Question

In order to estirmate the difftrence bcthween tht yrar y unomts of mar teting managers n the East and west of tre untedl States stnired Sttes the folong nformalron was qatincred. The values of 12 andt 18 are mean East n1 = 40 West n2 45 -72 (in $1,000)78 (in $1,000) s1#6(in S1,000) s2-8(in S1.000) I. Develop an interval estimate for the difference between the average yearly incomes of the marketing managers in the East and West. Use α = 0.05. Show your calculation for the degrees of freedom Show your calculation for the interv AL 95% confidence, use the p-value approach and test to determine managers in the East is significantly different from the West. Show the calculation of p. Should we reject or fail to reject the null hypothesis? Why? 1. 2. 2 al including the formula and how you plugged in the numbers. if the average yearly income of marketing Week 7 Question 1 Assume you have noted the following prices for paperback books and the number of pages that each book contains. Develop a least-squares estimated regression line. Book Pages x) Price (y) 500 $7.00

Answer 1

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Following is the 95% confidence interval :

Here we have 72,81-6, n-40 =78,S,-8,1,-45 X2 Level of significance a0.05 CL-0.95 Con fidence level: Since we can assume that variances are equal so degree of freedom oft test will be: The pooled standard deviation is: (n, -1)s21+ (n2-1)s2 -7.1305 Critical value: , =1.989 The standard error is: -+--7.1305" sart((1/40)+(1/45)) = 1.5495 Margon of error: ME- t SE 3.082 Confidence interval is: (4-5)±ME =(72-78)±3.082 -(-9.082, -2.918) Excel function for critical value:TINV(0.05,83) Hence, the required confidence interval is (-9.082,-2.918)

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