Question

Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln Tunnel connecting New York and New Jersey. The concentrations ( $\mathbf{\pm }$standard deviations) ofm- andp -xylene were

$$\begin{gathered} \mathbf{}\mathbf{Turnpike}\mathbf{:}\mathbf{}\mathbf{}\mathbf{31}\mathbf{.}\mathbf{4}\mathbf{\pm }\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{\mu g}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{32}\mathbf{}\mathbf{measurements}\mathbf{}\mathbf{)} \\ \mathbf{Tunnel}\mathbf{:}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{52}\mathbf{.}\mathbf{9}\mathbf{\pm }\mathbf{29}\mathbf{.}\mathbf{8}\mathbf{\mu g}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{(}\mathbf{32}\mathbf{}\mathbf{measurements}\mathbf{}\mathbf{)} \end{gathered}$$

Do these results differ at the 95%confidence level? At the 99%confidence level?

Short answer

The analyses done from the New Jersey Turnpike through the Lincoln Tunnel at 95% and 99% confidence intervals differ significantly in terms of their means but not significantly in terms of their standard deviations and variances.

Step-by-step solution

Perform F test

For comparing two different locations of analysis, use both the test and the test.

The F-test is done as shown below.

$$\begin{gathered} F_{calc}=\frac{s_1^2}{s_2^2} \\ =\frac{{\left(30.0\right)}^2}{{\left(29.8\right)}^2} \\ =1.01. \end{gathered}$$

$F_{table}\left(95\%CL\right)\approx 1.84.$

$F_{table}\left(99\%CL\right)\approx 2.35.$

Since $F_{calc}<F_{table}$at both 95% and 99% confidence levels, there is no significant difference between the variances and the standard deviations of the analysis done in two different locations at both confidence levels.

Perform t test

The t-test is done as shown below.

The conclusion in our F test result states that $F_{calc}<F_{table}$ at both confidence levels. Now, perform a case 2a of the -test.

$$\begin{gathered} s_{pooled}=\sqrt{\frac{s_1^2\left(n_1-1\right)+s_2^2\left(n_2-1\right)}{n_1+n_2-n_t}} \\ =\sqrt{\frac{{\left(30.0\right)}^2\left(32-1\right)+{\left(29.8\right)}^2\left(32-1\right)}{32+32-2}} \\ =29.90016722 \end{gathered}$$

$$\begin{gathered} t_{calc}=\frac{\left|{\overline{\mathrm{x}}}_1-{\overline{\mathrm{x}}}_2\right|}{{\mathrm{s}}_{\mathrm{pooled}}}\sqrt{\frac{{\mathrm{n}}_1{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}} \\ =\frac{\left|31.4-52.9\right|}{29.90016722}\sqrt{\frac{\left(32\right)\left(32\right)}{32+32}} \\ =2.88 \end{gathered}$$

$$\begin{gathered} t_{table}\left(95\%CL\right)\approx 2.042. \\ {t}_{table}\left(99\%CL\right)\approx 2.750. \end{gathered}$$

Since $t_{calc}>t_{table}$at both 95% and 99% confidence levels, there is a significant difference between the means of the analysis done in two different locations at both confidence levels.