Lerven's test of equality of variances shows: 1. The result is significant, so the variances are not different 2. The result is significant, so the variances are different 3. The result is non-significant, so the variances are not different 4. The result is non-significant, so the variances are different From the Levene’s test result, what might you conclude about the use of the t test? 1. The assumption of equal variances is met, so the use of the t test may be appropriate 2. The assumption of equal variances is not met, so the use of the t test may be inappropriate 3. The assumption of equal variances is not met, so the use of the t test may be appropriate 4. The assumption of equal variances is met, so the use of the t test may be inappropriate
Lerven's test of equality of variances shows: 1. The result is significant, so the variances are not different 2. The result is significant, so the variances are different 3. The result is non-significant, so the variances are not different 4. The result is non-significant, so the variances are different From the Levene’s test result, what might you conclude about the use of the t test? 1. The assumption of equal variances is met, so the use of the t test may be appropriate 2. The assumption of equal variances is not met, so the use of the t test may be inappropriate 3. The assumption of equal variances is not met, so the use of the t test may be appropriate 4. The assumption of equal variances is met, so the use of the t test may be inappropriate
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Lerven's test of equality of variances shows:
1. The result is significant, so the variances are not different
2. The result is significant, so the variances are different
3. The result is non-significant, so the variances are not different
4. The result is non-significant, so the variances are different
From the Levene’s test result, what might you conclude about the use of the t test?
1. The assumption of equal variances is met, so the use of the t test may be appropriate
2. The assumption of equal variances is not met, so the use of the t test may be inappropriate
3. The assumption of equal variances is not met, so the use of the t test may be appropriate
4. The assumption of equal variances is met, so the use of the t test may be inappropriate
![Group Statistics
Std. Error
Age in Years
N
Mean
Std. Deviation
Mean
Math Achievement
Lower than 13
556
50.7216
12.10574
.51340
Higher than 14
12.51362
.59454
443
50.9457
Independent Samples Test
Levene's Test for Equality of
Variances
t-test for Equality of Means
95% Confidence Interval of the
Difference
Significance
Mean
Std. Error
Sig.
t
df
One-Sided p
Two-Sided p
Difference
Difference
Lower
Upper
Math Achievement
Equal variances
.671
.413
-.286
997
.387
.775
-.22410
.78259
-1.75981
1.31160
assumed
Equal variances not
assumed
-.285
933.546
.388
.775
-.22410
.78553
-1.76571
1.31750
Independent Samples Effect Sizes
95% Confidence Interval
Point
Estimate
Standardizer
Lower
Upper
Math Achievement
Cohen's d
12.28823
-.018
-.143
.107
Hedges' correction
12.29749
-.018
-.143
.107
Glass's delta
12.51362
-.018
-143
.107
a. The denominator used in estimating the effect sizes.
Cohen's d uses the pooled standard deviation.
Hedges' correction uses the pooled standard deviation, plus a correction factor.
Glass's delta uses the sample standard deviation of the control group.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faa525447-a133-46b9-9d27-4fb7931f0eae%2F61452a89-31e3-423a-aef4-296520b11596%2Fygngpsi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Group Statistics
Std. Error
Age in Years
N
Mean
Std. Deviation
Mean
Math Achievement
Lower than 13
556
50.7216
12.10574
.51340
Higher than 14
12.51362
.59454
443
50.9457
Independent Samples Test
Levene's Test for Equality of
Variances
t-test for Equality of Means
95% Confidence Interval of the
Difference
Significance
Mean
Std. Error
Sig.
t
df
One-Sided p
Two-Sided p
Difference
Difference
Lower
Upper
Math Achievement
Equal variances
.671
.413
-.286
997
.387
.775
-.22410
.78259
-1.75981
1.31160
assumed
Equal variances not
assumed
-.285
933.546
.388
.775
-.22410
.78553
-1.76571
1.31750
Independent Samples Effect Sizes
95% Confidence Interval
Point
Estimate
Standardizer
Lower
Upper
Math Achievement
Cohen's d
12.28823
-.018
-.143
.107
Hedges' correction
12.29749
-.018
-.143
.107
Glass's delta
12.51362
-.018
-143
.107
a. The denominator used in estimating the effect sizes.
Cohen's d uses the pooled standard deviation.
Hedges' correction uses the pooled standard deviation, plus a correction factor.
Glass's delta uses the sample standard deviation of the control group.
![How would you report the results?
The derived t = 0.41 was significant at p = .05 with df = 997. Therefore, Ho was
rejected, and it was concluded that the mean math achievement score for the older
students (12.11) was significantly lower than the mean math achievement score for the
younger students (50.72), t(997) = 0.41, p < .001. In terms of the research question, it
appears that younger and older students are different in their level of math
achievement.
The derived t = 0.78 was significant at p = .05 with df = 997. Therefore, Ho was
rejected, and it was concluded that the mean math achievement score for the older
students (12.51) was significantly lower than the mean math achievement score for the
younger students (50.95), t(997) = 0.78, p < .01. In terms of the research question, it
appears that younger students are better at math than older students.
The derived t = -0.29 was significant at p = .05 with df = 997. Therefore, Ho was
rejected, and it was concluded that the mean math achievement score for the older
students (50.72) was significantly higher than the mean math achievement score for the
younger students (12.11), t(997) = -0.29, p < .000. In terms of the research question, it
appears that older students are better at math than younger students.
The derived t = 0.67 was not significant at p = .05 with df = 997. Therefore, Ho was not
rejected, and it was concluded that the mean math achievement score for the older
students (50.72) was not different than the mean math achievement score for the
younger students (50.95), t(997) = 0.67, p > .05. In terms of the research question, it
appears that younger and older students are not different in their level of math
achievement.
The derived t = -0.29 was not significant at p = .05 with df = 997. Therefore, Ho was
not rejected, and it was concluded that the mean math achievement score for the older
students (50.72) was not different than the mean math achievement score for the
younger students (50.95), t(997) = -0.29, p > .05. In terms of the research question, it
appears that younger and older students are not different in their level of math
achievement.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faa525447-a133-46b9-9d27-4fb7931f0eae%2F61452a89-31e3-423a-aef4-296520b11596%2F0hpgklq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:How would you report the results?
The derived t = 0.41 was significant at p = .05 with df = 997. Therefore, Ho was
rejected, and it was concluded that the mean math achievement score for the older
students (12.11) was significantly lower than the mean math achievement score for the
younger students (50.72), t(997) = 0.41, p < .001. In terms of the research question, it
appears that younger and older students are different in their level of math
achievement.
The derived t = 0.78 was significant at p = .05 with df = 997. Therefore, Ho was
rejected, and it was concluded that the mean math achievement score for the older
students (12.51) was significantly lower than the mean math achievement score for the
younger students (50.95), t(997) = 0.78, p < .01. In terms of the research question, it
appears that younger students are better at math than older students.
The derived t = -0.29 was significant at p = .05 with df = 997. Therefore, Ho was
rejected, and it was concluded that the mean math achievement score for the older
students (50.72) was significantly higher than the mean math achievement score for the
younger students (12.11), t(997) = -0.29, p < .000. In terms of the research question, it
appears that older students are better at math than younger students.
The derived t = 0.67 was not significant at p = .05 with df = 997. Therefore, Ho was not
rejected, and it was concluded that the mean math achievement score for the older
students (50.72) was not different than the mean math achievement score for the
younger students (50.95), t(997) = 0.67, p > .05. In terms of the research question, it
appears that younger and older students are not different in their level of math
achievement.
The derived t = -0.29 was not significant at p = .05 with df = 997. Therefore, Ho was
not rejected, and it was concluded that the mean math achievement score for the older
students (50.72) was not different than the mean math achievement score for the
younger students (50.95), t(997) = -0.29, p > .05. In terms of the research question, it
appears that younger and older students are not different in their level of math
achievement.
Expert Solution
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Step 1
Introduction:
Denote σ12, σ22 as the true population variances of the two groups of interest.
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