Question

A scrap metal dealer claims that the mean of his cash scales 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 purchases to be $91, with a standard deviation of $21. Assuming the population is approximately normally distributed, and using the 0.05 level of significance, what is the calculated value of test statistic? (Specify your answer to the 3rd decimal.)

Answer 1

Answer:

The p-value for this hypothesis test is P=0.015.

Step-by-step explanation:

In this case we have hypothesis test for the mean, with standard deviation of the population unknown.

The null hypothesis we want to test is

$H_0: \mu\leq80$

To work with this test we have a sample of size n=20, sample mean=91 and sample standard deviation=21.

First, we estimate the standard deviation of the population

$s_M=s/\sqrt{n}=21/\sqrt{20}=4.696$

Then, because we have an estimated standard deviation, we have to calculate the statistics t.

$t=\frac{M-\mu}{s_M}=\frac{91-80}{4.696} =2.342$

We can look up this value of t in a t-table to know the probability of this value, taking into account 19 degrees of freedom:

$df=n-1=20-1=19$

The p-value or the probability of P(t>2.342) is 0.01511.

This value P=0.0151 is compared to the significance level (0.05). Since the probability value (0.0151) is less than the significance level (0.05) the effect is statistically significant. Since the effect is significant, the null hypothesis is rejected.