Question
Chapter 10, Question 23 part a and part b.
Let d dexte te dange in price per share for company i whered,st quarter of 2 values to estimate the dollar amount a share of stock has changed during the Ist quare price per share of stock during the first quarter? Interpret this result a. mean of thee price per share minus the beginning of 2012 price per share. Use the sample meas of fhee b. What is the 95% confidence interval estimate of the population mean change in- 23. Bank of America's Consumer Spending Survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries. dining out, household expenses, home furnishings, apparel, and entertainment. Using data from a sample of 42 credit card accounts, assume that each account was used to identify the annual credit card charges for groceries (population I) and the annual credit card charge for dining out (population 2). Using the difference data, the sample mean difference was d-$850, and the sample standard deviation was ¾-S1123 a Formulate the null and alternative hypotheses to test for no difference betwoen he population mean credit card charges for groceries and the population mean credit cand charges for dining out b. Use a 05 level of significance. Can you conclude that the population meats dif- fer? What is the p-value? c. Which category, groceries or dining out, has a higher population mean annual credi card charge? What is the point estimate of the difference between the populatin means? What is the 95% confidence interval estimate of the difference between the population means? DATA24 : The Global Business Travel Association reported the domestic airfare for business travel for the current year and the previous year (INC. Magazine, Fsbraary 2012). Below is a 24· BusinessTravel sample of 12 flights with their domestic airfares shown for byears Previous Year 585 650 545 547 508 580 Current Year 345 526 420 216 285 Previous Year 315 Currest Year 635 710 462 206 275 432 605 517 570 610
Explanation / Answer
given n= 42
standard deviation =1123
mean =850
a).H0: (population mean)d = 0
Ha: (population mean)d 0
b) tstatstic = 850-0 / 1123/sqrt(42) =4.905
Test statistics is Degree of freedom: df =n-1=41
p-value of the test is : 0.00003
Since p-value is less than 0.05 so we reject the null.hypothesis
