Question

listed in the accompanying table are waiting times (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)

1. use a 0.01 significance level to test the claim that cars in two queues have a mean waiting time equal to that of cars in a single queue.

let population 1 correspond to the single waiting line and let population 2 correspond to two waiting lines. what are the null and alternative hypotheses?
a. ( h_0: mu_1 < mu_2 ) ( h_1: mu_1 = mu_2 )
b. ( h_0: mu_1 = mu_2 ) ( h_1: mu_1 > mu_2 )
c. ( h_0: mu_1
eq mu_2 ) ( h_1: mu_1 = mu_2 )
d. ( h_0: mu_1 = mu_2 ) ( h_1: mu_1
eq mu_2 )
calculate the test statistic.
t = \\(\square\\) (round to two decimal places as needed )

Explanation:

Step1: Identify Hypotheses

The claim is that the mean waiting time for two queues (population 2) is equal to that of a single queue (population 1). So, the null hypothesis \( H_0 \) is \( \mu_1 = \mu_2 \), and the alternative hypothesis \( H_1 \) (since we test for equality) is \( \mu_1 eq \mu_2 \). So the correct option is D: \( H_0: \mu_1 = \mu_2 \), \( H_1: \mu_1 eq \mu_2 \).

Step2: Calculate Sample Statistics (One Line)

First, find the mean (\( \bar{x}_1 \)) and standard deviation (\( s_1 \)) for the "One Line" data. The data points for One Line (single queue) are: 63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1, 733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8. Wait, actually, looking at the table, the "One Line" has two rows? Wait, no, the first two rows under "One Line" are: first row: 63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1; second row: 733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8. Wait, maybe I miscounted. Let's count the number of data points for One Line: first row has 14 points, second row has 14 points? Wait, no, looking at the table, "One Line" has two columns? Wait, no, the table is:

One Line (single queue) data:
Row 1: 63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1
Row 2: 733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8

Wait, that's 14 + 14 = 28 data points? Wait, no, maybe the first column is "One Line" and the second is "Two Lines". Wait, the table is:

Waiting Times:

One Line (single queue) data:
63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1,
733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8

Wait, maybe I made a mistake. Let's count the number of elements. Let's list them:

First set (top row of One Line): 63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1 (14 points)

Second set (bottom row of One Line): 733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8 (14 points)

Total for One Line: 14 + 14 = 28 data points.

Calculate \( \bar{x}_1 = \frac{\sum x_1}{n_1} \), \( s_1 = \sqrt{\frac{\sum (x_1 - \bar{x}_1)^2}{n_1 - 1}} \)

Similarly, for "Two Lines" (population 2):

Data points: 63.6, 215.7, 86.1, 340.1, 200.1, 630.3, 332.5, 329.2, 915.3, 553.2, 597.4,
885.3, 1090.1, 622.5, 517.6, 566.2, 267.7, 350.3, 94.5, 100.3, 162.6, 101.1

Wait, let's count: top row of Two Lines: 63.6, 215.7, 86.1, 340.1, 200.1, 630.3, 332.5, 329.2, 915.3, 553.2, 597.4 (11 points)

Bottom row of Two Lines: 885.3, 1090.1, 622.5, 517.6, 566.2, 267.7, 350.3, 94.5, 100.3, 162.6, 101.1 (11 points)

Total for Two Lines: 11 + 11 = 22 data points? Wait, maybe I'm misinterpreting the table. Alternatively, maybe the "One Line" has 14 data points (first column) and "Two Lines" has 14? Wait, the problem says "two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal". So we use the two-sample t-test with unequal variances (Welch-Satterthwaite approximation).

But maybe the actual data is:

One Line (single queue) data (n1 = 14? Wait, looking at the first column of One Line: 63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1 (14 points). Then the second column of One Line: 733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8 (14 points). So n1 = 28?

Two Lines data: first column: 63.6, 215.7, 86.1, 340.1, 200.1, 630.3, 332.5, 329.2, 915.3, 553.2, 597.4 (11 points). Second column: 885.3, 1090.1, 622.5, 517.6, 566.2, 267.7, 350.3, 94.5, 100.3, 162.6, 101.1 (11 points). So n2 = 22?

Wait, maybe the problem has n1 = 14 and n2 = 14? Maybe I misread the table. Let's check the original problem: "the data from two waiting lines are real observations, and the data from the single waiting line are modeled from those real observations". Maybe the "One Line" (single queue) has 14 data points, and "Two Lines" has 14? Let's re-express the table:

One Line (single queue) data (let's take the first 14 points from the top):

63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1 (n1 = 14)

Two Lines (two queues) data:

63.6, 215.7, 86.1, 340.1, 200.1, 630.3, 332.5, 329.2, 915.3, 553.2, 597.4, 885.3, 1090.1, 622.5 (n2 = 14)

Ah, maybe that's the case. Let's proceed with n1 = 14, n2 = 14.

Calculate \( \bar{x}_1 \) for One Line:

Sum of One Line data: 63.7 + 136.8 + 141.7 + 278.7 + 253.1 + 476.3 + 477.9 + 474.2 + 401.9 + 721.6 + 760.9 + 692.1 + 836.7 + 903.1

Let's compute step by step:

63.7 + 136.8 = 200.5

200.5 + 141.7 = 342.2

342.2 + 278.7 = 620.9

620.9 + 253.1 = 874

874 + 476.3 = 1350.3

1350.3 + 477.9 = 1828.2

1828.2 + 474.2 = 2302.4

2302.4 + 401.9 = 2704.3

2704.3 + 721.6 = 3425.9

3425.9 + 760.9 = 4186.8

4186.8 + 692.1 = 4878.9

4878.9 + 836.7 = 5715.6

5715.6 + 903.1 = 6618.7

So \( \bar{x}_1 = \frac{6618.7}{14} \approx 472.76 \)

Now, calculate \( s_1 \). First, find the sum of squared deviations:

For each data point \( x_1 \), compute \( (x_1 - 472.76)^2 \):

63.7: (63.7 - 472.76)^2 = (-409.06)^2 ≈ 167,330.08

136.8: (136.8 - 472.76)^2 = (-335.96)^2 ≈ 112,868.12

141.7: (141.7 - 472.76)^2 = (-331.06)^2 ≈ 109,600.72

278.7: (278.7 - 472.76)^2 = (-194.06)^2 ≈ 37,659.28

253.1: (253.1 - 472.76)^2 = (-219.66)^2 ≈ 48,250.52

476.3: (476.3 - 472.76)^2 = (3.54)^2 ≈ 12.53

477.9: (477.9 - 472.76)^2 = (5.14)^2 ≈ 26.42

474.2: (474.2 - 472.76)^2 = (1.44)^2 ≈ 2.07

401.9: (401.9 - 472.76)^2 = (-70.86)^2 ≈ 5,021.14

721.6: (721.6 - 472.76)^2 = (248.84)^2 ≈ 61,921.35

760.9: (760.9 - 472.76)^2 = (288.14)^2 ≈ 82,924.66

692.1: (692.1 - 472.76)^2 = (219.34)^2 ≈ 48,110.04

836.7: (836.7 - 472.76)^2 = (363.94)^2 ≈ 132,452.32

903.1: (903.1 - 472.76)^2 = (430.34)^2 ≈ 185,192.52

Sum these squared deviations:

167,330.08 + 112,868.12 = 280,198.2

+109,600.72 = 389,798.92

+37,659.28 = 427,458.2

+48,250.52 = 475,708.72

+12.53 = 475,721.25

+26.42 = 475,747.67

+2.07 = 475,749.74

+5,021.14 = 480,770.88

+61,921.35 = 542,692.23

+82,924.66 = 625,616.89

+48,110.04 = 673,726.93

+132,452.32 = 806,179.2

Answer:

Step1: Identify Hypotheses

The claim is that the mean waiting time for two queues (population 2) is equal to that of a single queue (population 1). So, the null hypothesis \( H_0 \) is \( \mu_1 = \mu_2 \), and the alternative hypothesis \( H_1 \) (since we test for equality) is \( \mu_1
eq \mu_2 \). So the correct option is D: \( H_0: \mu_1 = \mu_2 \), \( H_1: \mu_1
eq \mu_2 \).

Step2: Calculate Sample Statistics (One Line)

First, find the mean (\( \bar{x}_1 \)) and standard deviation (\( s_1 \)) for the "One Line" data. The data points for One Line (single queue) are: 63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1, 733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8. Wait, actually, looking at the table, the "One Line" has two rows? Wait, no, the first two rows under "One Line" are: first row: 63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1; second row: 733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8. Wait, maybe I miscounted. Let's count the number of data points for One Line: first row has 14 points, second row has 14 points? Wait, no, looking at the table, "One Line" has two columns? Wait, no, the table is:

One Line (single queue) data:
Row 1: 63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1
Row 2: 733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8

Wait, that's 14 + 14 = 28 data points? Wait, no, maybe the first column is "One Line" and the second is "Two Lines". Wait, the table is:

Waiting Times:

One Line (single queue) data:
63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1,
733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8

Wait, maybe I made a mistake. Let's count the number of elements. Let's list them:

First set (top row of One Line): 63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1 (14 points)

Second set (bottom row of One Line): 733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8 (14 points)

Total for One Line: 14 + 14 = 28 data points.

Calculate \( \bar{x}_1 = \frac{\sum x_1}{n_1} \), \( s_1 = \sqrt{\frac{\sum (x_1 - \bar{x}_1)^2}{n_1 - 1}} \)

Similarly, for "Two Lines" (population 2):

Data points: 63.6, 215.7, 86.1, 340.1, 200.1, 630.3, 332.5, 329.2, 915.3, 553.2, 597.4,
885.3, 1090.1, 622.5, 517.6, 566.2, 267.7, 350.3, 94.5, 100.3, 162.6, 101.1

Wait, let's count: top row of Two Lines: 63.6, 215.7, 86.1, 340.1, 200.1, 630.3, 332.5, 329.2, 915.3, 553.2, 597.4 (11 points)

Bottom row of Two Lines: 885.3, 1090.1, 622.5, 517.6, 566.2, 267.7, 350.3, 94.5, 100.3, 162.6, 101.1 (11 points)

Total for Two Lines: 11 + 11 = 22 data points? Wait, maybe I'm misinterpreting the table. Alternatively, maybe the "One Line" has 14 data points (first column) and "Two Lines" has 14? Wait, the problem says "two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal". So we use the two-sample t-test with unequal variances (Welch-Satterthwaite approximation).

But maybe the actual data is:

One Line (single queue) data (n1 = 14? Wait, looking at the first column of One Line: 63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1 (14 points). Then the second column of One Line: 733.7, 605.7, 267.7, 309.9, 128.9, 133.1, 122.3, 128.8, 233.3, 461.1, 481.7, 517.7, 509.4, 579.8 (14 points). So n1 = 28?

Two Lines data: first column: 63.6, 215.7, 86.1, 340.1, 200.1, 630.3, 332.5, 329.2, 915.3, 553.2, 597.4 (11 points). Second column: 885.3, 1090.1, 622.5, 517.6, 566.2, 267.7, 350.3, 94.5, 100.3, 162.6, 101.1 (11 points). So n2 = 22?

Wait, maybe the problem has n1 = 14 and n2 = 14? Maybe I misread the table. Let's check the original problem: "the data from two waiting lines are real observations, and the data from the single waiting line are modeled from those real observations". Maybe the "One Line" (single queue) has 14 data points, and "Two Lines" has 14? Let's re-express the table:

One Line (single queue) data (let's take the first 14 points from the top):

63.7, 136.8, 141.7, 278.7, 253.1, 476.3, 477.9, 474.2, 401.9, 721.6, 760.9, 692.1, 836.7, 903.1 (n1 = 14)

Two Lines (two queues) data:

63.6, 215.7, 86.1, 340.1, 200.1, 630.3, 332.5, 329.2, 915.3, 553.2, 597.4, 885.3, 1090.1, 622.5 (n2 = 14)

Ah, maybe that's the case. Let's proceed with n1 = 14, n2 = 14.

Calculate \( \bar{x}_1 \) for One Line:

Sum of One Line data: 63.7 + 136.8 + 141.7 + 278.7 + 253.1 + 476.3 + 477.9 + 474.2 + 401.9 + 721.6 + 760.9 + 692.1 + 836.7 + 903.1

Let's compute step by step:

63.7 + 136.8 = 200.5

200.5 + 141.7 = 342.2

342.2 + 278.7 = 620.9

620.9 + 253.1 = 874

874 + 476.3 = 1350.3

1350.3 + 477.9 = 1828.2

1828.2 + 474.2 = 2302.4

2302.4 + 401.9 = 2704.3

2704.3 + 721.6 = 3425.9

3425.9 + 760.9 = 4186.8

4186.8 + 692.1 = 4878.9

4878.9 + 836.7 = 5715.6

5715.6 + 903.1 = 6618.7

So \( \bar{x}_1 = \frac{6618.7}{14} \approx 472.76 \)

Now, calculate \( s_1 \). First, find the sum of squared deviations:

For each data point \( x_1 \), compute \( (x_1 - 472.76)^2 \):

63.7: (63.7 - 472.76)^2 = (-409.06)^2 ≈ 167,330.08

136.8: (136.8 - 472.76)^2 = (-335.96)^2 ≈ 112,868.12

141.7: (141.7 - 472.76)^2 = (-331.06)^2 ≈ 109,600.72

278.7: (278.7 - 472.76)^2 = (-194.06)^2 ≈ 37,659.28

253.1: (253.1 - 472.76)^2 = (-219.66)^2 ≈ 48,250.52

476.3: (476.3 - 472.76)^2 = (3.54)^2 ≈ 12.53

477.9: (477.9 - 472.76)^2 = (5.14)^2 ≈ 26.42

474.2: (474.2 - 472.76)^2 = (1.44)^2 ≈ 2.07

401.9: (401.9 - 472.76)^2 = (-70.86)^2 ≈ 5,021.14

721.6: (721.6 - 472.76)^2 = (248.84)^2 ≈ 61,921.35

760.9: (760.9 - 472.76)^2 = (288.14)^2 ≈ 82,924.66

692.1: (692.1 - 472.76)^2 = (219.34)^2 ≈ 48,110.04

836.7: (836.7 - 472.76)^2 = (363.94)^2 ≈ 132,452.32

903.1: (903.1 - 472.76)^2 = (430.34)^2 ≈ 185,192.52

Sum these squared deviations:

167,330.08 + 112,868.12 = 280,198.2

+109,600.72 = 389,798.92

+37,659.28 = 427,458.2

+48,250.52 = 475,708.72

+12.53 = 475,721.25

+26.42 = 475,747.67

+2.07 = 475,749.74

+5,021.14 = 480,770.88

+61,921.35 = 542,692.23

+82,924.66 = 625,616.89

+48,110.04 = 673,726.93

+132,452.32 = 806,179.2