Session Nine (Lab): Cluster Analysis
MART 307 Assignment Four: Cluster Analysis 1.

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When looking at the Agglomeration Schedule for Wards linkage for the last 10 clusters, the difference between coefficients of stage 162 and 16(Cluster #2) is 352.72. The difference between the coefficients of stage 161 and 160(Cluster#3) is 304.538. The difference between the coefficients of stage 160 and 159(Cluster#4) is 177.043. When looking at the chart, there is a biggest jump between clusters 3 and 4, indicating that there is a biggest difference between those two clusters. This is backed up by the Dendrogram as shown to the left, when putting a straight line through the longest horizontal lines; the line is cut by three clusters. Also, when looking at the Ward Scree Plot, the biggest kink is at 3 as shown by the arrow above which shows an abrupt change in angle (elbow.) Which indicates the 3rd cluster being more unique than the forth. The single linkage message also shows we should use 3 clusters, because looking at the Dendrogram, if we put a line through the longest horizontal distances it would be cut at 3 points. I would choose Wards method over Single Linkage because it is much clearer, the dendogram has much clearer clusters and there are fewer clusters. The agglomeration schedule is easier to figure out

2)

1 means not at all considered
2 unlikely to consider
3 would possibly consider
4 would actively consider
5 already do

As shown in the Initial Cluster Centers to the left, cluster 1 shows that every variable except cooking on gas, most the respondents would not at all consider the other 5 variables however, most respondents already do cook on gas. In cluster 2, it could be seen that the respondents already do all the variables except installing energy in efficient heating systems, which they would not at all consider. In cluster 3, the respondents would not consider applying hot water cylinder insulation, cook on gas and install an energy efficient washing machine. However, they already have installed an energy efficient refrigerator, heating pipe and energy efficient heating system.

Looking at the Final Cluster Centers, there is mostly 2’s(unlikely to consider) and 3’s(would possibly consider), which means they are less likely to consider energy saving behaviours. Cluster 2 has two statements which they already do, which they may think that the appliances are a big drain on their power bill. Also, there are 2 statements installing an energy efficient refrigerator and washing machine that respondents already do, this shows that these two items may take lots of energy and lead to a high electricity bill. For cluster 3, the statement cooking on gas, the respondents would not at all consider, showing that cooking in other ways may be for fast and effective. In cluster 1,2 and 3 the individuals are less likely to want to engage in energy saving behaviours, don’t actually do it.

Anova cannot be used, it should only be used for descriptive purposed. The clusters have been chosen to maximize the difference among cases in different clusters. As all the P-values are under 0.05, It indicates that all the variables contribute to the seperation of the clusters.

3) In the hierarchical clusters, it’s used to see the relationship between units and to narrow down how many clusters there are. In Hierarchical clusters the most common statistical packages use agglomerative method and the most popular agglomerative methods are single linkage (nearest neighbor), complete linkage (furthest neighbour), average linkage and Ward’s method. It also requires a distance or similarity matrix between all pairs of cases

In the non-hierarchical method (K-Means Clustering) a position in the measurement is taken as central place and distance is measured from such central point(useful to figure out central points), which means it doesn’t require computation of all possible distances. Identifying a right central position is a big challenge and hence non-hierarchical methods are less popular. You have to know in advance the number of clusters you want. You can’t get solutions for a range of cluster numbers unless you repeat the analysis for each different number of clusters. The algorithm repeatedly reassigns cases to clusters, so the same case can move from cluster to cluster during the analysis. Use when you have a hypothesis on how many cluters you want to use.

4) a) The six energy saving behaviours- Anova
-Ho: Variances are equal
Ha: Variances aren’t equal
-When looking at the summarized chart above, Levenes test of Homogenity shows that for applying hot water cylinder(0.603) and for insulation of heating pipes(0.134) the sig. level are greater than 0.05 therefore we accept the Ho and reject the Ha, because the variances are equal these two statements must do the Anova Test as shown above. For the statements cooking on gas, installing an energy efficient refrigerator, washing machine, and heating system the sig. levels of the statements are under 0.05(0.000,0.000,0.000,0.049) therefore all are significant. Therefore, for those statements Anova test should not be used; instead we use the Welch test, which is a more robust test of equality of means. According to the Anova test, the sig. of applying hot water cylinder insulation(F=10.269) and Insulation of heating pipes(F=3.408)) is 0.00 and 0.036 which is less than the 5% level, therefore different in means between income groups. For these two statements, turkey Hd cannot be used because the sample size needs to be within 25% of the harmonic mean and the Tamhane cannot be used because variances are equal. Therefore we must use bonferonni. For the other statements we will use the Welch test, The sig level of cooking on gas(Statistic=29.653), installing an energy efficient refrigerator(Stat:177.159), washing machine(97.357) and heating system(51.909) are all 0.000 which is less than the 5% level, therefore the ha is accepted and there is a difference between means. I did not use the Tukey HSD and Bonferroni test because they are used when equal variance are assumed. Therefore we are using the Tamhane test because variances aren’t equal

-Ho: No difference in the means between clusters Ha: Difference in the means between clusters
-There is difference between cluster 1 & 3 and clusters 2 and 3 for applying hot water cylinder insulation because the significance level is 0.005 and 0.000 which is less the sig. level therefore we accept the Ha and reject the Ho. However clusters 1&2 the Ho is accepted and the Ha is rejected because the sig. level is 0.262 which is greater than the 0.05
-There is difference between mean for insulations of heating pipes for clusters 2&3 because the sig. level is 0.043 which is less than 0.05 therefore we accept the Ha and reject the Ho. There is no difference between means for clusters 1&2 and !&3 because they are less than the sig. level.
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For the statement cooking on gas there is a differences between means for cluster 1&3 and 2&3 because the sign. Level is 0.00 which is less than 0.05 therefore we accept the Ha and reject the Ho.
For the statement energy efficient refrigerator there is a difference between means for all clusters 1&2, 2&3 and 1&3 because sig. is 0.00 which is less than 0.05 therefore accept Ha reject Ho
For the statement washing machine, all the clusters have a difference between means because the sig. level is less than 0.05, therefore accept Ha and reject Ho For the statement efficient heating system, there is a difference between means for clusters 1&3 and 3&3 because the sig. level is 0.00 which is less than the sig. level 0.05, therefore we accept the Ha. b) Age

Ho: No difference between mean ranks
Ha: Difference between mean ranks
Chi-Square is 2.350 and the sig. is 0.309 greater than 0.05, accept the Ho no difference between the mean ranks. This is shown by the mean rank of cluster 1, 2 and 3. When looking at the mean ranks(75.56,86.54 and 87.35), the ranks are not varied therefore there is not much difference between them and it supports the Ho.

c) Gender is Nominal; therefore we must use the crosstabs.

According to the cross tabulation above, the expected count is greater than 5. The total expected count of males is higher than the females.

The adjusted residual of males and females in cluster 1(2.1) is greater than 2, therefore there is a significance (association) between cluster one and gender. For the other 2 clusters, the residual is under 2, therefore there is no difference between them.

Ho: No association between gender and the clusters
Ha: There is an association between gender and the clusters
As shown by the chi-square test above, the Pearson Chi-square is x2 =0.105 which is greater than 0.05, therefore we accept the null hypothesis (Ho) and reject the Ha. Therefore the clusters do not vary by gender.

d) Income-

Ho: No difference between mean ranks
Ha: Difference between mean ranks
Chi-Square is 1.516 and the sig. is 0.469 which is greater than 0.05 accept the Ho and reject the Ha and there is no difference between the mean ranks. This is shown by the mean rank of cluster 1, 2 and 3. When looking at the mean ranks(74.71, 71.69 and 82.57), the ranks are not varied therefore there is not much difference between them and it supports the Ho.

e) The six environmental values questions

-Ho: Variances are equal
Ha: Variances aren’t equal
-When looking at the summarized chart above, Levenes test of Homogenity shows that for the balance of nature is very delicate(Stat:2.597), the earth is like a spaceship(stat: 2.059) and Humans were meant to rule(Stat:1.263) the sig. level are greater than 0.05(0.078, 0.131 and 0.286) therefore we accept the Ho and reject the Ha, because the variances are equal these two statements must do the Anova Test as shown above. For the statements Modifying the environment(Stat:5.773), Plants and animals(5.115) and Limits to economic growth the sig. levels of the statements are under 0.05(0.04,0.07 and 0.017) therefore all are significant. Therefore, for those statements Anova test should not be used; instead we use the Welch test, which is a more robust test of equality of means.
According to the Anova test, the sig. of The balance of nature(F=0.550), Earth is like a spaceship(F=2.890) and humans were meant to rule(F=0.891) are all greater than 0.05(0.578, 0.059,0.412, therefore there is no difference in means between the clusters. For these two statements, turkey Hd cannot be used because the sample size needs to be within 25% of the harmonic mean and the Tamhane cannot be used because variances are equal. Therefore we must use bonferonni. For the other statements we will use the Welch test, The sig level of Modifying the environment(3.857 and limits to economic growth(Stat: 3.636) is -.025 and 0.030 which is less than the 5% level, therefore the ha is accepted and there is a difference between clusters. I did not use the Tukey HSD and Bonferroni test because they are used when equal variance are assumed. Therefore we are using the Tamhane test because variances aren’t equal

There is no difference between clusters for all three statements

For the statement modifying the environment 1&3 because the significance level is 0.031 which is less than 0.05 therefore we accept the Ha and reject the Ho.
For the statement Plants and animals there are no difference between clusters
For the statement limits to economic growth, cluster 1&3 have a difference between means because the sig. level is less than 0.05, therefore accept Ha and reject Ho

f)

-Ho: Variances are equal
Ha: Variances aren’t equal
-When looking at the summarized chart above, Levenes test of Homogenity shows that for Peak times during (Stat:4.371), Off peak during(stat: 3.309) and Peaktimes before(stat: 3.122) the sig. level are less than 0.05(0.014, 0.039 and 0.047) therefore we accept the Ha and reject the Ho, therefore all are not significant. Therefore, for those statements Anova test should not be used; instead we use the Welch test, which is a more robust test of equality of means. For the statement off-peak times before(stat:2.413) the sig. Level is 0.093 which greater than 0.05, therefore reject the Ha and accept the Ho, because the variances are equal this statement must do the Anova Test
According to the Anova test, the sig. of Off-Peaktimes(F=1.150) is 0.319 which is greater than 0.05, therefore we reject the Ha and accept the Ho there is no difference in means between the clusters. For the statement, turkey Hd cannot be used because the sample size needs to be within 25% of the harmonic mean and the Tamhane cannot be used because variances are equal. Therefore we must use Bonferonni. For the other statements we will use the Welch test, The sig level of PeakTimes during (Stat:1.725), Off-Peak times during(Stat:1.108) and Peaktimes before(stat:2.815) are all greater than the 5% level(0.184,0.3340.443), therefore the Ho is rejected and there is no difference between clusters. I did not use the Tukey HSD and Bonferroni test because they are used when equal variance are assumed. Therefore we are using the Tamhane test because variances aren’t equal

There is no difference between clusters for the statement, because all sig. are greater than 0.05.

There is no difference between clusters for all three statements, because all the sig. are greater than 0.05, therefore we accept the Ho and reject the Ha.