STAT20028: Stem and leaf display for WPL and WOL - Statistics Assessment Answers

January 15, 2018
Author : Julia Miles

Solution Code: 1AEJB

Question:

This assignment falls under Statisticswhich was successfully solved by the assignment writing experts at My Assignment Services AU under assignment help service.

Statistics Assignment Help

The assignment file was solved by professionalstatistics assignment helpexperts and academic professionals at My Assignment Services AU. The solution file, as per the marking rubric, is of high quality and 100% original (as reported by Plagiarism). The assignment help was delivered to the student within the 2-3 days to submission.

Looking for a new solution for this exact same question? Our assignment help professionals can help you with that. With a clientele based in top Australian universities, My Assignment Services AU’s assignment writing service is aiding thousands of students to achieve good scores in their academics. OurBusiness Management System assignment experts are proficient with following the marking rubric and adhering to the referencing style guidelines.

Solution:

Question – 1

(a)

The stem and leaf display for WPL and WOL is given below

WPL Stem (Units = 10) WOR
0 5 6 7 9 9
1 0 0 5 8 8
2 1 6 8
9 4 4 3 0 1 2
9 5 0 4 0 3 4 8 9
3 5 2 4 9 9
3 0 6 1 1 3 4 6 8
3 7 0 0 6 6 7 7 7 8
8 0 2 2 3 3 4 4
2 9 1
9 8 7 7 6 3 3 2 2 1 1 0 10 6 7 9
6 6 5 4 4 3 2 11
9 9 8 6 5 5 3 2 0 12 9
8 7 5 5 3 2 13
9 1 14 1
3 3 15
16
4 17
2 18

 

The stem and leaf plot indicates that the open price of WOR is less when compared with WPL open price shares. In addition, the distribution of both WPL and WOR distribution has longer tail facing towards the left side of normal curve, which means that the distribution of WPL and WOR is skewed left

(b)

(c) Bar Chart

(d)

Proportion of Stock prices above $40 for WPL = 49/50 = 0.94

Proportion of Stock prices above $40 for WOR = 35/50 = 0.67

 

Question 2

(a)

The descriptive statistics is given below

Prices
Mean 153.3245
Median 141.63
First Quartile 95.0925
Third Quartile 216.73

(b)

The variation measures for Prices is given below

Prices
Standard Deviation 75.66799
Range 326.8
Coefficient of Variation 49.35152

(c) The box whisker plot is given below

(d) Required student last four digit id no S0256033

Ewen, Dale, Joan S Gary, and James E Trefzger. Technical Calculus. Upper Saddle River, N.J.: Pearson/Prentice Hall, 2005. Print. ISBN – 13: 978-0-130-48818-3

Margaret L. Andersen, 2016. Sociology: The Essentials. 9 Edition. Cengage Learning. ISBN – 13: 978-1-305-50308-3

Michael Swan, 2005. Practical English Usage. 3 Edition. Oxford University Press. ISBN – 13: 978-0-194-42098-3

Louis P. Pojman, 2013. Philosophy: The Quest For Truth. 9 Edition. Oxford University Press. ISBN – 13: 978-0-199-96108-3

Question 3

(a) The total sources of agricultural water is 9779.7 (000 ML), out of which 1163.9 (000 ML) was used from on farm dams or tanks. Therefore,

P (water used from on-farm dams or tanks) = 1163.9/9779.7 = 0.119

(b) P (Groundwater and located in Qld) = 631.1/9779.7 = 0.065

(c) It is found that 1414.4 out of 6174.3 (000 ML) amount of water from MDB is taken from Rivers, creeks or lakes

P (water taken from Rivers, creeks or lakes for farm located in MDB) = 1414.4/6174.3 = 0.2291

(d) It is found that 957 out of 3425.8 (000 ML) amount of water from NSW is not taken from Rivers, creeks or lakes or Irrigation channels or pipelines

P (water not taken from Rivers, creeks or lakes or Irrigation channels or pipelines) =

957/3425.8 = 0.2794

Question 4

(i) Here, ? = 1.51 (average of overall rainfall)

P (x = 0) = e-1.51 = 0.221

(ii) Here, ? = 78.60/52 = 1.51 (average of weekly rainfall)

P (x = 0) = e-1.51 * (1.51)0 /0! = 0.221

P (x = 1) = e-1.51 * (1.51)1 /1!= 0.3334

P(X<2) = 0.221+0.3334 = 0.5543

P (X >=2) = 1 – P (X < 2) = 1 – 0.5543 = 0.4457

Therefore, the probability that there will be 2 or more days of rainfall in a week is 0.4457

(b)

The mean weekly total amount of rainfall is 10.58 and its standard deviation is 14.61

= P (-0.3136 < Z < 0.0972)

= P (-? < Z < 0.0972) – P (-? < Z < -0.3136)

= 0.5387 – 0.3769

= 0.1618

(ii)

------- (1)

On referring normal table, it is observed that

P (Z < 0.8416) = 0.80 ----------------------------- (2)

From (1) and (2), we have

=0.8416

A = 12.29

Therefore, we conclude that for the rainfall 12.29mm, only 20 % of the weeks have that amount of rainfall or higher

Question 5

(a) Normal Probability Plots

The points fall close to the trend line, indicating that the distribution of Temperature follows normal distribution, approximately

The points fall close to the trend line, indicating that the distribution of Relative Humidity follows normal distribution, approximately. Also, we see that there exist outliers in the dataset

The points fall close to the trend line, indicating that the distribution of Wind (km/hr) follows normal distribution, approximately

The points fall far away from the trend line, indicating that the distribution of Area burned do not follow normal distribution.

(b)

The formula used to calculate the 95% confidence interval for mean is given below

Here, the table value of z corresponding to 5% level of significance and is 1.96

Therefore, the margin of error, E is computed using this table value of t along with standard deviation and sample size.

The table given below shows the workings of 95% confidence interval for mean values

Temperature Relative Humidity Wind Area Burned
Sample Standard Deviation 6.603110477 17.43006 1.777181326 77.71624
Sample Mean 47.40540541 21.54955 4.362162162 17.89532
Sample Size 140 140 140 140
Level of Confidence 95% 95% 95% 95%
Confidence Interval Workings
Se (Standard Error) 0.558064691 1.473109 0.150199236 6.568221
Degrees of Freedom 139 139 139 139
Z Value 1.9600 1.9600 1.9600 1.9600
Interval Half Width 1.0938 2.8873 0.2944 12.8737
Confidence Interval
Interval Lower Limit 46.31 18.66 4.07 5.02
Interval Upper Limit 48.50 24.44 4.66 30.77

The 95% confidence interval for the Temperature (0 Celsius) is (46.31, 48.50). This indicates that, when more number of samples are extracted from the same population, then there is a 95% chance (95 out of 100 times) the true mean Temperature (0 Celsius) will lie within this interval

The 95% confidence interval for the Relative Humidity (%) is (18.66, 24.44). This indicates that, when more number of samples are extracted from the same population, then there is a 95% chance (95 out of 100 times) the true mean Relative Humidity (%) will lie within this interval

The 95% confidence interval for the Wind (km/hr) is (4.07, 4.66). This indicates that, when more number of samples are extracted from the same population, then there is a 95% chance (95 out of 100 times) the true mean Wind (km/hr) will lie within this interval

The 95% confidence interval for the Area burned (ha) is (5.02, 30.77). This indicates that, when more number of samples are extracted from the same population, then there is a 95% chance (95 out of 100 times) the true mean Area burned (ha) will lie within this interval

(c) Hypothesis Testing

In order to determine whether more areas in the forest burn when the temperature is above 250 Celsius than when it is below 250 Celsius, we perform two mean z test. The null and alternate hypotheses are given below

Null Hypothesis: H0: µ1 = µ2

That is, the mean forest area burned when the temperature is above 250 Celsius is significantly not greater than the mean forest area burned when it is below 250 Celsius

Alternate Hypothesis: H0: µ1 > µ2

That is, the mean forest area burned when the temperature is above 250 Celsius is significantly greater than the mean forest area burned when it is below 250 Celsius

Level of Significance: The level of significance is set to be ? = 0.05

Test Statistic

The z test statistic is

Z=x1-x2s12n1+s22n2

The descriptive statistics for the two groups

Area burned (ha) > 25 degree Area burned (ha) <=25 degree
Average 29.0431 13.10816
Std dev 117.2704 36.2119
Sample Size 42 98

The table given below shows the workings of two mean z test

Data
Hypothesized Difference 0
Level of Significance 0.05
Area burned (ha) > 25 degree
Sample taken 42
Mean for the sample 29.0431
Standard deviation for the sample 117.2704
Area burned (ha) <=25 degree
Sample taken 98
Mean for the sample 13.10816
Standard deviation for the sample 36.2119
Testing Workings
Mean Difference 15.93494
Standard Error 18.4612
Z-Test Statistic 0.8632
Upper-Tail Test
Upper Critical Value 1.6449
p-Value 0.1940

The value of z test statistic is 0.8632 and its corresponding p – value is 0.1940 > 0.05. This indicates that, the chance of rejecting the null hypothesis is violated. Therefore, we fail to conclude that more areas in the forest burn when the temperature is above 250 Celsius than when it is below 250 Celsius

Thisstatisticsassignment sample was powered by the assignment writing experts of My Assignment Services AU. You can free download thisstatisticsassessment answer for reference. This solvedstatisticsassignment sample is only for reference purpose and not to be submitted to your university. For a fresh solution to this question, fill the form here and get our professional assignment help.

RELATED SOLUTIONS

Order Now

Request Callback

Tap to ChatGet instant assignment help

Get 500 Words FREE