Assessment item 2—

Assessment Item 2

Measurement of Central Tendency and Dispersal, and Determination of Confidence Intervals for Data

Student’s Name

Institution’s Name

Course Title

Tutor’s Name

Date

Question One

Year January April July October

2001 4.68 4.70 4.81 5.46

2002 5.98 5.81 6.36 5.84

2003 5.27 5.71 5.75 5.72

2004 6.23 6.82 6.96 7.02

2005 8.13 8.65 7.96 7.90

2006 7.71 7.28 7.15 6.69

2007 7.76 8.68 9.55 9.00

2008 9.48 8.07 7.01 8.65

2009 9.19 8.52 8.52 9.76

2010 11.50 11.26 11.89 12.00

2011 10.86 11.80 11.42 11.66

2012 11.51 12.48 13.36 13.69

2013 13.45 14.58 12.55 12.15

2014 12.03 11.03 9.42 8.70

2015 9.32 10.77 9.10 8.99

Stem Leaf

4 68 70 81

5 27 46 71 72 75 98 81 84 98

6 23 36 69 82 96

7 01 02 15 28 71 76 90 96

8 07 13 52 52 65 65 68 70 99

9 0 10 19 32 42 48 55 76

10 77 86

11 03 26 42 50 51 66 80 89

12 0 03 15 48 55

13 36 45 69

14 58

Legend

4I68 = 4.68

Relative Frequency Histogram

Opening Price range Frequency Relative frequency

4.0 – 5.99 11 0.18

6.0- 7.99 13 0.22

8.0 – 9.99 17 0.28

10.0 -11.99 10 0.17

12.0 – 13.99 8 0.13

14.0 – 15.99 1 0.02

The number of prices above 10 dollars were 19, which implies

Proportion of prices above $ 10= 1960=31.7 %Question

Computing the mean, median, first quartile, and third quartile for each capital city

Sydney

Arranging values in ascending order

420, 485, 500, 550, 580, 600, 630, 650, 658, 850, 850, 890, 950, 999

Mean= 420+485+500+550+580+600+630+650+658+850+850+890+950+99914=686.57Median

Median is the middle value

Median= 630+6502=640First Quartile Q1= n+14th= 14+14=3.75th termTo get Q1 we find the average between 3rd and 4th term

Q1= 500+5502=525Third Quartile Q3= 3(n+1)4th= 3(14+1)4=11.25th termTo get Q3 we find the average between 11th and 12th term

Q1= 850+8902=870Melbourne

Values for Melbourne in Ascending order

300, 330, 340, 340, 370, 370, 380, 413, 465, 480, 605, 645, 700, 813, 895

Mean= 300+330+340+340+370+370+380+413+465+480+605+645+700+813+89515=496.40Median= n+12thterm=15+12 = 8th term= 413

First Quartile Q1= n+14th= 15+14=4th termThus Q1 = 340

Third Quartile Q3= 3(n+1)4th= 3(15+1)4=12th termThus Q3 = 645

Brisbane

Arranging values in ascending order

390, 395, 415, 420, 420, 425, 450, 450, 480, 535, 610, 685, 700

Mean= 390+395+415+420+420+425+450+450+480+535+610+685+70013=490.38Median

Median is the middle value

Median= n+12th=13+12th=7th term=450First Quartile Q1= n+14th= 13+14=3.5th termTo get Q1 we find the average between 3rd and 4th term

Q1= 415+4202=417.5Third Quartile Q3= 3(n+1)4th= 3(13+1)4=10.5th termTo get Q3 we find the average between 10th and 11th term

Q1= 535+6102=572.5Perth

Arranging values in ascending order

340, 350, 380, 395, 395 400, 430, 440, 470, 715, 725, 800

Mean= 340+350+380+395+395+400+430+440+470+715+725+80012=486.67Median

Median is the middle value

Median= 400+4302=415First Quartile Q1= n+14th= 12+14=3.25th termTo get Q1 we find the average between 3rd and 4th term

Q1= 380+3952=387.5Third Quartile Q3= 3(n+1)4th= 3(12+1)4=9.75th termTo get Q3 we find the average between 11th and 12th term

Q1= 470+7152=592.5Perth Sydney Melbourne Brisbane

725 890 700 610

470 999 413 480

395 850 645 450

440 600 895 395

715 550 480 450

400 580 605 415

395 485 380 425

380 650 465 700

340 420 300 420

430 658 330 390

800 950 370 420

350 800 370 535

850 340 685

630 340 813 Mean 486.6667 708 496.4 490.384615

Standard Deviation 161.9951 189.7857 179.4397637 95.1036118

Min 340 420 300 390

Max 800 999 895 700

Coefficient of Variation 0.332867 0.268059 0.361482199 0.19393678

Box and Whisker Plot

Question 3:

Probability that an Australian household, randomly selected, uses solar as a source of energy

Total Number of Households in Australia from the provided data= 5,612.6

Number of Households using Solar= 959.3

Probability of an event happening = Number of ways it can happen Total number of outcomes

Probability of A household Using Solar Picked at Random=Number of Housholds Using SolarTotal Number of Housholds in Australia=959.35612.6=0.171Probability of a household picked at random to use mains gas and is from Australia

Probability of independent events P A and B= PAX P(B)

Let’s take Probability of Household using mains gas = P(A)

Let’s take Probability of Household coming from Victoria = P(B)

P A=3559.15612.6=0.634P B= 1607.95612.6=0.286P A and B= PAX PB= 0.634*0.286=0.181Probability of independent events P A and B= PAX P(B)

Let’s take Probability of Household using LPG = P(A)

Let’s take Probability of Household coming from South Australia = P(B)

P A=781.55612.6=0.139P B= 498.35612.6=0.089P A and B= PAX PB= 0.089*0.139=0.012Question Four

The probability that on any given week in a year there would be no rainfall

From the provided data, and beginning the first week on 29th Dec 2014, the number of whole weeks without rainfall = 13

The number of weeks in a year = 52

Probability No rainfall on any given Week in a year= 1352=0.25Probability that there will be 2 or more days of rainfall in a week

Number of weeks with 2 or more days of rainfall = 39 weeks

Number of weeks in a year = 52

Probability There will be 2 or more days with rainfall in a week= 3952=0.75Standard Deviation

The standard deviation is the square root of the variance, where the variance is obtained using the formula below

S2=1n-1t=1nx-x2Thus Standard Deviation, S

S=1n-1t=1nx-x2The tabulation below shows the calculation of x-x where the mean is subtracted from each observed value. x-x2 is then obtained for each value and the summation is computed.

Week Rainfall (mm) x-xx-x21 15.7 -0.19 0.0361

2 15.6 -0.29 0.0841

3 16.4 0.51 0.2601

4 16.1 0.21 0.0441

5 15.9 0.01 0.0001

6 15.8 -0.09 0.0081

7 16.1 0.21 0.0441

8 15.8 -0.09 0.0081

9 15.6 -0.29 0.0841

10 16.2 0.31 0.0961

11 16 0.11 0.0121

12 16 0.11 0.0121

13 15.8 -0.09 0.0081

14 16.3 0.41 0.1681

15 15.9 0.01 0.0001

16 15.8 -0.09 0.0081

17 15.7 -0.19 0.0361

18 16 0.11 0.0121

19 15.6 -0.29 0.0841

20 15.6 -0.29 0.0841

21 16 0.11 0.0121

22 16.1 0.21 0.0441

23 15.9 0.01 0.0001

24 15.7 -0.19 0.0361

25 15.8 -0.09 0.0081

26 15.3 -0.59 0.3481

27 16.2 0.31 0.0961

28 15.8 -0.09 0.0081

29 16.1 0.21 0.0441

30 15.9 0.01 0.0001

x-x221.687

Thus S2=1n-1t=1nx-x2=1.68730-1=0.58172 The standard deviation S= 0.58172 = 0.24119

Question 5

Constructing Normal Probability Plots for each variable of Good wine

Constructing 95% confidence Interval

Taking the obtained mean as the sample statistic (P’), P= 16.0 and n=30 we compute for the standard error as follows

Standard Error= P'(P-P’)n= 15.89(16.0-15.89)30=0.24From the standard normal distribution table, we take a multiply 2 for the 95% confidence interval, and compute the confidence interval

C.I=P’±2Standard Error=15.89 ±2*0.24=(15.65,16.37)With 95% confidence, we estimate that the true mean value of ounces for the wine bottles falls between 15.65 ounces and 16.37 ounces.

Reference

Motulsky, H. (2007), Statistics Guide, Version 5.0, GraphPad Software, Inc. Retrieved on September, 02, 2015 from < http://www.graphpad.com/downloads/docs/Prism5Stats.pdf>