the two conditions? Use a two-tailed test with a=.01. b) Compute Cohen's d to estimate the size of the treatment effect. 16. A researcher conducts an independent-measures study comparing two treatments and reports the t statistic as t(25)=2.071. a. How many individuals partipated in the entire study? B. Using a two-tailed test with an alpha= .05, is there a signifigant difference between the two treatments? C. Compute r2 (r squared) to measure the percentage of variance accounted for by the treatment effect. 18. Describe the homogeneity of variance assumption and explain why it is important for the independent measures t-test. 20. As noted on page 304, when the two population means are equal, the estimated standard error for the independent-measures t test provides a measure of how much difference to expect between two sample means. For each of the following situations, assume that u_1=u_2 and calculate how much difference should be expected between the two sample means. a. One sample has n = 6 scores with SS = 75 and the second sample has n = 10 scores with SS =135. b. One sample has n = 6 scores with SS = 310 and the second sample has n = 10 scores with SS = 530. c. In part b, the samples have larger variability (bigger SS values) than in part a, but the sample sizes are unchanged. How does larger variability affect the size of the standard error for the sample mean difference? 22. 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. A. Each sample has n=4 scores with = 68 for the first sample and = 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.) ___ (Options are: 4.24, 0.24, 8.48, 6.00) B. each sample has n=16 scores with = 68 for the first sample and (Options include: 0.12, 2.12, 4.24,3.00) = 76 for the second. ___ C. How does the sample size affect the size of the standard error for the sample mean difference? 2. Describe what is measured by the estimated standard error in the bottom of the independentmeasures t statistic. Total amount of error involved in using two sample means to approximate two population means 4. One sample has SS= 60 and a second sample has SS=48. a. If n=7 for both samples, find each of the sample variances and calculate the pooled variance. SS1 = 60 SS2 = 48 n1 = 7 n2 = 7 s 12= SS 1 60 60 = = =10 n 11 71 6 s 22= SS 2 48 48 = = =8 n 21 71 6 Since sample sizes are equal, Pooled variance will be given by 1 pooled s 2= (s12 +s 22) 2 1 = 2 ( 10+8 )=9 b. Assume n=7 for the first sample and n=5 for the second. Again, calculate the sample variances and the pooled variance. SS1 = 60 SS2 = 48 n1 = 7 n2 = 5 s1 = SS 1 60 60 = = =10 n 11 71 6 s 22= SS 2 48 48 = = =12 n 21 51 4 2 Since sample sizes are NOT equal, Pooled variance will be given by ( n11 ) s12 +( n 21 ) s 22 6 10+ 4 12 108 s= = = =10.8 2 n1 +n 22 10 10 6. Two separate samples receive different treatments. After treatment, the first sample has n=9 with SS=462 and the second has n=7 and the SS=420. a. Compute pooled variances for the two samples. n1 = 9 n2 = 7 SS1 = 462 SS2 = 420 SS 1+ SS 2 462+ 420 Pooled variance n1+n 22 = 9+72 =63 b. Calculate the estimated standard error for the sample mean difference. Estimated standard error for the sample mean difference = 63 s2 ( n1 + n1 ) 1 2 ( 19 + 17 )= 16=4 c. If the sample mean difference is 10 points, is this enough to reject the null hypothesis using a two-tailed test with an alpha = .05? t= sample mean difference 10 = =2.5 standard error 4 d.f=n1+n2-2=9+7-2=14 Critical value=t0.025, 14= 2.145 Computed value of t=2.5 is > 2.16 T is significant. The information is sufficient to reject the null hypothesis. 8. In an attempt to examine a phenomenon, a researcher surveys a group of college students to determine the negative life events they experienced in the past 5 years and their current feelings. For n=18 participants with 2 or fewer negative experiences, the average well being score M=42 with SS=398 and for n=16 participants with 5 to 10 negative experiences, average score is M=48.6 with SS=370. a. Is there a significant difference between the two populations? Use a two tailed test with an alpha= .01 H0: 1=2 Ha: 12 Pooled variance SS 1+ SS 2 398+370 = =24 n1+n 22 18+162 Standard error = s2 ( n1 + n1 ) 1 2 = ( 181 + 161 )=1.6833 24 Mean difference=48.6-42=6.6 Test statistics t= sample mean difference 6.6 = =3.92 standard error 1.6833 d.f=n1+n2-2=18+16-2=32 Critical value=t0.005, 32= 2.74 Computed value of t=3.92 is > 2.74 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there a significant difference between the two populations b. Compute Cohen's D to measure the size of the effect. 6.6 = = 24 1.3472 M 2M 1 Cohen's D= SD pooled c. Write a sentence demonstrating how the outcome of the hypothesis test and the measure of effect sixe would appear in a research report. Statistical analysis indicates significantly significant difference between the two populations t(32) = 3.92 , p .05, Cohen's d = 1.3472. 10. Estimated Speed Smashed Into Hit N=15 n=15 M=40.8 M=34.0 SS=510 SS=414 a. Do the results indicate a signifigantly higher estimated speed for the \"smashed into\" group? Use a one tailed test with alpha =.01 H0: 1=2 Ha: 1>2 SS 1+ SS 2 510+ 414 Pooled variance n1+n 22 = 15+152 =33 Standard error = s2 ( 1 1 + n1 n2 ) = Mean difference=40.8-34=6.8 Test statistics t= ( 151 + 151 )=2.0976 33 sample mean difference 6.8 = =3.2418 standard error 2 .0976 d.f=n1+n2-2=15+15-2=30 Critical value=t0.01, 30= 2.4573 Computed value of t=3.242 is > 2.4573 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is a significantly higher estimated speed for the \"smashed into\" group b. Compute the estimated value for Cohen's D to measure the size of the effect. 6.6 M 2M 1 = Cohen's D= SD pooled = 33 1.184 c. Write a sentence demonstrating how the outcome of the hypothesis test and the measure of effect size would appear in a research report. Statistical analysis indicates that there is a significantly higher estimated speed for the \"smashed into\" group t(30) = 3.2 42 , p .01, Cohen's d = 1.184 12. Recent research has demonstrated that music-based physical training for elderly people can improve balance and walking efficiency and reduce the risk of falls (Trombetti et al., 2011). As part of the training, participants walked in time to music and responded to changes in the music's rhythm during a 1-hour per week exercise program. After 6 months, participants in the training group increased their walking speed and their stride length compared to individuals in the control group. The following data are similar to the results obtained in the study. Exercise Group Control Group Stride length Stride length 24, 25, 22, 24 26, 23, 20, 23 26, 17, 21, 22 20, 16, 21, 17 22, 19, 24, 23 18, 23, 16, 20 23, 28, 25, 23 25, 19, 17, 16 a. Do the results indicate a significant difference in the stride length for the two groups? Use a two-tailed test with =.05 H0: 1=2 Ha: 12 Mean Variance Observations Exercise Control group group 23 20 6.933333 10.66667 16 10.66667+ 1 Pooled variance= 6.933333)=8.8 2 16 Standard error= 8.8( 1 1 + ) 16 16 =1.05 Mean difference=23-20=3 3 Test statistics= 1.05 =2.86 d.f=16+16-2=3 t-critical=2.042 Computed value of t=2.86is > 2.042 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is significant difference in the stride length for the two groups 14. In the Chapter Preview we presented a study showing that handling money reduces the perception pain (Zhou,Vohs, &Baumeisyer,2009).In the experiment, a group of college students was told that they were participating in a manual dexterity study. Half of the students were given a stack of money to count and the other half got a stack of blank pieces of paper. After the counting task, the participants were asked to dip their hands into bowls of very hot water (122 F) and rate how uncomfortable it was. The following data show rating of pain similar to the results obtained in the study. Counting Money 7 8 10 6 8 5 7 12 5 Counting Paper 9 11 13 10 11 9 15 14 10 a) Is there a significant differences in reported pain between the two conditions? Use a twotailed test with a=.01. H0: 1=2 Ha: 12 Mean Variance Counting Counting Money Paper 7.555556 11.33333 5.277778 4.75 Observations 9 9 1 5.2778 + 4.75)=5.014 Pooled variance= 2 Standard error= 1 1 5.014( + ) =1.056 9 9 Mean difference=11.33333- 7.555556=3.78 3.78 =3.58 Test statistics= 1.05 d.f=9+9-2=15 t-critical =2.12 Computed value of t=3.58 > 2.12 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is significant differences in reported pain between the two conditions b) Compute Cohen's d to estimate the size of the treatment effect. M 2M 1 Cohen's D= SD pooled = 3.78 = 1.69 5.014 16. A researcher conducts an independent-measures study comparing two treatments and reports the t statistic as t(25)=2.071. a. How many individuals partipated in the entire study? N1+n2=25+2=27 B. Using a two-tailed test with an alpha= .05, is there a signifigant difference between the two treatments? T-statistics= 2.060 Since t 2.16 T is significant. The information is sufficient to reject the null hypothesis. 8. In an attempt to examine a phenomenon, a researcher surveys a group of college students to determine the negative life events they experienced in the past 5 years and their current feelings. For n=18 participants with 2 or fewer negative experiences, the average well being score M=42 with SS=398 and for n=16 participants with 5 to 10 negative experiences, average score is M=48.6 with SS=370. a. Is there a significant difference between the two populations? Use a two tailed test with an alpha= .01 H0: 1=2 Ha: 12 Pooled variance SS 1+ SS 2 398+370 = =24 n1+n 22 18+162 Standard error = s2 ( n1 + n1 ) 1 2 = ( 181 + 161 )=1.6833 24 Mean difference=48.6-42=6.6 Test statistics t= sample mean difference 6.6 = =3.92 standard error 1.6833 d.f=n1+n2-2=18+16-2=32 Critical value=t0.005, 32= 2.74 Computed value of t=3.92 is > 2.74 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there a significant difference between the two populations b. Compute Cohen's D to measure the size of the effect. 6.6 = = 24 1.3472 M 2M 1 Cohen's D= SD pooled c. Write a sentence demonstrating how the outcome of the hypothesis test and the measure of effect sixe would appear in a research report. Statistical analysis indicates significantly significant difference between the two populations t(32) = 3.92 , p .05, Cohen's d = 1.3472. 10. Estimated Speed Smashed Into Hit N=15 n=15 M=40.8 M=34.0 SS=510 SS=414 a. Do the results indicate a signifigantly higher estimated speed for the \"smashed into\" group? Use a one tailed test with alpha =.01 H0: 1=2 Ha: 1>2 SS 1+ SS 2 510+ 414 Pooled variance n1+n 22 = 15+152 =33 Standard error = s2 ( 1 1 + n1 n2 ) = Mean difference=40.8-34=6.8 Test statistics t= ( 151 + 151 )=2.0976 33 sample mean difference 6.8 = =3.2418 standard error 2 .0976 d.f=n1+n2-2=15+15-2=30 Critical value=t0.01, 30= 2.4573 Computed value of t=3.242 is > 2.4573 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is a significantly higher estimated speed for the \"smashed into\" group b. Compute the estimated value for Cohen's D to measure the size of the effect. 6.6 M 2M 1 = Cohen's D= SD pooled = 33 1.184 c. Write a sentence demonstrating how the outcome of the hypothesis test and the measure of effect size would appear in a research report. Statistical analysis indicates that there is a significantly higher estimated speed for the \"smashed into\" group t(30) = 3.2 42 , p .01, Cohen's d = 1.184 12. Recent research has demonstrated that music-based physical training for elderly people can improve balance and walking efficiency and reduce the risk of falls (Trombetti et al., 2011). As part of the training, participants walked in time to music and responded to changes in the music's rhythm during a 1-hour per week exercise program. After 6 months, participants in the training group increased their walking speed and their stride length compared to individuals in the control group. The following data are similar to the results obtained in the study. Exercise Group Control Group Stride length Stride length 24, 25, 22, 24 26, 23, 20, 23 26, 17, 21, 22 20, 16, 21, 17 22, 19, 24, 23 18, 23, 16, 20 23, 28, 25, 23 25, 19, 17, 16 a. Do the results indicate a significant difference in the stride length for the two groups? Use a two-tailed test with =.05 H0: 1=2 Ha: 12 Mean Variance Observations Exercise Control group group 23 20 6.933333 10.66667 16 10.66667+ 1 Pooled variance= 6.933333)=8.8 2 16 Standard error= 8.8( 1 1 + ) 16 16 =1.05 Mean difference=23-20=3 3 Test statistics= 1.05 =2.86 d.f=16+16-2=3 t-critical=2.042 Computed value of t=2.86is > 2.042 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is significant difference in the stride length for the two groups 14. In the Chapter Preview we presented a study showing that handling money reduces the perception pain (Zhou,Vohs, &Baumeisyer,2009).In the experiment, a group of college students was told that they were participating in a manual dexterity study. Half of the students were given a stack of money to count and the other half got a stack of blank pieces of paper. After the counting task, the participants were asked to dip their hands into bowls of very hot water (122 F) and rate how uncomfortable it was. The following data show rating of pain similar to the results obtained in the study. Counting Money 7 8 10 6 8 5 7 12 5 Counting Paper 9 11 13 10 11 9 15 14 10 a) Is there a significant differences in reported pain between the two conditions? Use a twotailed test with a=.01. H0: 1=2 Ha: 12 Mean Variance Counting Counting Money Paper 7.555556 11.33333 5.277778 4.75 Observations 9 9 1 5.2778 + 4.75)=5.014 Pooled variance= 2 Standard error= 1 1 5.014( + ) =1.056 9 9 Mean difference=11.33333- 7.555556=3.78 3.78 =3.58 Test statistics= 1.05 d.f=9+9-2=15 t-critical =2.12 Computed value of t=3.58 > 2.12 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is significant differences in reported pain between the two conditions b) Compute Cohen's d to estimate the size of the treatment effect. M 2M 1 Cohen's D= SD pooled = 3.78 = 1.69 5.014 16. A researcher conducts an independent-measures study comparing two treatments and reports the t statistic as t(25)=2.071. a. How many individuals partipated in the entire study? N1+n2=25+2=27 B. Using a two-tailed test with an alpha= .05, is there a signifigant difference between the two treatments? T-statistics= 2.060 Since t 2.16 T is significant. The information is sufficient to reject the null hypothesis. 8. In an attempt to examine a phenomenon, a researcher surveys a group of college students to determine the negative life events they experienced in the past 5 years and their current feelings. For n=18 participants with 2 or fewer negative experiences, the average well being score M=42 with SS=398 and for n=16 participants with 5 to 10 negative experiences, average score is M=48.6 with SS=370. a. Is there a significant difference between the two populations? Use a two tailed test with an alpha= .01 H0: 1=2 Ha: 12 Pooled variance SS 1+ SS 2 398+370 = =24 n1+n 22 18+162 Standard error = s2 ( n1 + n1 ) 1 2 = ( 181 + 161 )=1.6833 24 Mean difference=48.6-42=6.6 Test statistics t= sample mean difference 6.6 = =3.92 standard error 1.6833 d.f=n1+n2-2=18+16-2=32 Critical value=t0.005, 32= 2.74 Computed value of t=3.92 is > 2.74 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there a significant difference between the two populations b. Compute Cohen's D to measure the size of the effect. 6.6 = = 24 1.3472 M 2M 1 Cohen's D= SD pooled c. Write a sentence demonstrating how the outcome of the hypothesis test and the measure of effect sixe would appear in a research report. Statistical analysis indicates significantly significant difference between the two populations t(32) = 3.92 , p .05, Cohen's d = 1.3472. 10. Estimated Speed Smashed Into Hit N=15 n=15 M=40.8 M=34.0 SS=510 SS=414 a. Do the results indicate a signifigantly higher estimated speed for the \"smashed into\" group? Use a one tailed test with alpha =.01 H0: 1=2 Ha: 1>2 SS 1+ SS 2 510+ 414 Pooled variance n1+n 22 = 15+152 =33 Standard error = s2 ( 1 1 + n1 n2 ) = Mean difference=40.8-34=6.8 Test statistics t= ( 151 + 151 )=2.0976 33 sample mean difference 6.8 = =3.2418 standard error 2 .0976 d.f=n1+n2-2=15+15-2=30 Critical value=t0.01, 30= 2.4573 Computed value of t=3.242 is > 2.4573 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is a significantly higher estimated speed for the \"smashed into\" group b. Compute the estimated value for Cohen's D to measure the size of the effect. 6.6 M 2M 1 = Cohen's D= SD pooled = 33 1.184 c. Write a sentence demonstrating how the outcome of the hypothesis test and the measure of effect size would appear in a research report. Statistical analysis indicates that there is a significantly higher estimated speed for the \"smashed into\" group t(30) = 3.2 42 , p .01, Cohen's d = 1.184 12. Recent research has demonstrated that music-based physical training for elderly people can improve balance and walking efficiency and reduce the risk of falls (Trombetti et al., 2011). As part of the training, participants walked in time to music and responded to changes in the music's rhythm during a 1-hour per week exercise program. After 6 months, participants in the training group increased their walking speed and their stride length compared to individuals in the control group. The following data are similar to the results obtained in the study. Exercise Group Control Group Stride length Stride length 24, 25, 22, 24 26, 23, 20, 23 26, 17, 21, 22 20, 16, 21, 17 22, 19, 24, 23 18, 23, 16, 20 23, 28, 25, 23 25, 19, 17, 16 a. Do the results indicate a significant difference in the stride length for the two groups? Use a two-tailed test with =.05 H0: 1=2 Ha: 12 Mean Variance Observations Exercise Control group group 23 20 6.933333 10.66667 16 10.66667+ 1 Pooled variance= 6.933333)=8.8 2 16 Standard error= 8.8( 1 1 + ) 16 16 =1.05 Mean difference=23-20=3 3 Test statistics= 1.05 =2.86 d.f=16+16-2=3 t-critical=2.042 Computed value of t=2.86is > 2.042 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is significant difference in the stride length for the two groups 14. In the Chapter Preview we presented a study showing that handling money reduces the perception pain (Zhou,Vohs, &Baumeisyer,2009).In the experiment, a group of college students was told that they were participating in a manual dexterity study. Half of the students were given a stack of money to count and the other half got a stack of blank pieces of paper. After the counting task, the participants were asked to dip their hands into bowls of very hot water (122 F) and rate how uncomfortable it was. The following data show rating of pain similar to the results obtained in the study. Counting Money 7 8 10 6 8 5 7 12 5 Counting Paper 9 11 13 10 11 9 15 14 10 a) Is there a significant differences in reported pain between the two conditions? Use a twotailed test with a=.01. H0: 1=2 Ha: 12 Mean Variance Counting Counting Money Paper 7.555556 11.33333 5.277778 4.75 Observations 9 9 1 5.2778 + 4.75)=5.014 Pooled variance= 2 Standard error= 1 1 5.014( + ) =1.056 9 9 Mean difference=11.33333- 7.555556=3.78 3.78 =3.58 Test statistics= 1.05 d.f=9+9-2=15 t-critical =2.12 Computed value of t=3.58 > 2.12 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is significant differences in reported pain between the two conditions b) Compute Cohen's d to estimate the size of the treatment effect. M 2M 1 Cohen's D= SD pooled = 3.78 = 1.69 5.014 16. A researcher conducts an independent-measures study comparing two treatments and reports the t statistic as t(25)=2.071. a. How many individuals partipated in the entire study? N1+n2=25+2=27 B. Using a two-tailed test with an alpha= .05, is there a signifigant difference between the two treatments? T-statistics= 2.060 Since t 2.16 T is significant. The information is sufficient to reject the null hypothesis. 8. In an attempt to examine a phenomenon, a researcher surveys a group of college students to determine the negative life events they experienced in the past 5 years and their current feelings. For n=18 participants with 2 or fewer negative experiences, the average well being score M=42 with SS=398 and for n=16 participants with 5 to 10 negative experiences, average score is M=48.6 with SS=370. a. Is there a significant difference between the two populations? Use a two tailed test with an alpha= .01 H0: 1=2 Ha: 12 Pooled variance SS 1+ SS 2 398+370 = =24 n1+n 22 18+162 Standard error = s2 ( n1 + n1 ) 1 2 = ( 181 + 161 )=1.6833 24 Mean difference=48.6-42=6.6 Test statistics t= sample mean difference 6.6 = =3.92 standard error 1.6833 d.f=n1+n2-2=18+16-2=32 Critical value=t0.005, 32= 2.74 Computed value of t=3.92 is > 2.74 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there a significant difference between the two populations b. Compute Cohen's D to measure the size of the effect. 6.6 = = 24 1.3472 M 2M 1 Cohen's D= SD pooled c. Write a sentence demonstrating how the outcome of the hypothesis test and the measure of effect sixe would appear in a research report. Statistical analysis indicates significantly significant difference between the two populations t(32) = 3.92 , p .05, Cohen's d = 1.3472. 10. Estimated Speed Smashed Into Hit N=15 n=15 M=40.8 M=34.0 SS=510 SS=414 a. Do the results indicate a signifigantly higher estimated speed for the \"smashed into\" group? Use a one tailed test with alpha =.01 H0: 1=2 Ha: 1>2 SS 1+ SS 2 510+ 414 Pooled variance n1+n 22 = 15+152 =33 Standard error = s2 ( 1 1 + n1 n2 ) = Mean difference=40.8-34=6.8 Test statistics t= ( 151 + 151 )=2.0976 33 sample mean difference 6.8 = =3.2418 standard error 2 .0976 d.f=n1+n2-2=15+15-2=30 Critical value=t0.01, 30= 2.4573 Computed value of t=3.242 is > 2.4573 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is a significantly higher estimated speed for the \"smashed into\" group b. Compute the estimated value for Cohen's D to measure the size of the effect. 6.6 M 2M 1 = Cohen's D= SD pooled = 33 1.184 c. Write a sentence demonstrating how the outcome of the hypothesis test and the measure of effect size would appear in a research report. Statistical analysis indicates that there is a significantly higher estimated speed for the \"smashed into\" group t(30) = 3.2 42 , p .01, Cohen's d = 1.184 12. Recent research has demonstrated that music-based physical training for elderly people can improve balance and walking efficiency and reduce the risk of falls (Trombetti et al., 2011). As part of the training, participants walked in time to music and responded to changes in the music's rhythm during a 1-hour per week exercise program. After 6 months, participants in the training group increased their walking speed and their stride length compared to individuals in the control group. The following data are similar to the results obtained in the study. Exercise Group Control Group Stride length Stride length 24, 25, 22, 24 26, 23, 20, 23 26, 17, 21, 22 20, 16, 21, 17 22, 19, 24, 23 18, 23, 16, 20 23, 28, 25, 23 25, 19, 17, 16 a. Do the results indicate a significant difference in the stride length for the two groups? Use a two-tailed test with =.05 H0: 1=2 Ha: 12 Mean Variance Observations Exercise Control group group 23 20 6.933333 10.66667 16 10.66667+ 1 Pooled variance= 6.933333)=8.8 2 16 Standard error= 8.8( 1 1 + ) 16 16 =1.05 Mean difference=23-20=3 3 Test statistics= 1.05 =2.86 d.f=16+16-2=3 t-critical=2.042 Computed value of t=2.86is > 2.042 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is significant difference in the stride length for the two groups 14. In the Chapter Preview we presented a study showing that handling money reduces the perception pain (Zhou,Vohs, &Baumeisyer,2009).In the experiment, a group of college students was told that they were participating in a manual dexterity study. Half of the students were given a stack of money to count and the other half got a stack of blank pieces of paper. After the counting task, the participants were asked to dip their hands into bowls of very hot water (122 F) and rate how uncomfortable it was. The following data show rating of pain similar to the results obtained in the study. Counting Money 7 8 10 6 8 5 7 12 5 Counting Paper 9 11 13 10 11 9 15 14 10 a) Is there a significant differences in reported pain between the two conditions? Use a twotailed test with a=.01. H0: 1=2 Ha: 12 Mean Variance Counting Counting Money Paper 7.555556 11.33333 5.277778 4.75 Observations 9 9 1 5.2778 + 4.75)=5.014 Pooled variance= 2 Standard error= 1 1 5.014( + ) =1.056 9 9 Mean difference=11.33333- 7.555556=3.78 3.78 =3.58 Test statistics= 1.05 d.f=9+9-2=15 t-critical =2.12 Computed value of t=3.58 > 2.12 T is significant. The information is sufficient to reject the null hypothesis. Conclude that there is significant differences in reported pain between the two conditions b) Compute Cohen's d to estimate the size of the treatment effect. M 2M 1 Cohen's D= SD pooled = 3.78 = 1.69 5.014 16. A researcher conducts an independent-measures study comparing two treatments and reports the t statistic as t(25)=2.071. a. How many individuals partipated in the entire study? N1+n2=25+2=27 B. Using a two-tailed test with an alpha= .05, is there a signifigant difference between the two treatments? T-statistics= 2.060 Since t