QAB 105 Quantitative Analysis forBusinessAssessment Task – 2[Pick the date]Student Id & Name
QUESTION 1 (a)The objective is to select a random sample of 200 observations. The retailers who areselling the company’s product would form the population. From this population, arandom sample of 200 retailers has to be drawn. Provide a unique numeric ID to each ofthe retailers who form part of the population. Then using particular computer softwarelike MS-Excel, generate 200 random numbers between the range of the unique IDs.Choose the retailers corresponding to the random numbers drawn and include the same inthe sample. This would result in derivation of a random sample where the odds ofselection of each retailer would be the same. Also, in order to avoid any bias, the processof random number selection is computerized. (b)Calculation of 95% confidence interval for proportion of retailers selling lower than therecommended price Sample size (n) = 200 Proportion (p) = 79/200 = 0.395 For 95% confidence interval, the z value = 1.96Lower limit of 95% confidence interval¿Mean−(zvalue∗√p(1−p)n)= (0.395)−{(1.96)×(√0.395(1−0.395)200)}=0.327Upper limit of 95% confidence interval¿Mean+(zvalue∗√p(1−p)n)= (0.395)+{(1.96)×(√0.395(1−0.395)200)}=0.463Hence, the 95% confidence interval for the proportion is [0.327 0.463]. This implies that thereis 95% that the proportion of retailers selling below the recommended price falls between 0.327and 0.463. 1
QUESTION 2 Probability distributionCalculation of probabilities S. No.CalculationProbability(a)P(X>0)¿{P(X=1)+P(X=2)}=(0.4+0.1)=0.50.5(b)P(X≤0)¿{P(X=0)+P(X=−4)}=(0.3+0.2)=0.50.5(c)P(x≤X≤1)¿{P(X=1)+P(X<0)+P(X←4)}=(0.4+0.3+0.2)=0.90.9(d)P(X=−2)=00(e)P(X=−4)=0.20.2(f)P(X<2)¿{P(X=1)+P(X=0)+P(X=−4)}=(0.4+0.3+0.2)=0.90.9QUESTION 3 Marijuana useCigarette SmokingHas usedNever usedTotalHas smoked488196684Never smoked30216861988Total79018822672The value of test statistics X2=770.6Step 1 The respective hypotheses are highlighted below:2
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