In: Statistics and Probability
For the question below, state what your null and alternative hypotheses are and select the appropriate test. Assess assumptions and do whatever needs to be done depending on your results (ie, doing a transformation of data or selecting a non-parametric if data violate assumptions). If you decide to run a Kruskal-Wallis test, you can assume the data satisfy those assumptions. Recall that it is generally considered ok to violate one assumption, especially if sample size is not too small. Perform the test and report the p value with the biological interpretation of your data. Finally, make a plot of the data that DOES NOT have a title, but does have proper x and y axis labels (can do it in excel or SPSS).
1)
You are interested in the connection between vitamin C and bone health. You expose guinea pigs to three different doses of vitamin C and then you record tooth growth following the doses. You collect the following data which are measures of tooth length in mm.
Dose 0.5 |
Dose 1.0 |
Dose 2.0 |
4.2 |
16.5 |
23.6 |
11.5 |
16.5 |
18.5 |
7.3 |
15.2 |
33.9 |
5.8 |
17.3 |
25.5 |
6.4 |
22.5 |
26.4 |
10 |
17.3 |
32.5 |
11.2 |
13.6 |
26.7 |
11.2 |
14.5 |
21.5 |
5.2 |
18.8 |
23.3 |
7 |
15.5 |
29.5 |
For the question below, state what your null and alternative hypotheses are and select the appropriate test. Assess assumptions and do whatever needs to be done depending on your results (ie, doing a transformation of data or selecting a non-parametric if data violate assumptions). If you decide to run a Kruskal-Wallis test, you can assume the data satisfy those assumptions. Recall that it is generally considered ok to violate one assumption, especially if sample size is not too small. Perform the test and report the p value with the biological interpretation of your data. Finally, make a plot of the data that DOES NOT have a title, but does have proper x and y axis labels (can do it in excel or SPSS).
Test for assumptions.
Test for normality shows that Kolmogorov-Smirnov test p values for the 3 groups are 0.20 which is > 0.05 level of significance. This shows that the normality assumption is not violated.
Tests of Normality |
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Group |
Kolmogorov-Smirnova |
Shapiro-Wilk |
|||||
Statistic |
df |
Sig. |
Statistic |
df |
Sig. |
||
Tooth length |
Dose 0.5 |
.198 |
10 |
.200* |
.890 |
10 |
.170 |
Dose 1.0 |
.217 |
10 |
.200* |
.908 |
10 |
.270 |
|
Dose 2.0 |
.154 |
10 |
.200* |
.973 |
10 |
.919 |
|
*. This is a lower bound of the true significance. |
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a. Lilliefors Significance Correction |
1)
You are interested in the connection between vitamin C and bone health. You expose guinea pigs to three different doses of vitamin C and then you record tooth growth following the doses. You collect the following data which are measures of tooth length in mm.
ANOVA used to compare the means of the three groups.
Descriptives |
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Tooth length |
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N |
Mean |
Std. Deviation |
Std. Error |
95% Confidence Interval for Mean |
Minimum |
Maximum |
||
Lower Bound |
Upper Bound |
|||||||
Dose 0.5 |
10 |
7.9800 |
2.74663 |
.86856 |
6.0152 |
9.9448 |
4.20 |
11.50 |
Dose 1.0 |
10 |
16.7700 |
2.51531 |
.79541 |
14.9707 |
18.5693 |
13.60 |
22.50 |
Dose 2.0 |
10 |
26.1400 |
4.79773 |
1.51718 |
22.7079 |
29.5721 |
18.50 |
33.90 |
Total |
30 |
16.9633 |
8.26603 |
1.50916 |
13.8767 |
20.0499 |
4.20 |
33.90 |
ANOVA |
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Tooth length |
|||||
Sum of Squares |
df |
Mean Square |
F |
Sig. |
|
Between Groups |
1649.489 |
2 |
824.744 |
67.072 |
.000 |
Within Groups |
332.001 |
27 |
12.296 |
||
Total |
1981.490 |
29 |
Calculated F(2,27) = 67.072, P=0.000 which is < 0.05 level of significance. Ho is rejected. We conclude that three group means are different. We conclude that there is a connection between vitamin C and bone health.