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 34 cardholders has a mean credit card balance of $5,081 and a standard deviation of $625. at \\( \alpha = 0.05 \\), can you support the claim? complete parts (a) through (e) below. assume the population is normally distributed
(a) write the claim mathematically and identify \\( h _ { 0 } \\) and \\( h _ { a } \\).
which of the following correctly states \\( h _ { 0 } \\) and \\( h _ { a } \\) ?
\\( \

$$\begin{array} { l l } { \\text { a. } h _ { 0 } : \\mu > 54,900 } & { \\text { b. } h _ { 0 } : \\mu \\leq 54,900 } \\\\ { h _ { a } : \\mu \\leq 54,900 } & { h _ { a } : \\mu > 54,900 } \\end{array}$$

\\)
\\( \

$$\begin{array} { l l } { \\text { c. } h _ { 0 } : \\mu \\geq 54,900 } & { } \\\\ { h _ { a } : \\mu < 54,900 } & { } \\end{array}$$

\\)
\\( \

$$\begin{array} { l l } { \\text { d. } h _ { 0 } : \\mu > 54,900 } & { \\text { e. } h _ { 0 } : \\mu = 54,900 } \\\\ { h _ { a } : \\mu \\leq 54,900 } & { h _ { a } : \\mu > 54,900 } \\end{array}$$

\\)
\\( \

$$\begin{array} { l l } { \\text { f. } h _ { 0 } : \\mu = 54,900 } & { } \\\\ { h _ { a } : \\mu \ eq 54,900 } & { } \\end{array}$$

\\)
(b) find the critical value(s) and identify the rejection region(s).
what is(are) the critical value(s), \\( t _ { 0 } \\) ?
\\( t _ { 0 } = 1.692 \\)
(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.)
\\( \

$$\begin{array} { l l } { \\text { a. } t < } & { \\text { b. } < t < } \\\\ { \\text { c. } t < \\text { and } t > } & { \\text { d. } t > 1.692 } \\end{array}$$

\\)
(c) find the standardized test statistic t.
\\( t = \square \\) (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 population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.

We are given that \(\bar{x} = 5081\), \(\mu=4900\), \(s = 625\), and \(n = 34\).

Step2: Substitute the values into the formula

First, calculate the denominator \(s/\sqrt{n}\):
\(\frac{s}{\sqrt{n}}=\frac{625}{\sqrt{34}}\approx\frac{625}{5.831}\approx107.19\)

Then, calculate the numerator \(\bar{x}-\mu\):
\(\bar{x}-\mu=5081 - 4900=181\)

Now, find the t - statistic:
\(t=\frac{181}{107.19}\approx1.69\)

Answer:

\(t\approx1.69\)