6. Birthweight of an infant has been hypothesized to be associated with smoking
ID: 3260092 • Letter: 6
Question
6. Birthweight of an infant has been hypothesized to be associated with smoking status of the mother during first trimester of pregnancy. This hypothesis is tested by recording birhweights of infants and smoking status of mother during pregnancy for all mothers who register at the prenatal clinic at a particular hospital within 1-month period. The mothers are divided into four groups according to smoking habit, and the sample of birthweights in pounds within each group is as follows Group 1: Mother is nonsmoker (NON): 7.5 6.2 6.9 7.4 9.2 8.3 7.6 Group 2: Mother is ex-smoker (smoked at some time before pregnancy but not during pregnancy) (EX): 5.8 7.3 8.2 7.1 7.8 Group 3: Mother is current smoker and smokes less than 1 pack per day (CUR < 1): 5.9 6.2 5.8 4.7 8.3 7.2 6.2 Group 4: Mother is current smoker and smokes less than 1 pack per day or more (CUR < 1): 6.2 6.8 5.7 4.9 6.2 7.1 5.8 5.4 Birthweights in pounds of infants and smoking status of mother during pregnancy NON EX CUR<1 CUR>=1 7.5 6.2 6.9 7.4 9.2 8.3 7.6 5.8 7.3 8.2 7.1 7.8 5.9 6.2 5.8 4.7 8.3 7.2 6.2 6.2 6.8 5.7 4.9 6.2 7.1 5.8 5.4 7 53.1 7.586 0.962 5 36.2 7.240 0.913 7 44.3 6.329 1.140 8 48.1 6.012 0.720 a. Compute Sum of Squares: SS SS among (between) groups: SSA SS within group: SSWb. Compute Mean Squares: MS Mean Squares among (between) groups: MSA Mean Squares within group: MSW
c. ConductANOVAtest
At = 0.05 level of significance, use the analysis of variance hypothesis testing to determine if a difference existed in the mean birthweights among the four different groups. In other words, test the hypothesis that the five groups (NON , EX , CUR < 1 , CUR >= 1) have the same average birthweights. I. Hypotheses: II. Calculations: The test statistic: The critical value: The p-value: III. Decision and Conclusion: Decision: Conclusion: d. Complete the following ANOVA Table: ANOVA summary Table based on my calculations: Source of Variance df SS MS F Among groups Within groups Total e. IfyourejectH0inpart(c),conductamultiplecomparisontesttoidentify which means differ. Use the following Minitab partial output. Partial Minitab Output Grouping Information Using Tukey Method n Mean Grouping NON 7 7.5857 A EX 5 7.2400 A B CUR<1 7 6.3286 A B CUR>=1 8 6.0125 B Means that do not share a letter are significantly different. 6. Birthweight of an infant has been hypothesized to be associated with smoking status of the mother during first trimester of pregnancy. This hypothesis is tested by recording birhweights of infants and smoking status of mother during pregnancy for all mothers who register at the prenatal clinic at a particular hospital within 1-month period. The mothers are divided into four groups according to smoking habit, and the sample of birthweights in pounds within each group is as follows Group 1: Mother is nonsmoker (NON): 7.5 6.2 6.9 7.4 9.2 8.3 7.6 Group 2: Mother is ex-smoker (smoked at some time before pregnancy but not during pregnancy) (EX): 5.8 7.3 8.2 7.1 7.8 Group 3: Mother is current smoker and smokes less than 1 pack per day (CUR < 1): 5.9 6.2 5.8 4.7 8.3 7.2 6.2 Group 4: Mother is current smoker and smokes less than 1 pack per day or more (CUR < 1): 6.2 6.8 5.7 4.9 6.2 7.1 5.8 5.4 Birthweights in pounds of infants and smoking status of mother during pregnancy NON EX CUR<1 CUR>=1 7.5 6.2 6.9 7.4 9.2 8.3 7.6 5.8 7.3 8.2 7.1 7.8 5.9 6.2 5.8 4.7 8.3 7.2 6.2 6.2 6.8 5.7 4.9 6.2 7.1 5.8 5.4 7 53.1 7.586 0.962 5 36.2 7.240 0.913 7 44.3 6.329 1.140 8 48.1 6.012 0.720 a. Compute Sum of Squares: SS SS among (between) groups: SSA SS within group: SSW
b. Compute Mean Squares: MS Mean Squares among (between) groups: MSA Mean Squares within group: MSW
c. ConductANOVAtest
At = 0.05 level of significance, use the analysis of variance hypothesis testing to determine if a difference existed in the mean birthweights among the four different groups. In other words, test the hypothesis that the five groups (NON , EX , CUR < 1 , CUR >= 1) have the same average birthweights. I. Hypotheses: II. Calculations: The test statistic: The critical value: The p-value: III. Decision and Conclusion: Decision: Conclusion: d. Complete the following ANOVA Table: ANOVA summary Table based on my calculations: Source of Variance df SS MS F Among groups Within groups Total e. IfyourejectH0inpart(c),conductamultiplecomparisontesttoidentify which means differ. Use the following Minitab partial output. Partial Minitab Output Grouping Information Using Tukey Method n Mean Grouping NON 7 7.5857 A EX 5 7.2400 A B CUR<1 7 6.3286 A B CUR>=1 8 6.0125 B Means that do not share a letter are significantly different. Group 1: Mother is nonsmoker (NON): 7.5 6.2 6.9 7.4 9.2 8.3 7.6 Group 2: Mother is ex-smoker (smoked at some time before pregnancy but not during pregnancy) (EX): 5.8 7.3 8.2 7.1 7.8 Group 3: Mother is current smoker and smokes less than 1 pack per day (CUR < 1): 5.9 6.2 5.8 4.7 8.3 7.2 6.2 Group 4: Mother is current smoker and smokes less than 1 pack per day or more (CUR < 1): 6.2 6.8 5.7 4.9 6.2 7.1 5.8 5.4 Birthweights in pounds of infants and smoking status of mother during pregnancy Group 1: Mother is nonsmoker (NON): 7.5 6.2 6.9 7.4 9.2 8.3 7.6 Group 2: Mother is ex-smoker (smoked at some time before pregnancy but not during pregnancy) (EX): 5.8 7.3 8.2 7.1 7.8 Group 3: Mother is current smoker and smokes less than 1 pack per day (CUR < 1): 5.9 6.2 5.8 4.7 8.3 7.2 6.2 Group 4: Mother is current smoker and smokes less than 1 pack per day or more (CUR < 1): 6.2 6.8 5.7 4.9 6.2 7.1 5.8 5.4 Birthweights in pounds of infants and smoking status of mother during pregnancy NON EX CUR<1 CUR>=1 7.5 6.2 6.9 7.4 9.2 8.3 7.6 5.8 7.3 8.2 7.1 7.8 5.9 6.2 5.8 4.7 8.3 7.2 6.2 6.2 6.8 5.7 4.9 6.2 7.1 5.8 5.4 7 53.1 7.586 0.962 5 36.2 7.240 0.913 7 44.3 6.329 1.140 8 48.1 6.012 0.720 NON EX CUR<1 CUR>=1 7.5 6.2 6.9 7.4 9.2 8.3 7.6 5.8 7.3 8.2 7.1 7.8 5.9 6.2 5.8 4.7 8.3 7.2 6.2 6.2 6.8 5.7 4.9 6.2 7.1 5.8 5.4 7 53.1 7.586 0.962 5 36.2 7.240 0.913 7 44.3 6.329 1.140 8 48.1 6.012 0.720 a. Compute Sum of Squares: SS SS among (between) groups: SSA a. Compute Sum of Squares: SS SS among (between) groups: SSA SS within group: SSW
b. Compute Mean Squares: MS Mean Squares among (between) groups: MSA b. Compute Mean Squares: MS Mean Squares among (between) groups: MSA Mean Squares within group: MSW
c. ConductANOVAtest
At = 0.05 level of significance, use the analysis of variance hypothesis testing to determine if a difference existed in the mean birthweights among the four different groups. In other words, test the hypothesis that the five groups (NON , EX , CUR < 1 , CUR >= 1) have the same average birthweights. I. Hypotheses: II. Calculations: The test statistic: At = 0.05 level of significance, use the analysis of variance hypothesis testing to determine if a difference existed in the mean birthweights among the four different groups. In other words, test the hypothesis that the five groups (NON , EX , CUR < 1 , CUR >= 1) have the same average birthweights. I. Hypotheses: II. Calculations: The test statistic: The critical value: The p-value: The critical value: The p-value: III. Decision and Conclusion: Decision: Conclusion: d. Complete the following ANOVA Table: ANOVA summary Table based on my calculations: III. Decision and Conclusion: Decision: Conclusion: d. Complete the following ANOVA Table: ANOVA summary Table based on my calculations: Source of Variance df SS MS F Among groups Within groups Total e. IfyourejectH0inpart(c),conductamultiplecomparisontesttoidentify which means differ. Use the following Minitab partial output. Partial Minitab Output Grouping Information Using Tukey Method n Mean Grouping NON 7 7.5857 A EX 5 7.2400 A B CUR<1 7 6.3286 A B CUR>=1 8 6.0125 B Means that do not share a letter are significantly different. Source of Variance df SS MS F Among groups Within groups Total e. IfyourejectH0inpart(c),conductamultiplecomparisontesttoidentify which means differ. Use the following Minitab partial output. Partial Minitab Output Grouping Information Using Tukey Method n Mean Grouping NON 7 7.5857 A EX 5 7.2400 A B CUR<1 7 6.3286 A B CUR>=1 8 6.0125 B Means that do not share a letter are significantly different.
Explanation / Answer
Answer:
One-way ANOVA: NON, EX, CUR<1, CUR>=1
Method
Null hypothesis
All means are equal
Alternative hypothesis
Not all means are equal
Significance level
= 0.05
Equal variances were assumed for the analysis.
Factor Information
Factor
Levels
Values
Factor
4
NON, EX, CUR<1, CUR>=1
Analysis of Variance
Source
DF
Adj SS
Adj MS
F-Value
P-Value
Factor
3
11.67
3.8909
4.41
0.014
Error
23
20.30
0.8828
Total
26
31.98
Model Summary
S
R-sq
R-sq(adj)
R-sq(pred)
0.939556
36.50%
28.22%
12.10%
Means
Factor
N
Mean
StDev
95% CI
NON
7
7.586
0.962
(6.851, 8.320)
EX
5
7.240
0.913
(6.371, 8.109)
CUR<1
7
6.329
1.140
(5.594, 7.063)
CUR>=1
8
6.012
0.720
(5.325, 6.700)
Pooled StDev = 0.939556
Tukey Pairwise Comparisons
Grouping Information Using the Tukey Method and 95% Confidence
Factor
N
Mean
Grouping
NON
7
7.586
A
EX
5
7.240
A
B
CUR<1
7
6.329
A
B
CUR>=1
8
6.012
B
Means that do not share a letter are significantly different.
a. Compute Sum of Squares: SS=31.98
SS among (between) groups: SSA=11.67
SS within group: SSW=20.30
b. Compute Mean Squares: MS
Mean Squares among (between) groups: MSA=3.8909
Mean Squares within group: MSW=0.8828
c. ConductANOVAtest
At = 0.05 level of significance, use the analysis of variance hypothesis testing to determine if a difference existed in the mean birthweights among the four different groups.
In other words, test the hypothesis that the five groups (NON , EX , CUR < 1 , CUR >= 1) have the same average birthweights.
H_0: mu_1 = mu_2= mu_3 = mu_4
H1: At least one pair of population means are different
II. Calculations: The test statistic: F=4.41
The critical value: F(3,23) =3.028
The p-value:P=0.014
III. Decision and Conclusion: Decision:
Conclusion: Reject HO.
We conclude that there is a difference existed in the mean birth weights among the four different groups
e. If you reject H0 inpart(c),conduct a multiple comparison test to identify which means differ.
Multiple comparison result shows that Group NON is significant with Group CUR>=1. All other pairs of groups are significant.
Null hypothesis
All means are equal
Alternative hypothesis
Not all means are equal
Significance level
= 0.05
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.