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Statistical Investigation Assignment

   

Added on  2021-04-17

10 Pages1353 Words275 Views
Running head: STATISTICAL INVESTIGATIONSTATISTICAL INVESTIGATIONName of the student:Name of the university:Author’s note:
Statistical Investigation Assignment_1
1STATISTICAL INVESTIGATIONQ1. (a)Given, μ = population mean = 11.6 and σ = population standard deviation = 14.8.Here,x̅ = μ and s = (σ/√n) since population follows normal distribution. Therefore,Sample mean or x̅=11.6 andPopulation standard deviation = σ = 14.8.(b) 95% z - score value of normal distribution is 1.96 when n is large. Again, 90% z - scorevalue of normal distribution is 1.644 with large n (Ross 2014). Interval gets wider with theincrease in confidence level. Therefore, 90% confidence interval in narrower from 95%confidence interval.(c) Let n denote the required sample size. Then,n = [(zα/2)2 * σ * (1- σ)]/(margin of error)2. Where, zα/2 = z – score at α/2, σ = population standard deviation and margin of error is thepermitted error level. Given, σ = 14.8, margin of error is 0.01 and 99% Z – score is 2.576. Therefore, n = mod [{2.5762 * 14.8 * (1-14.8)}/ (0.01)2] = [6.635776*(-13.8)*14.8]/(0.00001)=13552909.The required sample size is 13552909.
Statistical Investigation Assignment_2
2STATISTICAL INVESTIGATION(d) The statement is not correct. In case of normal distribution, 95% confidence interval of μ is given by {μ± zα/2* (σ/√n)}. Standard error multiplied with confidence co-efficient is to be added or subtracted hererather than only the standard deviation (Leon-Garcia 2017). Therefore, we cannot claim with95% probability that μ will fall in between 11.6 and 14.8. (e) The statement is not correct. We can say with confidence that 68% of the observation will fallin the interval of 11.6 ± 14.8.A normal curve is a bell shaped curve with its mean as the central line of it (Grimmett2018). The curve has half of its area in the right tail of the curve and the rest of the half in the lefttail. Standard deviation denotes the range of dispersion from mean. It has been seen that 68% ofthe value falls within μ± σ i.e. P[μ- σ<X<μ+ σ] ≈ 0.068. (f) Let μ1 be the calculated mean and μbe the hypothesized mean . We are to test:H0: μ = 12.5 vs H1: μ > 12.5.Testing statistic is : t = (μ1 - μ)/(σ/√n)So, calculated t is (14.5 – 12.5)/(14.8/√30) = 0.740 and tabulated t is 1.96. Therefore, calculated t < tabulated t and null hypothesis is accepted and it can be said that theaverage alcohol content is 12.5%.
Statistical Investigation Assignment_3

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