A credit card company claims that the mean credit card debt for individuals is greater than $5,100.
You want to test this claim. You find that a random sample of 27 cardholders has a mean credit card balance of $ 5,270 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.

(a) Write the claim mathematically and identify H0 and Ha.

(b) Find the critical​ value(s) and identify the rejection​ region(s).

What​ is(are) the critical​ value(s), t0​?

(c) Find the standardized test statistic t.

​(d) Decide whether to reject or fail to reject the null hypothesis.

​(e) Interpret the decision in the context of the original claim.

Respuesta :

Answer:

We conclude that that there is enough evidence to support the claim mean credit card debt for individuals is greater than $5,100.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = $5,100

Sample mean, [tex]\bar{x}[/tex] = $ 5,270

Sample size, n = 27

Alpha, α = 0.10

Sample standard deviation, s = $550

a) First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 5100\text{ dollars}\\H_A: \mu > 5100\text{ dollars}[/tex]

We use one-tailed t test to perform this hypothesis.

b) Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{5270 - 5100}{\frac{550}{\sqrt{27}} } =1.606[/tex]

c) Now,

[tex]t_{critical} \text{ at 0.10 level of significance, 26 degree of freedom } = 1.314[/tex]

Since,                    

[tex]t_{stat} > t_{critical}[/tex]

d) We fail to accept the null hypothesis, and accept the alternate hypothesis.

e) We conclude that that there is enough evidence to support the claim mean credit card debt for individuals is greater than $5,100.

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