Question

a credit card company claims that the mean credit card debt for individuals is greater than $4,900. you want to test this claim. you find that a random sample of 35 cardholders has a mean credit card balance of $5,154 and a standard deviation of $550. at α = 0.10, can you support the claim? complete parts (a) through (e) below. assume the population is normally distributed.

ha: μ < $4,900
ha: μ ≤ $4,900
ha: μ > $4,900
d. h0: μ = $4,900
e. h0: μ = $4,900
f. h0: μ > $4,900
ha: μ ≠ $4,900
ha: μ > $4,900
ha: μ ≤ $4,900
(b) find the critical value(s) and identify the rejection region(s).
what is(are) the critical value(s), t0?
t0 = 1.307
(use a comma to separate answers as needed. round to three decimal places as needed.)
determine the rejection region(s). select the correct choice below and fill in the answer box(es) within your choice.
(round to three decimal places as needed.)
a. t <
b. < t <
c. t < and t >
d. t > 1.307
(c) find the standardized test statistic t
t = (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the t - statistic

The formula for the t - statistic in a one - sample t - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step2: Identify the values

We are given that \(\bar{x} = 5154\), \(\mu=4900\), \(s = 550\), and \(n = 35\).

Step3: Substitute the values into the formula

First, calculate \(s/\sqrt{n}\):
\(\frac{s}{\sqrt{n}}=\frac{550}{\sqrt{35}}\approx\frac{550}{5.916}\approx93\)
Then, calculate \(t\):
\(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=\frac{5154 - 4900}{93}=\frac{254}{93}\approx2.73\)

Answer:

\(t = 2.73\)