This is an assignment on Statistics for Business which includes topics like covariance, correlation coefficient, probability, hypothesis testing and more. The assignment also includes references for further reading.
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Question 1 Variable description X = Years of experiences Y = Salary ($’000) (a)Covariance X bar = 44/8 = 5.5 Y bar = 148/8 = 18.5 Covariance=1 n−1∑¿¿ 1
(b)The covariance is negative primarily because the sample observations may belong to a profession where the salaries are inversely proportional to the age. (c)Calculation of correlation coefficient Correlationcoefficient=∑¿¿¿ Interpretation: The value of correlation coefficient implies that a perfect negative correlation is present between the variables. The strength of association is very strong which indicates that as the year of experiences (x) is increased then the salary (y) would be decreased (Eriksson and Kovalainen, 2015). The relationship can also be seen from the scatter plot shown below. 12345678910 0 5 10 15 20 25 30 Scatter Plot Year of experience Salary ($'000) 2
(d)The reason behind the negative relationship is due to the fact that as the experience in years tends to rise, there is a drop in the salary levels (Hillier, 2016). Question 2 a)Value of ƛ Here,∝=3 ƛ=1 3 b)Proportion of customers that hold more than 1.5 min will hang up before placing an order P(x>1.5)=e−0.5=0.6065 c)Waiting time at which only 10% of customers will continue to hold P(X>x)=e−ƛx e −x 3=0.10 x=6.908minutes d)Probability that a randomly selected caller would be placed on hold for 3 to 6 min P(3<x<6)=e −3 3-e −6 3=e−1−e−2=0.2325 Question 3 a)Probability of type II error, whenμ=1000and alpha = 0.10 3
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β=P(884.2<x<1015.8)=P(−2.9<z<0.40) β=0.6535 b)Power of testwhenμ=1000and alpha = 0.10 Here, Power=1−β=1−0.6535=0.3465 c)Power of test refers to the underlying probability associated with correctly rejecting a false null hypothesis. Hence, in the given case, there is 0.3465 probability of the correct rejection of null hypothesis (i.e. life is 950 hours) (Flick, 2015). d)Theβwill decrease as n increases and this conclusion is also supported from the given data. Question 4 Standard deviation = 6 ounces Mean filling weight = 47 ounces Sample size = 36 containers Level of significance = 5% Test whether the sample mean filling weight is 48.6 ounces or not. Hypothesis testing Hypotheses 4
NullhypothesisH0:μ=48.6 AlternativehypothesisH1:μ≠48.6 Test statistic The value of test statisticz=47−48.6 6 √36 =−1.6 The p value The p value = 2 NORMSDIST (z value) =2 NORMSDIST (-1.6) = 2* 0.0547 = 0.10959 It can be seen that p value is higher than level of significance and thus, it can be said that null hypothesis would not be rejected and thus, alternative hypothesis would not be accepted (Koch, 2013). Therefore, the sample mean filling weight is 48.6 ounces. References Eriksson, P. and Kovalainen, A. (2015)Quantitative methods in business research. 3rd ed. London: Sage Publications. Flick, U. (2015)Introducing research methodology: A beginner's guide to doing a research project.4th ed. New York: Sage Publications. Hillier, F. (2016)Introduction to Operations Research.6th ed.New York: McGraw Hill Publications. Koch, K.R. (2013)Parameter Estimation and Hypothesis Testing in Linear Models.2nd ed. London: Springer Science & Business Media. 5