In a sample of 7 cars, each car was tested for nitrogen-oxide emissions and the following results were obtained: 0.06, 0.11, 0.16, 0.15, 0.14, 0.08, 0.15. Cconstruct a 98\% confidence interval estimate for the mean amount of nitrogen-oxide emission for all cars.

UkusakazaL 2021-08-06
In a sample of 7 cars, each car was tested for nitrogen-oxide emissions and the following results were obtained: 0.06, 0.11, 0.16, 0.15, 0.14, 0.08, 0.15. Assuming that the sample is representative of all cars in use and that they are normally distributed, construct a \(\displaystyle{98}\%\) confidence interval estimate for the mean amount of nitrogen-oxide emission for all cars.
Use the sample data from #4 to construct a \(\displaystyle{98}\%\) confidence interval estimate for the standard deviation of the amount nitrogen-oxide emission for all cars.

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Answered 2021-08-16 Author has 28674 answers

Step 1
Determine the t-value by looking in the row starting with degrees of freedom \(\displaystyle{n}-{1}={7}-{1}={6}\) in table \(\displaystyle{a}-{3}\):
\(\displaystyle{t}_{{\frac{\alpha}{{2}}}}={t}_{{{0.01}}}={3.143}\)
The margin of error is then:
\(\displaystyle{E}={t}_{{\frac{\alpha}{{2}}}}\cdot{\frac{{{s}}}{{\sqrt{{{n}}}}}}={3.143}\cdot{\frac{{{0.0389}}}{{\sqrt{{{7}}}}}}\approx{0.0462}\)
The confidence interval is:
\(\displaystyle{0.0752}={0.1214}-{0.0462}=\overline{{{x}}}-{E}{<}\mu{<}\overline{{{x}}}+{E}={0.1214}+{0.0462}={0.1676}\)
Step 2
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