Answer to Question #173544 in Statistics and Probability for rylan

Question

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

Answer 1

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

We have that

$n = 20$

$\bar x =91$

$s = 21$

$\alpha=0.05$

$H_0: \mu \le80$

$H_a:\mu>80$

The hypothesis test is right-tailed.

Since the population standard deviation is unknown we use the t-test.

The critical value at the 5% significance level is and 19 df is 1.73

(degrees of freedom df = n – 1 = 20 – 1 = 19)

The critical region is t > 1.73

Test statistic:

$$t=\frac{\bar x-\mu}{\frac{s}{\sqrt n}}=\frac{91-80}{\frac{21}{\sqrt 20}}=2.34$$

Since 2.34 > 1.73 thus t falls into rejection region therefore we reject the null hypothesis.

At the 5% significance level the data do provide sufficient evidence to confirm the agent’s suspicion. We are 95% confident to conclude that the mean cash sales are more than $80.