Listed in the accompanying table are waiting times (in seconds) of observed cars at a Delaware inspection station. The data from two waiting lines are real observations, and the data from the single waiting line are modeled from those real observations. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b).

Let population 1 correspond to the single waiting line, and let population 2 correspond to the two waiting lines.

Hypotheses:

A. [tex]$ H_0: \mu_1 = \mu_2 $[/tex]
[tex]$ H_1: \mu_1 \neq \mu_2 $[/tex]

C. [tex]$ H_0: \mu_1 = \mu_2 $[/tex]
[tex]$ H_1: \mu_1 \ \textgreater \ \mu_2 $[/tex]

Calculate the test statistic:

[tex]\[ t = 0.16 \][/tex]
(Round to two decimal places as needed.)

Find the P-value:

P-value [tex]$ = 0.874 $[/tex]
(Round to three decimal places as needed.)

Conclusion about the null hypothesis:

Fail to reject [tex]$ H_0 $[/tex] because the P-value is greater than the significance level.

Final Conclusion:

The waiting time of cars in two lines is equal to that of cars in a single line.

(b) Construct the confidence interval suitable for testing the claim in part (a).

[tex]\[ \square \ \textless \ \mu_1 - \mu_2 \ \textless \ \square \][/tex]
(Round to one decimal place as needed.)

Waiting Times Table:

[tex]\[
\begin{array}{|cc|cc|}
\hline
\multicolumn{2}{c|}{\text{One Line}} & \multicolumn{2}{c|}{\text{Two Lines}} \\
\hline
63.6 & 733.6 & 63.5 & 865.4 \\
157.3 & 606.2 & 215.5 & 1089.8 \\
141.9 & 267.9 & 85.7 & 663.1 \\
278.5 & 310.2 & 339.9 & 518.2 \\
252.7 & 129.2 & 199.7 & 565.8 \\
476.3 & 133.2 & 630.3 & 267.9 \\
477.9 & 122.2 & 332.7 & 350.1 \\
473.6 & 128.8 & 328.6 & 94.8 \\
401.5 & 233.3 & 915.3 & 99.8 \\
721.6 & 460.7 & 552.6 & 162.6 \\
761.3 & 481.6 & 596.7 & 100.6 \\
692.3 & 518.1 & & \\
837.2 & 508.9 & & \\
902.7 & 579.9 & & \\
\hline
\end{array}
\][/tex]

To solve this problem, we'll go through the steps provided and understand the results derived from the problem statement. We are comparing waiting times from two different setups at an inspection station: a single waiting line and two waiting lines. Here's how to interpret the problem:

### Part (a): Hypothesis Testing

1. Hypotheses:
- Null Hypothesis ([tex]$H_0$[/tex]): The means of the two populations ([tex]$\mu_1$[/tex] and [tex]$\mu_2$[/tex]) are equal ([tex]$\mu_1 = \mu_2$[/tex]).
- Alternative Hypothesis ([tex]$H_1$[/tex]): The means of the two populations are not equal ([tex]$\mu_1 \neq \mu_2$[/tex]).

2. Test Statistic:
- The test statistic calculated is approximately [tex]$t = 0.16$[/tex]. This statistic measures the standardized difference between the means of the two samples.

3. P-value:
- The P-value found is approximately [tex]$0.874$[/tex]. The P-value helps us determine the strength of the evidence against the null hypothesis.

4. Conclusion:
- We typically set a significance level (commonly 0.05 for 95% confidence) to decide whether to reject the null hypothesis.
- Since the P-value is [tex]$0.874$[/tex], which is greater than the significance level of 0.05, we fail to reject the null hypothesis.
- Conclusion: There is not enough evidence to conclude that the waiting times for a single line are different from two waiting lines.

### Part (b): Confidence Interval

1. Confidence Interval for the Difference in Means ([tex]$\mu_1 - \mu_2$[/tex]):
- A 95% confidence interval for the difference in means is calculated to further interpret the results.
- The confidence interval is: [tex]$-140.5 < \mu_1 - \mu_2 < 165.4$[/tex].

2. Interpretation of the Confidence Interval:
- This interval includes zero, suggesting that there is no significant difference in the average waiting times between the two setups.

By going through these steps, we've analyzed the waiting time data to understand the differences (or lack thereof) between having a single line versus two lines at an inspection station.