3. We will choose SBP as the response. The two variables QUET and SMK will be co
ID: 3179372 • Letter: 3
Question
3. We will choose SBP as the response. The two variables QUET and SMK will be considered. The data has columns as: ID, SBP, QUET, age, and SMK. Can you please write out the hypothesis test for the two problems and give conclusions?
A.) (Interaction effect) Does the effect of QUET on the SBP depend on the status of smoking? You need to write the hypotheses, report R output and show your conclusion.
B.) Assuming that the interaction is not significant, for every QUET, does the smoking status will cause any difference in SBP? Set up appropriate hypotheses and report your conclusion.
Data set:
1 135 2.876 45 0
2 122 3.251 41 0
3 130 3.100 49 0
4 148 3.768 52 0
5 146 2.979 54 1
6 129 2.790 47 1
7 162 3.668 60 1
8 160 3.612 48 1
9 144 2.368 44 1
10 180 4.637 64 1
11 166 3.877 59 1
12 138 4.032 51 1
13 152 4.116 64 0
14 138 3.673 56 0
15 140 3.562 54 1
16 134 2.998 50 1
17 145 3.360 49 1
18 142 3.024 46 1
19 135 3.171 57 0
20 142 3.401 56 0
21 150 3.628 56 1
22 144 3.751 58 0
23 137 3.296 53 0
24 132 3.210 50 0
25 149 3.301 54 1
26 132 3.017 48 1
27 120 2.789 43 0
28 126 2.956 43 1
29 161 3.800 63 0
30 170 4.132 63 1
31 152 3.962 62 0
32 164 4.010 65 0
Explanation / Answer
Result:
3. We will choose SBP as the response. The two variables QUET and SMK will be considered. The data has columns as: ID, SBP, QUET, age, and SMK. Can you please write out the hypothesis test for the two problems and give conclusions?
A.) (Interaction effect) Does the effect of QUET on the SBP depend on the status of smoking? You need to write the hypotheses, report R output and show your conclusion.
H0: The interaction between QUET and SMK is not significant
H1: The interaction between QUET and SMK is significant
R output
Call:
lm(formula = SBP ~ QUET + SMK + QUET * SMK, data = mydata)
Residuals:
Min 1Q Median 3Q Max
-22.3713 -5.5705 -0.6357 7.4972 17.1051
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 49.312 19.972 2.469 0.0199 *
QUET 26.303 5.703 4.612 8.01e-05 ***
SMK 29.944 24.164 1.239 0.2256
QUET:SMK -6.185 6.932 -0.892 0.3799
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 8.948 on 28 degrees of freedom
Multiple R-squared: 0.6511, Adjusted R-squared: 0.6137
F-statistic: 17.42 on 3 and 28 DF, p-value: 1.408e-06
Calculated t= -0.892, P=0.3799 which is > 0.05 level of significance.
Ho is not rejected.
Interaction is not significant.
B.) Assuming that the interaction is not significant, for every QUET, does the smoking status will cause any difference in SBP? Set up appropriate hypotheses and report your conclusion.
Ho: smoking status is not significant
H1: smoking status is significant
anov2 <-lm(SBP~QUET+SMK, data = mydata)
> summary(anov2)
Call:
lm(formula = SBP ~ QUET + SMK, data = mydata)
Residuals:
Min 1Q Median 3Q Max
-23.6172 -4.8496 0.3664 6.5083 19.1832
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 63.876 11.468 5.570 5.21e-06 ***
QUET 22.116 3.230 6.847 1.61e-07 ***
SMK 8.571 3.167 2.707 0.0113 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 8.916 on 29 degrees of freedom
Multiple R-squared: 0.6412, Adjusted R-squared: 0.6165
F-statistic: 25.91 on 2 and 29 DF, p-value: 3.51e-07
Calculated t= 2.707, P=0.0113 which is < 0.05 level of significance.
Ho is rejected.
Smoking status is significant.
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