Question

CH 10.

1.

The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 46 male consumers was $135.67, and the average expenditure in a sample survey of 34 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $38, and the standard deviation for female consumers is assumed to be $20.

1. What is the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females (to 2 decimals)?
   
2. At 99% confidence, what is the margin of error (to 2 decimals)?
   
3. Develop a 99% confidence interval for the difference between the two population means (to 2 decimals). Use z-table.
   (  ,   )

2

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 (U.S. Airways Attache, December 2003). 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 1) and the annual credit card charges for dining out (population 2). Using the difference data, the sample mean difference was    = $827, and the sample standard deviation was sd= $1,151. Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out. H0:  d Selectgreater than or equal to 0greater than 0less than or equal to 0less than 0equal to 0not equal to 0Item 1 Ha:  d Selectgreater than or equal to 0greater than 0less than or equal to 0less than 0equal to 0not equal to 0Item 2 Use a .05 level of significance. What is the p-value? The p-value is Selectless than .01between .01 and .02between .02 and .05between .05 and .10between .10 and .20between .20 and .40greater than .40Item 3 Can you conclude that the population means differ? SelectThere is a difference between the annual mean expendituresCannot conclude there is a difference between the annual mean expendituresItem 4 Which category, groceries or dining out, has a higher population mean annual credit card charge? SelectGroceriesDining outItem 5 What is the point estimate of the difference between the population means? $ What is the 95% confidence interval estimate of the difference between the population means (to the nearest whole number)? (  ,  )

Answer 1

##Using R codes :

> ##1)
> mu=100.89
> nm=46;xm=135.67;nf=34;xf=68.64
> sm=38;sf=20
> #a) Point estimate of difference between population mean expenditure for males & population mean expenditure for females is
> PE=xm-xf;PE
[1] 67.03
> #b) Margin of error
> ME=round(qnorm(0.99)*sqrt((sm^2/nm)+(sf^2)/nf),2);ME
[1] 15.28
> #c) 99% C.I.
> c(PE-ME,PE+ME)
[1] 51.75 82.31
> ##2)
> n=42
> md=827
> sd=1151
> #a) Null & alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.
> #H0 : equal to 0 v/s Ha : not equal to 0
> #b)
> ts=md/(sd/sqrt(n));ts #test statistic
[1] 4.656449
> ttab=qnorm(0.95);ttab #table value
[1] 1.644854
> ##ts>ttab, reject H0 at 5 % l.o.s.
> pval=1-pnorm(ts);pval #p-value
[1] 1.608551e-06
> ##So, p-value < 0.01.
> ## Reject H0 at 5 % l.o.s.
> ##There is a difference between the annual mean expenditures.
> #c)
> ## The point estimate of the difference between the population means is md=827
> ##95% C.I.estimate of the difference between the population means is
> c(md-(qnorm(0.95)*sd/sqrt(n)),md+(qnorm(0.95)*sd/sqrt(n)))
[1] 534.8688 1119.1312