STAT20028: Stem and leaf display for WPL and WOL

Solution Code: 1AEJB

Question:

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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)

(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

(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

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