Question 1
Researchers are concerned with the decline in the proportion of koalas in the wild who are juveniles, as this will impact the future adult koala population. Historically, the proportion of koalas in South East Queensland who are juveniles is 20%. Use the information in the dataset koalas17.sav to answer the following questions:
Solution
Table 1: Frequency distribution of age
Frequency |
Percent |
Valid Percent |
Cumulative Percent |
||
Valid |
Adult |
313 |
86.5 |
86.5 |
86.5 |
Juvenile |
49 |
13.5 |
13.5 |
100.0 |
|
Total |
362 |
100.0 |
100.0 |
From table 1 above, the proportion of koalas in the study who are juveniles is 13.5%
Solution
Solution
Because of the central limit theorem (large sample size), the sampling distribution of p is normally distributed
Solution
Solution
The p-value associated with the computed z score value is 0.000997.
Solution
The p-value is less than 5% level of significance we therefore reject the null hypothesis and conclude that that the proportion of koalas who are juveniles is significantly less than 20%.
Question 2
Use the information in the dataset koalas17.sav to answer the following questions. You should use SPSS to calculate the sample statistics you will need to do this question, but for parts (b) and (c) you are required to do all other calculations by hand, using a calculator. Regardless of your answer to part (a), complete all parts of this question.
Solution
Normality Test
One of the key assumptions is related to the normality of the data. So we checked whether the variable height of the trees in which juvenile koalas are sighted in South East Queensland is normally distributed. Results are in table 2 below;
Table 2: Tests of Normality
Age |
Kolmogorov-Smirnova |
Shapiro-Wilk |
|||||
Statistic |
df |
Sig. |
Statistic |
df |
Sig. |
||
heightoftree |
Adult |
.074 |
313 |
.000 |
.960 |
313 |
.000 |
Juvenile |
.101 |
49 |
.200* |
.954 |
49 |
.052 |
|
*. This is a lower bound of the true significance. |
|||||||
a. Lilliefors Significance Correction |
Using either Kolmogorov-Smirnov test or Shapiro-Wilk test, results shows that the variable height of the trees in which juvenile koalas are sighted in South East Queensland is indeed normally distributed (p-value > 0.05).
Test of homogeneity of variance
Using Levene’s test, we checked whether the variable height has equal variance. As can be seen in table 3 below, the assumption on homogeneity is met (p-value > 0.05). The two populations have equal variance.
Table 3: Test of Homogeneity of Variances
heightoftree |
|||
Levene Statistic |
df1 |
df2 |
Sig. |
.416 |
1 |
360 |
.519 |
Other conditions that we found to have been met include
Solution
First we obtain the sample statistics.
Table 4: Descriptive Statistics
Age |
N |
Minimum |
Maximum |
Mean |
Std. Deviation |
|
Adult |
heightoftree |
313 |
5 |
50 |
17.43 |
5.390 |
Valid N (listwise) |
313 |
|||||
Juvenile |
heightoftree |
49 |
6 |
23 |
15.86 |
4.578 |
Valid N (listwise) |
49 |
Confidence interval;
Lower limit: 15.86 – 1.28184 = 14.57816
Upper limit: 15.86 + 1.28184 = 17.14184
Thus the 95% confidence interval is between 14.57816 and 17.14184
Solution
Solution
Solution
The p-value associated with the z score value of 1.314985 is 0.094272.
Solution
Since the p-value is greater than the 5% level of significance, we fail to reject the null hypothesis and conclude that there is no enough statistically significant evidence to conclude that the mean height of trees in which juvenile koalas sighted in South East Queensland is more than 15 metres.
Question 3
Use the information in the dataset koalas17.sav to answer the following questions. You should use SPSS to calculate any sample statistics you will need to do this question, but for parts b(ii) and (c) you are required to do all other calculations by hand, using a calculator. Koalas of two broad age groups were sighted in South East Queensland – adult and juvenile. As a researcher you are interested to see whether there is a difference in the mean height of trees in which these two groups of koalas are sighted. Regardless of your answer to part (a), complete all parts of this question.
Solution
Results are in table 3 below;
Table 2: Tests of Normality
Age |
Kolmogorov-Smirnova |
Shapiro-Wilk |
|||||
Statistic |
df |
Sig. |
Statistic |
df |
Sig. |
||
heightoftree |
Adult |
.074 |
313 |
.000 |
.960 |
313 |
.000 |
Juvenile |
.101 |
49 |
.200* |
.954 |
49 |
.052 |
|
*. This is a lower bound of the true significance. |
|||||||
a. Lilliefors Significance Correction |
Using Kolmogorov-Smirnov test, results shows that the variable height of the trees in which either adult or juvenile koalas are sighted in South East Queensland is indeed normally distributed (p-value > 0.05).
Using Levene’s test, we checked whether the variable height has equal variance. As can be seen in table 3 below, the assumption on homogeneity is met (p-value > 0.05). The two populations have equal variance.
Table 3: Test of Homogeneity of Variances
heightoftree |
|||
Levene Statistic |
df1 |
df2 |
Sig. |
.416 |
1 |
360 |
.519 |
Solution
Solution
Table 5: Descriptive Statistics
Age |
N |
Minimum |
Maximum |
Mean |
Std. Deviation |
|
Adult |
heightoftree |
313 |
5 |
50 |
17.43 |
5.390 |
Valid N (listwise) |
313 |
|||||
Juvenile |
heightoftree |
49 |
6 |
23 |
15.86 |
4.578 |
Valid N (listwise) |
49 |
Solution
Solution
The p-value associated with the z score value of 2.1761 is 0.0148. This value is less than the 5% level of signicance. We therefore reject the null hypothesis and conclude that there is a significant difference in the height of trees in which the two groups of koalas, based on age.
Solution
Using SPSS we obtain the following;
Group Statistics |
|||||
Age |
N |
Mean |
Std. Deviation |
Std. Error Mean |
|
heightoftree |
Adult |
313 |
17.43 |
5.390 |
.305 |
Juvenile |
49 |
15.86 |
4.578 |
.654 |
Independent Samples Test |
|||||||||||
Levene’s Test for Equality of Variances |
t-test for Equality of Means |
||||||||||
F |
Sig. |
t |
df |
Sig. (2-tailed) |
Mean Difference |
Std. Error Difference |
95% Confidence Interval of the Difference |
||||
Lower |
Upper |
||||||||||
heightoftree |
Equal variances assumed |
.416 |
.519 |
1.933 |
360 |
.054 |
1.571 |
.813 |
-.027 |
3.169 |
|
Equal variances not assumed |
2.177 |
70.583 |
.033 |
1.571 |
.721 |
.132 |
3.010 |
An independent samples t-test was done to compare the mean height of trees for the adults and for the juvenile. Results showed that the average height of trees for the adults (M = 17.43, SD = 5.39, N = 313) had significant difference with the height of trees for the juvenile (M = 15.86, SD = 4.578, N = 49), t (320) = 1.93, p < .05, two-tailed.
Solution
The test statistics from the SPSS is the same from that obtained from the hand calculations.
Solution
Lower limit: 1.57 – 1.186836 = 0.383164
Upper limit: 1.57 + 1.186836 = 2.756836
Therefore the 90% confidence interval for the population mean difference in height of trees in which the two groups of koalas, based on age is between 0.3832 and 2.7568.
Question 4
Use the information in the dataset koalas17.sav to answer the following questions. Researchers have theorized that juvenile koalas, when feeding, move down the tree rather than up. In order to test this theory, the height from the ground at which each juvenile koala was located sleeping within a tree was initially measured when the koala was sighted (positionintree) and then again three hours later (laterpositionintree). It now needs to be determined whether, on average, the height of the juvenile koalas in the trees has decreased after three hours.
Solution
Where,
Solution
Solution
Descriptive statistics
Descriptive Statistics |
||||||
Age |
N |
Minimum |
Maximum |
Mean |
Std. Deviation |
|
Adult |
Position of koala in tree (m) |
313 |
4 |
29 |
13.14 |
4.926 |
Position of koala in tree 3h after first sighting (m) |
313 |
0 |
21 |
11.02 |
4.263 |
|
Valid N (listwise) |
313 |
|||||
Juvenile |
Position of koala in tree (m) |
49 |
4 |
21 |
12.28 |
4.276 |
Position of koala in tree 3h after first sighting (m) |
49 |
1 |
19 |
11.02 |
4.323 |
|
Valid N (listwise) |
49 |
Solution
The p-value related to the test statistics is 0.001759.
Solution
The given p-value is greater than the 5% level of significance we therefore reject the null hypothesis and conclude that on average, the height of the juvenile koalas in the trees has decreased after three hours (John , 2006).
Solution
Using SPSS we obtained the following results;
Paired Samples Statistics |
|||||
Mean |
N |
Std. Deviation |
Std. Error Mean |
||
Pair 1 |
Position of koala in tree (m) |
12.28 |
49 |
4.276 |
.611 |
Position of koala in tree 3h after first sighting (m) |
11.02 |
49 |
4.323 |
.618 |
Paired Samples Correlations |
||||
N |
Correlation |
Sig. |
||
Pair 1 |
Position of koala in tree (m) & Position of koala in tree 3h after first sighting (m) |
49 |
.779 |
.000 |
Paired Samples Test |
|||||||||
Paired Differences |
t |
df |
Sig. (2-tailed) |
||||||
Mean |
Std. Deviation |
Std. Error Mean |
95% Confidence Interval of the Difference |
||||||
Lower |
Upper |
||||||||
Pair 1 |
Position of koala in tree (m) – Position of koala in tree 3h after first sighting (m) |
1.255 |
2.862 |
.409 |
.433 |
2.077 |
3.070 |
48 |
.004 |
Yes the above results agree with those obtained by hand.
Solution
Wilcoxon Signed Rank Test
Solution
The only condition for this test is that the values need to be numerical. It however does not rely on the assumptions that parametric tests need to have.
Solution
The test statistics is given as;
Solution
The main benefit of using Wilcoxon Signed Rank Test is the fact that it does not depends on the parent distribution nor on the parent parameters (Kerby, 2017). The assumptions on the distribution shape in not necessary when using this test.
Solution
This test can be used or applied anytime that the population of a given data cannot be assumed to be distributed normally..
Solution
Ranks |
||||
N |
Mean Rank |
Sum of Ranks |
||
Position of koala in tree 3h after first sighting (m) – Position of koala in tree (m) |
Negative Ranks |
31a |
24.71 |
766.00 |
Positive Ranks |
14b |
19.21 |
269.00 |
|
Ties |
4c |
|||
Total |
49 |
|||
a. Position of koala in tree 3h after first sighting (m) < Position of koala in tree (m) |
||||
b. Position of koala in tree 3h after first sighting (m) > Position of koala in tree (m) |
||||
c. Position of koala in tree 3h after first sighting (m) = Position of koala in tree (m) |
Test Statisticsa |
|
Position of koala in tree 3h after first sighting (m) – Position of koala in tree (m) |
|
Z |
-2.821b |
Asymp. Sig. (2-tailed) |
.005 |
a. Wilcoxon Signed Ranks Test |
|
b. Based on positive ranks. |
Solution
The p-value is 0.005, this value is less than 5% level of significance hence the null hypothesis is rejected and by rejecting the null hypothesis we conclude that the median height of the juvenile koalas in the trees has decreased after three hours.
References
John, R. A. (2006). Mathematical Statistics and Data Analysis. Journal of Statistical Computing, 42-53.
Kerby, D. S. (2017). Comparing Two Samples from an Individual Likert Question. Journal of Statistical Theory and Practice, 56-66.
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