Question

One-hour carbon monoxide concentrations in 43 air samples from a section of a city showed an average of 11.6 ppm and a standard deviation of 7.08. after a traffic control strategy was put into place, 19 air samples showed an average carbon monoxide concentration of 6.4 ppm and a standard deviation of 6.94. it is known that carbon monoxide concentrations are normally distributed. the state will adopt the traffic control strategy on a large scale if there is evidence that it reduces carbon monoxide concentrations by at least 2 ppm. with h_0:\mu_1-\mu_2=2; \ \ \ h_a:\mu_1-\mu_2>2 h 0 : μ 1 − μ 2 = 2 ; h
a.μ 1 − μ 2 > 2 the appropriate test statistic is . at the 10% level of significance, we , which means that .

Answer 1

The test statistic for difference between two means is given by:

$z= \frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{ \sqrt{ \frac{\sigma^2_1}{n_1} +\frac{\sigma^2_2}{n_2}} }$

Thus, the appropriate test statistic is given by:

$z= \frac{(11.6-6.4)-2}{ \sqrt{ \frac{(7.08)^2}{43} +\frac{(6.94)^2}{19}} } \\ \\ = \frac{5.2-2}{ \sqrt{ \frac{50.1264}{43} + \frac{48.1636}{19} } } = \frac{3.2}{ \sqrt{1.1657+2.5349} } \\ \\ = \frac{3.2}{ \sqrt{3.7007} } = \frac{3.2}{1.9237} =1.66$

At 10% level of significance, the rejection region is given by

$z\geq z_0$

where: $z_0$ for 10% level of significance is 1.65

Since z = 1.66 > $z_0=1.65$, we reject the null hypothesis, which means that the traffic control strategy does not reduce carbon monoxide concentrations by at least 2 ppm.