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At one vehicle inspection station, 13 of 52 trucks and 11 of 88 cars failed the emissions test. assuming these vehicles were representative of the cars and trucks in that area, what is the standard error of the difference in the percentages of all cars and trucks that are not in compliance with air quality regulations?

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nobillionaireNobley
The standard error (SE) of the sampling distribution difference between two proportions is given by:

[tex]SE=\sqrt{p(1-p)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}[/tex]

where p is the pooled sample proportion, [tex]n_1[/tex] is the size of sample 1, and [tex]n_2[/tex] is the size of sample 2.

[tex]p= \frac{p_1n_1+p_2n_2}{n_1+n_2} [/tex]

Given that at one vehicle inspection station, 13 of 52 trucks and 11 of 88 cars failed the emissions test.

[tex]p_1= \frac{13}{52} =0.25 \\ \\ p_2= \frac{11}{88} =0.125 \\ \\ n_1=52 \\ \\ n_2=88 \\ \\ p= \frac{0.25(52)+0.125(88)}{52+88} \\ \\ = \frac{13+11}{140} = \frac{24}{140} =0.1714[/tex]

[tex]SE=\sqrt{0.1714(1-0.1714)\left(\frac{1}{52}+\frac{1}{88}\right)} \\ \\ = \sqrt{0.1714(0.8286)(0.0192+0.0114)} \\ \\ = \sqrt{0.1714(0.8386)(0.0306)} = \sqrt{0.0044} \\ \\ =0.0663[/tex]
Mamasu
The two given proportions are
     [tex]p_1=\frac{13}{52}=0.25[/tex] and [tex]p_2=\frac{11}{88}=0.125[tex].

The standard error is given by the formula
     [tex]SE=\sqrt{\frac{p_1\left(1-p_1\right)}{n_1}+\frac{p_2\left(1-p_2\right)}{n_2}}[/tex]

Substituting the given values, we have
     [tex]SE=\sqrt{\frac{0.25\left(1-0.25\right)}{52}+\frac{0.125\left(1-0.125\right)}{88}=0.0696}[/tex]

Therefore, the standard error is 0.0696.

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