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.

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.