Team:SZU-China/Statistic Model

Statistic Model

Introduction

Statistic analysis can give us a clear and scientific understanding of our complex data, that was the reason why we construct a statistical model for our experimental data.

The data from simulating room were subjected to statistic analysis to determine whether there are significantly different in migration rate, mortality and gnawing rate using One-way ANOVA. Levene’s test to determine whether each group is homogeneity, then T-test for pooled data in pairwise gave us the value and finally to evaluate significant difference.

Levene’s test

Levene’s test is used to assess the equality of variances for a variable from two or more groups. In our model, we have four independent groups. Since in the following procedures for T-test, we assume that the variances of the populations from four samples are equal, so we need to test whether the following variances can be assume to be equal.

Assumptions:

1. Due to the advantage of Levene’s test, we do not need to assume the normal data

2. Null hypothesis that all populations from samples have the same variances $$ H_{0}:\sigma _{1}=\sigma _{2}=\cdots \sigma _{k}, \quad H_{A}:\sigma _{i}\neq \sigma _{j}, \quad \alpha =0.05 $$

The Levene Statistic (W) of each groups is calculated by: $$ W{=}\dfrac {\left( N-k\right) }{\left( k-1\right) } \dfrac {\sum ^{k}_{i=1}N_{i}\left( Z_{i\cdot}-Z_{i\cdot\cdot}\right) ^{2}} {\sum ^{k}_{i=1}\sum ^{N_{i}}_{j=1}\left( Z_{ij}-Z_{i\cdot}\right) ^{2}} \sim F\left( \alpha ,k-1,N-k\right) $$

Where

  • $k$ is the number of groups which need to be tested, equal to 4 in our data
  • $N_{i}$ is the number of cases in the ith group
  • $N$ is the total number of cases in all groups, equal to 12 in our data
  • $Y_{ij}$ is the value of the measured variable for the $i$th case from the $j$th group
  • $Z_{ij}=\left| Y_{ij}-\overline {Y_{i}}\right|$, $\overline {Y_{i}}$ is the mean of the $i$th group
  • $\overline {Z_{i\cdot}}$ is the mean of the $Z_{ij}$ for group $i$
  • $\overline {Z_{\cdot\cdot}}$ is the mean of all data

After calculate Levene Statistic of each group with SPSS, the W value are shown below:

Table1. Test of Homogeneity of Variances
Levene Statistic df1 df2 Sig.
Migration Rate 2.070 3 8 .183
Mortality 2.836 3 8 .088
Gnawing Rate .000 3 3 1.000

Fortunately, all $ W < F\left( 0.05,3,8\right)=4.066 $ or Sig.>0.05, indicating the null hypothesis is accepted, meaning $ \sigma _{1},\sigma _{2},\sigma _{3},\sigma _{4} $, do not have statistically different, so we can say they have the same variance σ2, and we can continues our procedure.

T-test for pooled data

The independent T-test for pooled data is used to determine whether the mean difference between two groups is statistically significantly different

Assumptions:

1.The sample groups are pairwise independent

2.All populations from samples have the same variances

The T-value (t) of eachpair is calculated by:$$t_{n_{1}+n_{2}-2}=\dfrac { \overline {x_{1}}-\overline {x_{2}}} {\sqrt {\dfrac {\left( n_{1}-1\right) s^{2}_{1}+\left( n_{2}-1\right) s^{2}_{2}} {\left( n_{1}-1\right) + \left( n_{2}-1\right) }\left( \dfrac {1}{n_{1}}+\dfrac {1}{n_{2}}\right) }}$$

When $ n_{1}=n_{2}=n=3 $ in this case, it can be simplify as: $$ t_{2\left( n-1\right) }=\dfrac {\overline {x_{1}}-\overline {x_{2}}}{\sqrt {\dfrac {s^{2}_{1}+s^{2}_{2}}{n}}} $$

By using two-tailed test, if p-value is lower than 0.05 or $ t > t_{4,0.05\left( two-tailed \right) }=2.776 $, we can say two mean have statistically significantly different.

Results

Levene’s test, gave us results that each group is homogeneity at P-value of 0.05, indicating we can assume they have equal variances.T-test for pooled data gave us following results:

The graphs in figure 1 divided in matrix cell $Z_{ij}$ show the visualized results of above procedures, t value lager than 2.776 showing red means $Z_{i}$ statistically significantly higher than $Z_{j}$, while lower than -2.776 showing green means $Z_{i}$ statistically significantly lower than $Z_{j}$. More red or green means different more significantly. Where, $Z_{1}$, $Z_{2}$, $Z_{3}$, $Z_{4}$ represent cockroach killing chalk, BAYER-Premise, M.anisopliae emulsifiable powder and blank group respectively.

Figure 1: Results of independent T-test for migration rate, mortality and gnawing rate. Matrix cell $Z_{ij}$ means $Z_{i}$ compared to $Z_{j}$. More red or green means different more significantly.

We can obviously see that cockroaches will behave strong migration tendency after people using cockroach killing chalk or BAYER-Premise. Migration rate infected by M.anisopliae emulsifiable powder was significantly lower than that by cockroach killing chalk or BAYER-Premise (P <0.05), even lower than blank group. While, there was no significant difference in migration rate between cockroach killing chalk and BAYER-Premise (P>0.05).

Similarly, gnawing rate in M.anisopliae group was significantly higher than that in cockroach killing chalk or BAYER-Premise group (P <0.05), while between the latter two groups, there was no significant difference(P>0.05). Thus, M.anisopliae will spread easier in cockroaches population causing an epidemic-like disease. As for mortality within three days, M.anisopliae showed it drawback, it did not have high mortality compared with two other chemical product, but it can still effective after several days and show similar mortality finally.

The results above show that M.anisopliae emulsifiable powder will not cause high migration rate compared with other traditional cockroach-killing methods, though the mortality within three days is not high, it can reach to high level after several days. What’s more, it can produce an epidemic-like disease and be long-term effective in an environmental friendly way.

References

[1] Levene, Howard (1960). “Robust tests for equality of variances”. In Ingram Olkin; Harold Hotelling; et al. Contributions to Probability and Statistics: Essays in Honor of Harold Hotelling. Stanford University Press. pp. 278–292.

[2] Rice, John A. (2006). Mathematical Statistics and Data Analysis (3rd ed.). Duxbury Advanced