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A scrap metal dealer claims that the mean of his sales is "no more than $80," but an Internal Revenue Service agent believes the dealer is untruthful. Observing a sample of 20 cash customers, the agent finds the mean purchase to be $91, with a standard deviation of $21. Assuming the population is approximately normally distributed and using the 0.05 level of significance, is the agent’s suspicion confirmed?

Answer :

Answer:

There is enough not evident to support dealer's claim that his sales is no more than $80.                              

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = $80

Sample mean, [tex]\bar{x}[/tex] = $91

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = $21

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu \leq 80\text{ dollars}\\H_A: \mu > 80\text{ dollard}[/tex]

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

Formula:

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

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{91 - 80}{\frac{21}{\sqrt{20}} } = 2.3425[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = 1.729[/tex]

Since,                        

The calculated test statistic is greater than the critical value, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

There is enough not evident to support dealer's claim that his sales is no more than $80.

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