Hello, Please assist me with confirming the following data is accurate, and answering the questions made bold.  Attached are 2 photos of the calculation for the charts being filled out. First-born children tend to develop language skills faster than their younger siblings. One possible explanation for this phenomenon is that first-born children have undivided attention from their parents. If this explanation is correct, then it is also reasonable that twins should show slower language development than single children and that triplets should be even slower. In 1937, Davis found exactly this result. The following hypothetical data demonstrate this relationship. The dependent variable is a measure of language skill at age three. Do the data provide evidence for significant differences between the three populations? Test at the .05 level. Single Child   9 8 9 7 9 Twin Child 7 6 7 4 9 Triplet Child 3 5 8 1 3 State the null and alternative hypothesis. Fill in the following table of descriptive statistics. Descriptive Statistics                                                       N        Mean            SD Single                                            5          8.4               .8944 Twin                                               5          6.6               1.8166 Triplet                                            5           4                 2.6458 Also, complete the following ANOVA table. Analysis of Variance of Sociability Data Source                                            SS             df         MS                     F   Between Groups                        48.9333         2          24.4667               6.61261 Within Groups                           44.4               12         3.7 If the F is statistically significant at p < .05 place an asterisk next to it and go on to compute a post-hoc test to determine which pairs of means are significantly different. Write a brief conclusion (interpretation) of the results

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Hello, Please assist me with confirming the following data is accurate, and answering the questions made bold.  Attached are 2 photos of the calculation for the charts being filled out.

First-born children tend to develop language skills faster than their younger siblings. One
possible explanation for this phenomenon is that first-born children have undivided attention from their parents. If this explanation is correct, then it is also reasonable that twins should show slower language development than single children and that triplets should be even slower. In 1937, Davis found exactly this result. The following hypothetical data demonstrate this relationship. The dependent variable is a measure of language skill at age three. Do the data provide evidence for significant differences between the three populations? Test at the .05 level.


Single Child  
9
8
9
7
9

Twin Child
7
6
7
4
9

Triplet Child

3
5
8
1
3


State the null and alternative hypothesis.




Fill in the following table of descriptive statistics.


Descriptive Statistics
                                                      N        Mean            SD
Single                                            5          8.4               .8944
Twin                                               5          6.6               1.8166
Triplet                                            5           4                 2.6458

Also, complete the following ANOVA table.


Analysis of Variance of Sociability Data

Source                                            SS             df         MS                     F  

Between Groups                        48.9333         2          24.4667               6.61261
Within Groups                           44.4               12         3.7


If the F is statistically significant at p < .05 place an asterisk next to it and go on to compute a post-hoc test to determine which pairs of means are significantly different.




Write a brief conclusion (interpretation) of the results.

The fratio value is 6.61261. The p-value is .01159. The result is significant at p<.05.
Post Hoc Tukey HSD (beta)
The Tukey's HSD (honestly significant difference) procedure facilitates pairwise comparisons within your
ANOVA data. The F statistic (above) tells you whether there is an overall difference between your sample
means. Tukey's HSD test allows you to determine between which of the various pairs of means - if any of
them - there is a signficant difference.
A couple of things to note. First, a blue value for Q (below) indicates a significant result. Second, it's worth
bearing in mind that there is some disagreement about whether Tukey's HSD is appropriate if the F-ratio
score has not reached significance.
HSD 05 = 3.2456
HSD 01 = 4.3406
Pairwise Comparisons
Q.05 = 3.7729 Q01 = 5.0459
M1 = 8.40
1.80
Q = 2.09 (p = .33442)
M2 = 6.60
M1 = 8.40
4.40
Q = 5.11 (p = .00916)
M3 = 4.00
M2 = 6.60
T2:T3
2.60
Q = 3.02 (p = .12377)
M3 = 4.00
Transcribed Image Text:The fratio value is 6.61261. The p-value is .01159. The result is significant at p<.05. Post Hoc Tukey HSD (beta) The Tukey's HSD (honestly significant difference) procedure facilitates pairwise comparisons within your ANOVA data. The F statistic (above) tells you whether there is an overall difference between your sample means. Tukey's HSD test allows you to determine between which of the various pairs of means - if any of them - there is a signficant difference. A couple of things to note. First, a blue value for Q (below) indicates a significant result. Second, it's worth bearing in mind that there is some disagreement about whether Tukey's HSD is appropriate if the F-ratio score has not reached significance. HSD 05 = 3.2456 HSD 01 = 4.3406 Pairwise Comparisons Q.05 = 3.7729 Q01 = 5.0459 M1 = 8.40 1.80 Q = 2.09 (p = .33442) M2 = 6.60 M1 = 8.40 4.40 Q = 5.11 (p = .00916) M3 = 4.00 M2 = 6.60 T2:T3 2.60 Q = 3.02 (p = .12377) M3 = 4.00
Summary of Data
Treatments
1
2
Total
5
5
5
15
ΣΧ
42
33
20
95
Mean
8.4
6.6
4
6.333
356
231
108
695
Std.Dev.
0.8944
1.8166
2.6458
2.582
Result Details
Source
df
MS
Between-
48.9333
24.4667
F= 6.61261
treatments
Within-
44.4
12
3.7
treatments
Total
93.3333
14
The fratio value is 6.61261. The p-value is .01159. The result is significant at p<.05.
4.
3.
2.
Transcribed Image Text:Summary of Data Treatments 1 2 Total 5 5 5 15 ΣΧ 42 33 20 95 Mean 8.4 6.6 4 6.333 356 231 108 695 Std.Dev. 0.8944 1.8166 2.6458 2.582 Result Details Source df MS Between- 48.9333 24.4667 F= 6.61261 treatments Within- 44.4 12 3.7 treatments Total 93.3333 14 The fratio value is 6.61261. The p-value is .01159. The result is significant at p<.05. 4. 3. 2.
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