Question

2. (6 pts) A credit bureau analysis of undergraduate students credit records found that the average number of credit cards in an undergraduate's wallet was 4.09 ("Undergraduate Students and Credit Cards in 2004," Nellie Mae, May 2005). It was also reported that in a random sample of 132 undergraduates, the sample mean number of credit cards that the students said they carried was 2.6. The sample standard deviation was not reported, but for purposes of this exercise, suppose that it was 1.2. Is there convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of 4.09? (a) (1 pt) What is the variable? Is it categorical or quantitative? (b) (1 pt) Define the parameter of interest in the context of the problem. (c) (1 pt) State the null and alternative hypotheses the participants in the (d) (1 pt) Is it reasonable to conclude the responses fro: sample are independent? Assuming the null hypothesis is true, is it reasonable to conclude the sampling distribution of X is approximately bell-shaped? (e) (1 pt) Determine the test statistic and p-value. 1 (e) ( pt) Conclude in the context of the problem

Answer 1

(a) The variable is quantitative

Note : quantitative variables are variable that are numeric and can be measured .

categorical data are qualitative data that are of types such as good /bad , old/new etc.

(b) Parameter : Population mean number of credit cards in an undergraduate's wallet.

(c)

The null and alternative hypothesis

$H0: \mu =4.09, Ha: \mu< 4.09$

(d) Yes sample observations are independent of each other , as number of credit card an undergraduate possesses is independent of what another one possesses.

Though distribution of population is known , as sample size is large (132) , the sampling distribution of sample mean is approximately normal that is bell shaped.

(e) Though sample size is large we cannot use z test ss population standard deviation is not known . We use t test

Test statistic

X - u t=sn

  $=\frac{2.6-4.09 }{1.2 /\sqrt{132}}= -14.27$

df = 132-1 =131

P value is less than 0.0001

Note : P value using excel"=T.DIST.RT(14.27,131)"

(e) Since P value is very small , the result is significant

We reject H0.

There is sufficient evidence to conclude that mean number of credit cards that the undergraduates report carrying is less than 4.09.