Respuesta :
Answer:
D. Yes, since the p-value of the one-sided test is less than 5%.
Step-by-step 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:
[tex]H_0: p = 0.3[/tex]
At the alternate hypothesis, we test if this proportion is less than 30%, that is:
[tex]H_1: p < 0.3[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.3 is tested at the null hypothesis:
This means that [tex]\mu = 0.3, \sigma = \sqrt{0.3*0.7}[/tex]
Of 350 randomly selected trucks, 87 were carrying hazardous materials.
This means that [tex]n = 350, X = \frac{87}{350} = 0.2486[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.2486 - 0.3}{\frac{\sqrt{0.3*0.7}}{\sqrt{350}}}[/tex]
[tex]z = -2.1[/tex]
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.
