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Quantitative Skills for Business | Assignment

   

Added on  2020-03-16

15 Pages2195 Words46 Views
Quantitative skills for business NameUniversity9th October 2017

Part A(a)Using the sample data attached, calculate the sample mean and standard deviation for the variables: - i)distribution cost ($000) (2 marks) SolutionMean=∑xin=1710.2924=71.2621Standarddeviation=√∑(xi− ́x)2n−1=√3845.1323=√167.1796=12.9298ii)sales ($000) (2 marks) SolutionMean=∑xin=1095624=456.5Standarddeviation=√∑(xi− ́x)2n−1=√15289223=√6647.478=81.5321iii)number of orders received (2 marks) SolutionMean=∑xin=10543224=4393Standarddeviation=√∑(xi− ́x)2n−1=√1249562623=√543288.1=737.0808(b)Using the sample data on number of orders, calculate the sample proportion of all orders that exceed 4,000 and the standard deviation of the distribution of the sample proportion. (3 marks)Solution

^p=1724=0.7083standarddeviation=√^p¿¿¿(c)Set up and interpret the following confidence intervals – i)A 95% confidence interval for the true population distribution cost in ($000) (5 marks) SolutionC.I: ́x±M.EM.E=zα/2σ√n=1.96∗12.9298√24=5.1730Lower limit: 71,262.1−5,173.0=66,089.1Upper limit: 71,262.1+5,173.0=76,435.1C.I: [66,089.1, 76,435.1]ii)A 99% confidence interval for the true population sales figures in ($000). (5 marks) SolutionC.I: ́x±M.EM.E=zα/2σ√n=2.576∗81.5321√24=42.8715Lower limit: 456,500−42,871.5=413,628.5Upper limit: 456,500+42,871.5=499,371.5C.I: [413,628.5, 499,371.5]iii)A 90% confidence interval for the true population proportion of all orders that exceed 4,000. (5 marks) Solution

C.I: ́x±M.EM.E=zα/2∗√^p¿¿¿Lower limit: 0.7083−0.1526=0.5557Upper limit: 0.7083+0.1526=0.8609(d)A follow-up study will provide a point estimate of the population proportion of orders that exceed 4,000. The study must provide 90% confidence that the point estimate is within 0.10 of the population proportion. If no previous proportion estimate is available (not even that calculated in (d) above), how large a sample would you recommend for this study?Solutionn=z2p(1−p)e2=1.6452∗0.5∗0.50.12=67.65n≈68(e)It is claimed that the average monthly warehouse distribution cost is more than $65,000. Do the data provide significant support for this claim? Use a 5% significance level and the critical value approach (classical approach) to test this claim. SolutionHypothesis tested:H0:μ=65,000HA:μ>65,000Tested at α = 0.05z= ́x−μσ√n=71262.1−6500012929.8√24=2.3727

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