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Descriptive Statistics for Company Revenue | Assignment

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Added on  2019-11-20

Descriptive Statistics for Company Revenue | Assignment

   Added on 2019-11-20

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Question 1a)Descriptive Statistics for company revenueNMeanStd. Deviationrevenue1500$2,496.4463$1,015.66544Valid N (listwise)1500b) Constructing confidence interval (C.I)C.I = μ±Zσ√nwhere μ=¿$2,496.4463 and std deviation = $1,015.66544 The value of Z from the table at 90% confidence interval is 1.645Standard error (S.E) = 1015.66544√1500=26.2243689Marginal error (M.E) = 1.645∗26.2243689=43.13908684C.I = 2,496.4463±43.13908694C.I = (2453.307213 - 2539.585387)c) Some of the assumptions and conditions necessary for the constructed confidence interval as stated by (Zhang & Zhang, 2014) are; The method of selection of the elements in a sample should be in such a way that the randomization condition is met. This is one of the best method of selection in the sample that also help to deal with issues of biasness to uphold reliable data. Independence assumptions is also another fundamental consideration. The involved event shouldbe independent of one another in their occurrence and having no influence in the outcome of other events. In this case the sample size that was used was relatively large i.e. (n=1500) that conforms to the sample size condition in the construction of the confidence interval. In order to determine the behavior of the sample means, the normal model should be used as suggested by the central limittheorem and thus the sample size should be sufficiently large.
Descriptive Statistics for Company Revenue | Assignment_1
d) From the calculated confidence interval, we can conclude that 90% of the population distribution is contained in the confidence interval (2453.307213 - 2539.585387) and that we are 90% confidence that the confidence interval has a 0.90 probability of covering the population mean.e) HypothesisH0: μ>2450H1: μ<=2450We shall carryout the Z-test μ= 2,496.4463std deviation = $1,015.66544Test statisticsZ=x−μσZ>2450−2496.44631015.66544>−0.04572992f) P (Z>-0.04572992) = 0.4801The P-value of the test is 0.4801 and the P-critical value is 0.05. Since the tested P-value 0.4801 is greater than critical P-value (0.05), we then fail to reject the null hypothesis and conclude that the mean revenue of the company is indeed greater than $2450.
Descriptive Statistics for Company Revenue | Assignment_2
Question 2a)The variable of interest for the researcher was age since the percentage (i.e. 40% and the frequency 44 out of 80) were both referring to the quantity arsonists whose ages were below 21 years.b)P=0.4n=80x=44μ= np = 80*0.4=32σ=√np(1−p)σ=√80∗0.4(1−0.4)=√19.2 = 4.38178046H0: Most of the arsonists are under 21 years of ageH1: Most of the arsonists are not under 21 years of age.Teste statisticZ=x−μσ=44−324.38178046=2.738612788P (Z=2.738612788) = 0.9969From the test, the sample data is seen to support their belief. The tested P-value (0.9969) is greater than the 0.05, we thus fail to reject the null hypothesis and conclude that most of the arsonists were under 21 years of age as claimed.c)Marginal error (M.E) = Zs√n = 0.05Z= 1.96S= 4.382n= 800.05 = 1.96*4.382√n√n=1.96∗4.3820.05=171.7744Squaring both sides we getn=29506.4445 Thus the sample size required is approximately equal to 29,506.
Descriptive Statistics for Company Revenue | Assignment_3

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