# Wilcoxon Signed Rank Test Example

VerifiedAdded on 2020/05/16

|20

|3290

|133

AI Summary

This assignment explores the Wilcoxon Signed Rank Test, a non-parametric test used to compare paired samples. It explains the test's conditions, statistic calculation, advantages, and when to prefer it over parametric tests. The assignment includes an SPSS analysis of koala height data, presents the output, interprets the p-value, and concludes on the significance of the results.

## Contribute Materials

Your contribution can guide someone’s learning journey. Share your
documents today.

Data analysis

Student Name:

University

23rd January 2018

Student Name:

University

23rd January 2018

## Secure Best Marks with AI Grader

Need help grading? Try our AI Grader for instant feedback on your assignments.

Question 1 (20 marks)

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:

(a) (2 mark) Using SPSS, calculate the proportion of koalas in the study who are juveniles.

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%

(b) (18 marks) Without using SPSS, determine whether there is evidence to support the

theory that the proportion of koalas who are juveniles is less than 20%. Perform a

hypothesis test to statistically justify your answer by completing the following:

i) State the appropriate hypotheses (define any symbols used).

Solution

H0 : P=0.2

H A : P<0.2

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:

(a) (2 mark) Using SPSS, calculate the proportion of koalas in the study who are juveniles.

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%

(b) (18 marks) Without using SPSS, determine whether there is evidence to support the

theory that the proportion of koalas who are juveniles is less than 20%. Perform a

hypothesis test to statistically justify your answer by completing the following:

i) State the appropriate hypotheses (define any symbols used).

Solution

H0 : P=0.2

H A : P<0.2

ii) Check the conditions and assumptions for this test.

Solution

Because of the central limit theorem (large sample size), the sampling distribution

of p is normally distributed

iii) Calculate the test statistic for this test.

Solution

z= ^p− p0

√ p0 (1−p0)

n

p0=0.2

^p= 49

362 =0.135

n=362

z= ^p− p0

√ p0 (1−p0)

n

= 0.135−0.2

√ 0.2(1−0.2)

362

=−3.09177

iv) Calculate the P-value for this test.

Solution

The p-value associated with the computed z score value is 0.000997.

v) Interpret the P-value and write a meaningful conclusion in the context of this

situation.

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%.

Solution

Because of the central limit theorem (large sample size), the sampling distribution

of p is normally distributed

iii) Calculate the test statistic for this test.

Solution

z= ^p− p0

√ p0 (1−p0)

n

p0=0.2

^p= 49

362 =0.135

n=362

z= ^p− p0

√ p0 (1−p0)

n

= 0.135−0.2

√ 0.2(1−0.2)

362

=−3.09177

iv) Calculate the P-value for this test.

Solution

The p-value associated with the computed z score value is 0.000997.

v) Interpret the P-value and write a meaningful conclusion in the context of this

situation.

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 (22 marks)

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.

(a) (7 marks) Check the appropriate conditions and assumptions needed to calculate either a

confidence interval or hypothesis test in relation to the population mean height of the trees in

which juvenile koalas are sighted in South East Queensland. Comment on what these checks

indicate about the appropriateness of proceeding with the analysis. Include an appropriate

graph to support your answer.

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

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.

(a) (7 marks) Check the appropriate conditions and assumptions needed to calculate either a

confidence interval or hypothesis test in relation to the population mean height of the trees in

which juvenile koalas are sighted in South East Queensland. Comment on what these checks

indicate about the appropriateness of proceeding with the analysis. Include an appropriate

graph to support your answer.

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

## Paraphrase This Document

Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser

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).

Figure 1: Histogram of height

The histogram in figure 1 above further confirms that the variable height is normally distributed.

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

height of the trees in which juvenile koalas are sighted in South East Queensland is indeed

normally distributed (p-value > 0.05).

Figure 1: Histogram of height

The histogram in figure 1 above further confirms that the variable height is normally distributed.

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

Each value in the sample was sampled independently from each other value

(b) (6 marks) Estimate the population mean height of the trees in which juvenile koalas are

sighted in South East Queensland, using a 95% confidence interval (show all working).

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;

x ± z s

√n

x=15.86

s=4.578

n=49

x ± z s

√ n → 15.86 ±1.96 4.578

√ 49

15.86 ±1.96 4.578

√ 49 =15.86± 1.96 4.578

√49 =15.86 ± 1.28184

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

(c) (9 marks) From historical data, it is known that the mean height of trees in which juvenile

koalas have been sighted is 15 metres. Perform a hypothesis test to see if there is evidence to

(b) (6 marks) Estimate the population mean height of the trees in which juvenile koalas are

sighted in South East Queensland, using a 95% confidence interval (show all working).

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;

x ± z s

√n

x=15.86

s=4.578

n=49

x ± z s

√ n → 15.86 ±1.96 4.578

√ 49

15.86 ±1.96 4.578

√ 49 =15.86± 1.96 4.578

√49 =15.86 ± 1.28184

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

(c) (9 marks) From historical data, it is known that the mean height of trees in which juvenile

koalas have been sighted is 15 metres. Perform a hypothesis test to see if there is evidence to

support a suspicion that the mean height of trees in which juvenile koalas are sighted in

South East Queensland is more than this. In performing this test:

i) State appropriate hypotheses (define any symbols used).

Solution

H0 : μ=15

H A : μ>15

μ isthe mean height of trees∈which juvenile koalas have been sighted

ii) Calculate the value of a suitable test statistic for this test.

Solution

z= x−μ

s / √n

x=15.86 ; μ=15; s=4.578 ; n=49

z= x−μ

s / √n = 15.86−15

4.578/ √49 =1.314985

iii) Calculate the P-value of this test.

Solution

The p-value associated with the z score value of 1.314985 is 0.094272.

iv) Write a meaningful conclusion at the 5% level of significance.

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.

South East Queensland is more than this. In performing this test:

i) State appropriate hypotheses (define any symbols used).

Solution

H0 : μ=15

H A : μ>15

μ isthe mean height of trees∈which juvenile koalas have been sighted

ii) Calculate the value of a suitable test statistic for this test.

Solution

z= x−μ

s / √n

x=15.86 ; μ=15; s=4.578 ; n=49

z= x−μ

s / √n = 15.86−15

4.578/ √49 =1.314985

iii) Calculate the P-value of this test.

Solution

The p-value associated with the z score value of 1.314985 is 0.094272.

iv) Write a meaningful conclusion at the 5% level of significance.

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.

## Secure Best Marks with AI Grader

Need help grading? Try our AI Grader for instant feedback on your assignments.

Question 3 (26 marks)

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.

(a) (6 marks) Check the appropriate conditions and assumptions needed to perform a

hypothesis test comparing the population mean heights of trees in which the two groups

of koalas, based on age, are sighted. Comment on what these checks indicate about the

appropriateness of proceeding with the analysis. Include an appropriate graph to support

your answer.

Solution

The very first assumption that needs to be made is in regard to the scale of measurement

that is applied on the collected data should be continuous or may be ordinal scale. The

variable height is a continuous variable hence this assumption was met.

The second assumption is in relation to randomness of the sample, the idea is that the

collected data should be a representative data that is selected randomly from a portion of

the total population. This assumption was also met.

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.

(a) (6 marks) Check the appropriate conditions and assumptions needed to perform a

hypothesis test comparing the population mean heights of trees in which the two groups

of koalas, based on age, are sighted. Comment on what these checks indicate about the

appropriateness of proceeding with the analysis. Include an appropriate graph to support

your answer.

Solution

The very first assumption that needs to be made is in regard to the scale of measurement

that is applied on the collected data should be continuous or may be ordinal scale. The

variable height is a continuous variable hence this assumption was met.

The second assumption is in relation to randomness of the sample, the idea is that the

collected data should be a representative data that is selected randomly from a portion of

the total population. This assumption was also met.

The third assumption is that the data follows a normal distribution, bell-shaped

distribution curve.

Results are in table 3 below;

Table 5: 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).

Figure 2: Histogram of height

distribution curve.

Results are in table 3 below;

Table 5: 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).

Figure 2: Histogram of height

The other assumption is in relation to a reasonably large sample size. As could be seen,

the sample size was 362 which is reasonably large hence this assumption was met.

The last assumption is of equal variance i.e. homogeneity of variance. The data need to

have equal variances.

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 6: Test of Homogeneity of Variances

heightoftree

Levene Statistic df1 df2 Sig.

.416 1 360 .519

(b) (14 marks) Using an appropriate statistical test, determine if, on average, there is a

difference in the height of trees in which the two groups of koalas, based on age, are

sighted in South East Queensland. In performing the test, include:

i) State appropriate hypotheses, clearly defining all symbols.

Solution

H0 : μA =μJ

H A : μA ≠ μJ

μA =Mean height for the adults

μJ =Mean height for the juvenile

ii) Calculate a suitable test statistic (you can use the results from part (a) in this

calculation.

Solution

the sample size was 362 which is reasonably large hence this assumption was met.

The last assumption is of equal variance i.e. homogeneity of variance. The data need to

have equal variances.

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 6: Test of Homogeneity of Variances

heightoftree

Levene Statistic df1 df2 Sig.

.416 1 360 .519

(b) (14 marks) Using an appropriate statistical test, determine if, on average, there is a

difference in the height of trees in which the two groups of koalas, based on age, are

sighted in South East Queensland. In performing the test, include:

i) State appropriate hypotheses, clearly defining all symbols.

Solution

H0 : μA =μJ

H A : μA ≠ μJ

μA =Mean height for the adults

μJ =Mean height for the juvenile

ii) Calculate a suitable test statistic (you can use the results from part (a) in this

calculation.

Solution

## Paraphrase This Document

Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser

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

z= x A−xJ −0

√ sA

2

nA

+ sJ

2

nJ

= 17.43−15.86−0

√ 5.392

313 + 4.5782

49

=2.1761

iii) Find the P-value of the test (and include the degrees of freedom).

Solution

df =nA +nJ −2

df =313+49−2=360

iv) Interpret the P-value and write a meaningful conclusion in the context of the

question.

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.

v) Now use SPSS to check your results for this hypothesis test. Attach or copy and

paste the relevant output from SPSS for this test to your assignment.

Solution

Using SPSS we obtain the following;

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

z= x A−xJ −0

√ sA

2

nA

+ sJ

2

nJ

= 17.43−15.86−0

√ 5.392

313 + 4.5782

49

=2.1761

iii) Find the P-value of the test (and include the degrees of freedom).

Solution

df =nA +nJ −2

df =313+49−2=360

iv) Interpret the P-value and write a meaningful conclusion in the context of the

question.

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.

v) Now use SPSS to check your results for this hypothesis test. Attach or copy and

paste the relevant output from SPSS for this test to your assignment.

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

Differen

ce

Std.

Error

Differen

ce

95% Confidence

Interval of the

Difference

Lower Upper

heightoftre

e

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.

vi) Briefly comment on how the test statistic and P-value from SPSS output are

similar to or differ from your hand calculations.

Solution

The test statistics from the SPSS is the same from that obtained from the hand

calculations.

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

Differen

ce

Std.

Error

Differen

ce

95% Confidence

Interval of the

Difference

Lower Upper

heightoftre

e

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.

vi) Briefly comment on how the test statistic and P-value from SPSS output are

similar to or differ from your hand calculations.

Solution

The test statistics from the SPSS is the same from that obtained from the hand

calculations.

(c) (6 marks) Estimate, with 90% confidence, the population mean difference in height of

trees in which the two groups of koalas, based on age, are sighted in South East

Queensland. Ensure you explain the confidence interval in the context of the question.

Solution

( x A −xJ ) ± z √ s A

2

nA

+ sJ

2

nJ

( 17.43−15.86 ) ± 1.645 √ 5.392

313 + 4.5782

49

1.57 ± 1.186836

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.

trees in which the two groups of koalas, based on age, are sighted in South East

Queensland. Ensure you explain the confidence interval in the context of the question.

Solution

( x A −xJ ) ± z √ s A

2

nA

+ sJ

2

nJ

( 17.43−15.86 ) ± 1.645 √ 5.392

313 + 4.5782

49

1.57 ± 1.186836

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.

## Secure Best Marks with AI Grader

Need help grading? Try our AI Grader for instant feedback on your assignments.

Question 4 (32 marks)

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.

(a) (16 marks) Use a parametric test to answer this question by completing the following

(parts i. to v. are to be completed without the aid of SPSS, although summary statistics,

i.e. mean and standard deviation, required for the test can be found using SPSS):

i) State appropriate hypotheses (define any symbols used).

Solution

H0 : μ1=μ2

H0 : μ1>μ2

Where,

μ1=meanheight of the juvenile koala when sighted

μ1=meanheight of the juvenile koala after 3 hours

ii) State (but do not check) the assumptions for carrying out this test. Describe the

assumptions in the context of this question.

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.

(a) (16 marks) Use a parametric test to answer this question by completing the following

(parts i. to v. are to be completed without the aid of SPSS, although summary statistics,

i.e. mean and standard deviation, required for the test can be found using SPSS):

i) State appropriate hypotheses (define any symbols used).

Solution

H0 : μ1=μ2

H0 : μ1>μ2

Where,

μ1=meanheight of the juvenile koala when sighted

μ1=meanheight of the juvenile koala after 3 hours

ii) State (but do not check) the assumptions for carrying out this test. Describe the

assumptions in the context of this question.

Solution

o The very first assumption that needs to be made is in regard to the scale of

measurement that is applied on the collected data should be continuous or may be

ordinal scale. The variable height is a continuous variable hence this assumption

was met.

o The second assumption is in relation to randomness of the sample, the idea is that

the collected data should be a representative data that is selected randomly from a

portion of the total population. This assumption was also met.

o The third assumption is that the data follows a normal distribution, bell-shaped

distribution curve.

o The other assumption is in relation to a reasonably large sample size. As could be

seen, the sample size was 362 which is reasonably large hence this assumption

was met.

o The last assumption is of equal variance i.e. homogeneity of variance. The data

need to have equal variances.

iii) Calculate the value of a suitable test statistic for this test.

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

o The very first assumption that needs to be made is in regard to the scale of

measurement that is applied on the collected data should be continuous or may be

ordinal scale. The variable height is a continuous variable hence this assumption

was met.

o The second assumption is in relation to randomness of the sample, the idea is that

the collected data should be a representative data that is selected randomly from a

portion of the total population. This assumption was also met.

o The third assumption is that the data follows a normal distribution, bell-shaped

distribution curve.

o The other assumption is in relation to a reasonably large sample size. As could be

seen, the sample size was 362 which is reasonably large hence this assumption

was met.

o The last assumption is of equal variance i.e. homogeneity of variance. The data

need to have equal variances.

iii) Calculate the value of a suitable test statistic for this test.

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

t= xA −xJ −0

√ sA

2

nA

+ sJ

2

nJ

= 12.28−11.02−0

√ 4.2762

49 + 4.3232

49

=3.07

iv) Calculate the P-value of this test.

Solution

The p-value related to the test statistics is 0.001759.

v) Interpret the P-value and describe the outcome of the test in the context of this

question.

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).

vi) Now use SPSS to carry out the analysis. Copy and paste the relevant SPSS output

to your assignment. Do these results agree with those found in part iv? (Hint:

comment on the pvalue).

Solution

Using SPSS we obtained the following results;

Paired Samples Statistics

Mean N Std. Deviation Std. Error Mean

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

t= xA −xJ −0

√ sA

2

nA

+ sJ

2

nJ

= 12.28−11.02−0

√ 4.2762

49 + 4.3232

49

=3.07

iv) Calculate the P-value of this test.

Solution

The p-value related to the test statistics is 0.001759.

v) Interpret the P-value and describe the outcome of the test in the context of this

question.

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).

vi) Now use SPSS to carry out the analysis. Copy and paste the relevant SPSS output

to your assignment. Do these results agree with those found in part iv? (Hint:

comment on the pvalue).

Solution

Using SPSS we obtained the following results;

Paired Samples Statistics

Mean N Std. Deviation Std. Error Mean

## Paraphrase This Document

Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser

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.

(b) (16 marks) Describe an alternative statistical test that could be used to answer this

question. Include in your answer:

i) the name of the test,

Solution

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.

(b) (16 marks) Describe an alternative statistical test that could be used to answer this

question. Include in your answer:

i) the name of the test,

Solution

Wilcoxon Signed Rank Test

ii) the conditions/assumptions required for this test (in the context of the question),

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.

iii) a definition of the test statistic that would need to be calculated to perform this

test,

Solution

The test statistics is given as;

z= W

σW

= √ Nr (N r +1)(2 Nr +1)

6

iv) the relative advantages and/or disadvantages of this test compared with the test

you conducted to answer part (a),

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.

v) the circumstances under which you would use this test in preference to the one

used in part (a).

Solution

This test can be used or applied anytime that the population of a given data cannot

be assumed to be distributed normally..

ii) the conditions/assumptions required for this test (in the context of the question),

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.

iii) a definition of the test statistic that would need to be calculated to perform this

test,

Solution

The test statistics is given as;

z= W

σW

= √ Nr (N r +1)(2 Nr +1)

6

iv) the relative advantages and/or disadvantages of this test compared with the test

you conducted to answer part (a),

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.

v) the circumstances under which you would use this test in preference to the one

used in part (a).

Solution

This test can be used or applied anytime that the population of a given data cannot

be assumed to be distributed normally..

vi) Now use SPSS to carry out the analysis. Copy and paste the relevant SPSS output

into your assignment.

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.

vii) State and interpret the P-value from the SPSS output and describe the outcome of

the test in the context of this question.

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

into your assignment.

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.

vii) State and interpret the P-value from the SPSS output and describe the outcome of

the test in the context of this question.

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

## Secure Best Marks with AI Grader

Need help grading? Try our AI Grader for instant feedback on your assignments.

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.

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.

1 out of 20

### Related Documents

Your All-in-One AI-Powered Toolkit for Academic Success.

##### +13062052269

##### info@desklib.com

Available 24*7 on WhatsApp / Email

Unlock your academic potential

© 2024 | Zucol Services PVT LTD | All rights reserved.