Step 1

Given:

\(\displaystyle{n}={S}{a}\mp\le\ {s}{i}{z}{e}={50}\)

\(\displaystyle\overline{{{x}}}={S}{a}\mp\le\ {m}{e}{a}{n}={654.1}\)

\(\displaystyle{s}={S}{a}\mp\le\ {s}{t}{a}{b}{d}{a}{r}{d}\ {d}{e}{v}{i}{a}{t}{i}{o}{n}={311.7}\)

a) \(\displaystyle{c}={C}{o}{n}{f}{i}{d}{e}{n}{c}{e}\ {c}{o}{e}{f}{f}{i}{c}{i}{e}{n}{t}={95}\%={0.95}\)

Since the sample size n is large (at least 30), it is appropriate to estimate the population standard deviation by the sample standard deviation.

\(\displaystyle\sigma\ \approx\ {s}\)

For confidence level \(\displaystyle{1}\ -\ \alpha\ {0.95},\) determine

\(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={z}_{{{0.025}}}\) using the normal probability table in the appendix (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):

\(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={1.96}\)

The margine of error is then:

\(\displaystyle{E}={z}_{{\frac{\alpha}{{2}}}}\ \times\ {\frac{{\sigma}}{{\sqrt{{{n}}}}}}\ \approx\ {z}_{{\frac{\alpha}{{2}}}}\ \times\ {\frac{{{s}}}{{\sqrt{{{n}}}}}}={1.96}\ \times\ {\frac{{{311.7}}}{{\sqrt{{{30}}}}}}\ \approx\ {111.5404}\)

The boundaries of the confidence interval then become:

\(\displaystyle\overline{{{x}}}\ -\ {E}={654.1}\ -\ {111.5404}={542.5596}\)

\(\displaystyle\overline{{{x}}}\ +\ {E}={645.1}\ +\ {111.5404}={765.6404}\)

We are \(\displaystyle{95}\%\) confident that the improvement in traffic flow due to the new system is between 542.5596 vehicles per hour and 765.6404 vehicles per hour.

Step 2

b) \(\displaystyle{c}={C}{o}{n}{f}{i}{d}{e}{n}{c}{e}\ {c}{o}{e}{f}{f}{i}{c}{i}{e}{n}{t}={98}\%={0.98}\)

Since the sample size n is large (at leasr 30), it is appropriate to estimate the population standard deviation by the sample standard deviation.

\(\displaystyle\sigma\ \approx\ {s}\)

For confidence level \(\displaystyle{1}\ -\ \alpha={0.98},\ {\det{{e}}}{r}\min{e}\ {z}_{{\frac{\alpha}{{2}}}}={z}_{{{0.01}}}\) using the normal probability table in the appendix (look up 0.01 in the table, the z-score is then the found z-score with opposite sing):

\(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={2.33}\)

The margin of error is then:

\(\displaystyle{E}={z}_{{\frac{\alpha}{{2}}}}\ \times\ {\frac{{\sigma}}{{\sqrt{{{n}}}}}}\ \approx\ {z}_{{\frac{\alpha}{{2}}}}\ \times\ {\frac{{{s}}}{{\sqrt{{{n}}}}}}={2.33}\ \times\ {\frac{{{311.7}}}{{\sqrt{{{30}}}}}}\ \approx\ {132.5965}\)

The boudaries of the confidenceinterval then become:

\(\displaystyle\overline{{{x}}}\ -\ {E}={654.1}\ -\ {132.5965}={521.5035}\)

\(\displaystyle\overline{{{x}}}\ +\ {E}={654.1}\ +\ {132.5965}={786.6965}\)

We are \(\displaystyle{98}\%\) confident that the improvement in traffic flow due to the new system is between 521.5035 vehicles per hour and 786.6965 vehicles per hour.

Step 3

c) Given confidence interval: \(\displaystyle{\left({581.6},\ {726.6}\right)}\)

The sampling distribution of the sample mean \(\displaystyle\overline{{{x}}}\) has mean

\(\displaystyle\mu\) and standard deviation

\(\frac{\sigma}{\sqrt{n}}.\)

The z-score is the value decreased by the mean, divided by the standard deviation

\(\displaystyle{z}={\frac{{\overline{{{x}}}\ -\ \mu_{{\overline{{{x}}}}}}}{{\sigma_{{\overline{{{x}}}}}}}}={\frac{{{581.6}\ -\ {654.1}}}{{\frac{{311.7}}{\sqrt{{{50}}}}}}}\ \approx\ -{1.64}\)

\(\displaystyle{z}={\frac{{\overline{{{x}}}\ -\ \mu_{{\overline{{{x}}}}}}}{{\sigma_{{\overline{{{x}}}}}}}}={\frac{{{726.6}\ -\ {654.1}}}{{\frac{{311.7}}{\sqrt{{{50}}}}}}}\ \approx\ {1.64}\)

The confidence level is the probability between the two z-scores. Determine the probability using the normal probability table:

\(\displaystyle{P}{\left(-{1.64}\ {<}\ {Z}\ {<}\ {1.64}\right)}={P}{\left({Z}\ {<}\ {1.64}\right)}\ -\ {P}{\left({Z}\ {<}\ -{1.64}\right)}\)

\(\displaystyle={0.9495}\ -\ {0.0505}\)

\(\displaystyle={0.8990}\)

Thus the confidence level is about \(\displaystyle{90}\%\)

Step 4

d) Given:

\(\displaystyle{c}={C}{o}{n}{f}{i}{d}{e}{n}{c}{e}\ lev{e}{l}={95}\%={0.95}\)

\(E=\text{Margin of error} =50\)

For confidence level \(1-\alpha=0.95,\ \text{determine}\ z_{\frac{\alpha}{2}}=1_{0.025}\) using the normal probability table in the appendix (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):

\(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={1.96}\)

Formula for the margin of error:

\(\displaystyle{E}={z}_{{\frac{\alpha}{{2}}}}\sigma\ \overline{{{x}}}={z}_{{\frac{\alpha}{{2}}}}\ {\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

Multiply each side of the previous equation by \(\displaystyle\sqrt{{{n}}}\):

\(\displaystyle{E}\sqrt{{{n}}}={z}_{{\frac{\alpha}{{2}}}}{s}\)

Divide each side by ME:

\(\displaystyle\sqrt{{{n}}}={\frac{{{z}_{{\frac{\alpha}{{2}}}}{s}}}{{{E}}}}\)

Square each side:

\(\displaystyle{n}={\left({\frac{{{z}_{{\frac{\alpha}{{2}}}}\ \cdot\ {s}}}{{{E}}}}\right)}\)

Fill in known values and evaluate (round up to the nearest integer!):

\(\displaystyle{n}={\left({\frac{{{1.96}\ \cdot\ {311.7}}}{{{50}}}}\right)}^{{{2}}}\ \approx\ {150}\)

Step 5 e) Given:

\(\displaystyle{c}={C}{o}{n}{f}{i}{d}{e}{n}{c}{e}\ le{v}{e}{l}=\frac{{98}}{\%}={0.98}\)

\(E=\text{Margin of error}=50\)

For confidence level \(1-\alpha=0.98,\ \text{determine}\ z_{\frac{\alpha}{2}}=1_{0.01}\) using the normal probability table in the appendix (look up 0.01 in the table, the z-score is then the found z-score with opposite sign):

\(\displaystyle{Z}_{{\frac{\alpha}{{2}}}}={2.33}\)

Formula for the margin of error:

\(\displaystyle{E}={z}_{{\frac{\alpha}{{2}}}}\sigma\ \overline{{{x}}}={z}_{{\frac{\alpha}{{2}}}}\ {\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

Multiply each side of the previous equation by \(\displaystyle\sqrt{{{n}}}\):

\(\displaystyle{E}\sqrt{{{n}}}={z}_{{\frac{\alpha}{{2}}}}{s}\)

Divide each side by ME:

\(\displaystyle\sqrt{{{n}}}={\frac{{{z}_{{\frac{\alpha}{{2}}}}{s}}}{{{E}}}}\)

Square each side:

\(\displaystyle{n}={\left({\frac{{{z}_{{\frac{\alpha}{{2}}}}\ \cdot\ {s}}}{{{E}}}}\right)}^{{{2}}}\)

Fill in the known values and evaluate (round up the nearest integer!):

\(\displaystyle{n}={\left({\frac{{{2.33}\ \cdot\ {311.7}}}{{{50}}}}\right)}^{{{2}}}\ \approx\ {211}\)