Question

A firm is hired by a small town to estimate the proportion of large trucks that are carrying hazardous materials on the main highway through the town. Of 350 randomly selected trucks, 87 were carrying hazardous materials. Is there evidence at the 5% level that the proportion of large trucks carrying hazardous materials is less than 30%?

A. No, since the p-value of the one-sided test is less than 5%.
B. Yes, since the sample proportion is less than 30%
C. The sample size is too small to make a determination.
D. Yes, since the p-value of the one-sided test is less than 5%.
E. No, since the sample proportion is larger than 5%

Answer 1

D. Yes, since the p-value of the one-sided test is less than 5%.

Explanation:

Test if there is evidence that the proportion of large trucks carrying hazardous materials is less than 30%.

At the null hypothesis, we test that this proportion is of 30%, that is:

`H_0: p = 0.3`

At the alternate hypothesis, we test if this proportion is less than 30%, that is:

`H_1: p < 0.3`

The test statistic is:

`z = (X - \mu)/((\sigma)/(√(n)))`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis,
`\sigma` is the standard deviation and n is the size of the sample.

0.3 is tested at the null hypothesis:

This means that
`\mu = 0.3, \sigma = √(0.3*0.7)`

Of 350 randomly selected trucks, 87 were carrying hazardous materials.

This means that
`n = 350, X = (87)/(350) = 0.2486`

Value of the test statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (0.2486 - 0.3)/((√(0.3*0.7))/(√(350)))`

`z = -2.1`

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.2486, which is the p-value of z = -2.1.

Looking at the z-table, z = -2.1 has a p-value of 0.0179.

The p-value of the test is 0.0179 < 0.05, which means that there is evidence at the 5% level that the proportion of large trucks carrying hazardous materials is less than 30%. Thus, the correct answer is given by option D.