The article “Modeling of Urban Area Stop-andGo Traffic Noise” presents measurements of traffic noise, in dBA, from 10 locations in Bangkok, Thailand. Measurements, presented in the following table, were made at each location, in both the acceleration and deceleration lanes. begin{array}{|c|c|}hline text{Location} & text{Acceleration} & text{Decelaration} hline 1 & 78.1 & 78.6 hline 2 & 78.1 & 80.0 hline 3 & 79.6 & 79.3 hline 4 & 81.0 & 79.1 hline 5 & 78.7 & 78.2 hline 6 & 78.1 & 78.0 hline 7 & 78.6 & 78.6 hline 8 & 78.5 & 78.8 hline 9 & 78.4 & 78.0 hline 10 & 79.6 & 78.4 hline end{array} Can you conclude that there is a difference in the mean noise levels between acceleration and deceleration lanes?

Question
Modeling
asked 2021-01-22
The article “Modeling of Urban Area Stop-andGo Traffic Noise” presents measurements of traffic noise, in dBA, from 10 locations in Bangkok, Thailand. Measurements, presented in the following table, were made at each location, in both the acceleration and deceleration lanes.
\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Location}&\text{Acceleration}&\text{Decelaration}\backslash{h}{l}\in{e}{1}&{78.1}&{78.6}\backslash{h}{l}\in{e}{2}&{78.1}&{80.0}\backslash{h}{l}\in{e}{3}&{79.6}&{79.3}\backslash{h}{l}\in{e}{4}&{81.0}&{79.1}\backslash{h}{l}\in{e}{5}&{78.7}&{78.2}\backslash{h}{l}\in{e}{6}&{78.1}&{78.0}\backslash{h}{l}\in{e}{7}&{78.6}&{78.6}\backslash{h}{l}\in{e}{8}&{78.5}&{78.8}\backslash{h}{l}\in{e}{9}&{78.4}&{78.0}\backslash{h}{l}\in{e}{10}&{79.6}&{78.4}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)
Can you conclude that there is a difference in the mean noise levels between acceleration and deceleration lanes?

Answers (1)

2021-01-23
Step 1
Given: \(\displaystyle{n}={10}\)
Let us assume: \(\displaystyle\alpha={5}\%={0.05}\)
Given claim: Difference \(\displaystyle\mu_{{{d}}}\ \ne{q}\ {0}\)
The claim is either the null hypothesis or the alternative hypothesis. The null hypothesisi and the alternative hypothesis state the opposite of each other.
\(\displaystyle{H}_{{{0}}}\ :\ \mu_{{{d}}}={0}\)
\(\displaystyle{H}_{{\alpha}}\ :\ \mu_{{{d}}}\ \ne{q}\ {0}\)
Determine the difference in value of each pair.
\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Sample 1}&\text{Sample 2}&\text{Difference D}\backslash{h}{l}\in{e}{78.1}&{78.6}&-{0.5}\backslash{h}{l}\in{e}{78.1}&{80}&-{1.9}\backslash{h}{l}\in{e}{79.6}&{79.3}&{0.3}\backslash{h}{l}\in{e}{81}&{79.1}&{1.9}\backslash{h}{l}\in{e}{78.7}&{78.2}&{0.5}\backslash{h}{l}\in{e}{78.1}&{78}&{0.1}\backslash{h}{l}\in{e}{78.6}&{78.6}&{0}\backslash{h}{l}\in{e}{78.5}&{78.8}&-{0.3}\backslash{h}{l}\in{e}{78.4}&{78}&{0.4}\backslash{h}{l}\in{e}{79.6}&{78.4}&{1.2}\backslash{h}{l}\in{e}\text{Mean}&&{0.17}\backslash{h}{l}\in{e}\text{sd}&&{1.0122}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)
Step 2
Determine the sample mean of the differences. The mean is sum of all values divided by number of values.
\(\displaystyle\overline{{{d}}}={\frac{{-{0.5}\ -\ {1.9}\ +\ {0.3}\ +\ \cdots\ -{0.3}\ +\ {0.4}\ +\ {1.2}}}{{{10}}}}\ \approx\ {0.17}\)
Determine the sample standard deviation of the differences:
\(\displaystyle{s}_{{{d}}}=\sqrt{{{\frac{{{\left(-{0.5}\ -\ {0.17}\right)}^{{{2}}}\ +\ \cdots\ +\ {\left({1.2}\ -\ {0.17}\right)}^{{{2}}}}}{{{10}\ -\ {1}}}}}}\ \approx\ {1.0122}\)
Determine the value of the test statistic:
\(\displaystyle{t}={\frac{{\overline{{{d}}}}}{{\frac{{s}_{{{d}}}}{\sqrt{{{n}}}}}}}={\frac{{{0.17}}}{{\frac{{1.0122}}{\sqrt{{{10}}}}}}}\ \approx\ {0.531}\)
The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The P-value is the number (or interval) in the column title of the Students T distribution in the appendix containing the t-value in the row \(\displaystyle{d}{f}={n}\ -\ {1}={10}\ -\ {1}={9}\) (Note: We double the boundaries, because the test is two-tailed).
\(\displaystyle{0.50}={2}\ \times\ {0.25}\ {<}\ {P}\ {<}\ {2}\ \times\ {0.40}={0.80}\)</span>
If the P-value is less than the significance level, reject the null hypothesis.
\(\displaystyle{P}\ {>}\ {0.10}={10}\%\ \Rightarrow\ {F}{a}{i}{l}\ \to\ {r}{e}{j}{e}{c}{t}\ {H}_{{{0}}}\)
There is sufficient evidence to support the claim that is a difference in the mean noise levels between accelaration and decelaration lanes.
0

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State the null and alternate hypotheses.
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\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}\)
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\(\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}\)
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value \(\displaystyle>{0.250}\)
\(\displaystyle{0.125}<{P}-\text{value}<{0},{250}\)
\(\displaystyle{0},{050}<{P}-\text{value}<{0},{125}\)
\(\displaystyle{0},{025}<{P}-\text{value}<{0},{050}\)
\(\displaystyle{0},{005}<{P}-\text{value}<{0},{025}\)
P-value \(\displaystyle<{0.005}\)
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
...