Major Questions in Stat Exam - Statistics Project Example

1. Given that Mean Arrival Rate = µ = 2.5 per hour Now p (x) = e ^- µ * µ^x x! a. Probability exactly arriving = p e^-2.5 * 2.5 1! = .205 b. Probability of less than 5 customers arriving in an hour = p (1) + p(2) + p(3) + p(4) = e^-2.5 * (2.5^1/1! +2.5^2/2! +2.5^3/3! +2.5^4/4!) = .891 c. Expected number of customers in any given 2 hour period = 2* Mean arrival rate per hour = 2*2.5 =5 d. Probability of 10 or more customers arriving in the next 2 hours = 1- Probability of less than 10 customers arriving in next 2 hours = p(1) +p(2)+……p(10) where Mean arrival rate per 2 hours =5 This can be done manually or obtained using Poisson function in excel The value comes out as: p(>=10) = 1-p( µ Ha: x bar .05, we fail to reject the null hypothesis and hence it can be concluded that the average age of cars driven to work by plant’s employees is greater than the national average. 4. For Peoria, Sample Mean = x1 bar = $86900 Sample Standard deviation = s1 = $2300 Sample size = n1 = 21 For Evansville, Sample Mean = x2 bar = $84000 Sample Standard deviation = $1750 Sample size = n2 = 26 The hypotheses can be formulated as: Ho: x1 bar = x2 bar Ha: x1 bar is not equal to x2 bar Now using t-test for difference in means, T statistic = (x1 bar –x2 bar)/[(s1^2/n1 + s2^2/n2)]^.5 = (86900-84000)/[(2300^2/21 + 1750^2/26)]^.5 = 4.77 Degrees of Freedom = (s1^2/n1 + s2^2/n2)^2/(s1^2/n1)^2/(n1-1)+(s2^2/n2)^2/(n2-1) = 36.67 = 37 p-value for the given degrees of freedom and significance level = .000029 Since p-value is less than .05, the null hypothesis is rejected and it can be concluded that there is a significant difference in the midrange prices of two cities. 5. The given data can be tabulated as shown below. The suitable hypothesis test for this would be paired t-test Given that, Level of significance =.05   2004 Q1 sales 2004 Q1 Market share 2003 Market share (Difference in market shares)^2 (Difference in market shares)^2/2003 Market share Toyota Camry 480 40.00 37.00 9.00 0.243243243 Honda Accord 390 32.50 34.00 2.25 0.066176471 Ford Taurus 330 27.50 29.00 2.25 0.077586207 Total 0.387005921 The hypothesis can be formulated as: Ho: The market shares of 2004 Q1 are equal to market shares of 2003 for all companies Ha: The market shares of 2004 Q1 are not equal to market shares of 2003 for all companies From the table chi-square statistic = .387 Here n=3 Degrees of freedom here = 3-1=2 Therefore, p-value of the chi-square statistic = .824 As p-value > .05, the null hypothesis is accepted and it can be concluded that there is not a significant difference between market shares of the companies in 2003 and 2004 Quarter 1. 6. The following contingency table can be used for calculating the chi-square statistic as 135. The hypothesis can be formulated as: Ho: The type of car purchased is independent of the city in which the purchaser lives Ha: The type of car purchased is not independent of the city in which the purchaser lives. Degrees of freedom =(m-1)(n-1) = (2-1)(3-1) =2 p-value for this chi-square statistic at .05 significance level =.0000 Since p-value .05, null hypothesis is rejected and fee is a significant variable 11. References The Anova Table for Regression Retrieved August 12, 2011 from Read More