Question

According to a lending​ institution, students graduating from college have an average credit card debt of

​$4100.

A random sample of

40

graduating seniors was​ selected, and their average credit card debt was found to be

​$4428.

Assume the standard deviation for student credit card debt is

​$1,300.

Using

alphaαequals=0.01​,

complete parts a through c.

​a) Does this sample provide enough evidence to challenge the findings by the lending​ institution?

Determine the null and alternative hypotheses.

Upper H 0H0​:

muμ

▼

greater than or equals≥

not equals≠

less than<

less than or equals≤

greater than>

equals=

4100

Upper H 1H1​:

muμ

▼

greater than or equals≥

not equals≠

less than or equals≤

greater than>

less than<

4100

​a) The​ z-test statistic is

​(Round to two decimal places as​ needed.)

The critical​ z-score(s) is(are)

​(Round to two decimal places as needed. Use a comma to separate answers as​ needed.)

Reject or fail to reject why or why not

​b) Determine the​ p-value for this test.

The​ p-value is

​(Round to four decimal places as​ needed.)

​c) Identify the critical sample mean or means for this problem.

The critical sample​ mean(s) is(are)

​(Round to two decimal places as needed. Use a comma to separate answers as​ needed.)

Answer 1

H0: mu = 4100
Ha: mu not equals 4100

n = 40, xbar = 4428
sigma = 1300

a)
Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (4428 - 4100)/(1300/sqrt(40))
z = 1.6

critical values of z are -2.58 and 2.58

Fail to reject

b)
p-value = 0.1096

c)
critical sample means are
lower limit = 4100 - 2.58*1300/sqrt(40) = 3569.69
upper limit = 4100 + 2.58*1300/sqrt(40) = 4630.31

(3569.69 , 4630.31)