The service life in kilometer of Goody tires is assumed to follow a normal distribution with a standard deviation of 5,000 km. A random sample of 25 tires yielded a mean service life of 30,000 km. 1) Find the 95% confidence interval for the true mean service life. 2) 2. Find a 75% confidence interval for the true mean service life. 3) Calculate the widths of the intervals found in 1 and 2. How do these widths change as the confidence level decreases?

Question
Confidence intervals
asked 2021-02-09
The service life in kilometer of Goody tires is assumed to follow a normal distribution with a standard deviation of 5,000 km. A random sample of 25 tires yielded a mean service life of 30,000 km. 1) Find the \(\displaystyle{95}\%\) confidence interval for the true mean service life. 2) 2. Find a \(\displaystyle{75}\%\) confidence interval for the true mean service life. 3) Calculate the widths of the intervals found in 1 and 2. How do these widths change as the confidence level decreases?

Answers (1)

2021-02-10
Step 1 Given information Sample size \(\displaystyle{\left({n}\right)}={25}\) Sample mean \(\displaystyle\overline{{{x}}}={30},{000}\ {k}{m}\) Population standard deviation \(\displaystyle={5},{000}\ {k}{m}\) 3) Significance level \(\displaystyle{\left(\alpha\right)}={1}\ -\ {0.95}={0.05}\) \(\displaystyle{95}\%\) Confidence interval is given by \(\displaystyle{C}.{I}=\ \overline{{{x}}}\ \pm\ {Z}_{{{\frac{{\alpha}}{{{2}}}}}}\ \times\ {\frac{{\sigma}}{{\sqrt{{{n}}}}}}\)
\(\displaystyle={30.000}\ \pm\ {1.9600}\ \times\ {\frac{{{5000}}}{{\sqrt{{{25}}}}}}={\left({28040},\ {31390}\right)}\)
\(\displaystyle{Z}_{{{\frac{{{0.05}}}{{{2}}}}}}={Z}_{{{0.025}}}=\ \pm\ {1.9600}\ \text{(From Excel = NORM.S.INV(0.025))}\) Step 2 2) Significance level \(\displaystyle{\left(\alpha\right)}={1}\ -\ {0.75}={0.25}\)
\(\displaystyle{75}\%\) Confidence interval is given by \(\displaystyle{C}.{I}=\ \overline{{{x}}}\ \pm\ {Z}_{{{\frac{{\alpha}}{{{2}}}}}}\ \times\ {\frac{{{5000}}}{{\sqrt{{25}}}}}={\left({28849.7},\ {31150.3}\right)}\)
\(\displaystyle{Z}_{{{\frac{{{0.25}}}{{{2}}}}}}={Z}_{{{0.125}}}=\ \pm\ {1.1503}\ \text{(From Excel = NORM. S. INV(0.125))}\) Step 3 3) Width of interval for \(\displaystyle{95}\%\) confidence interval \(\displaystyle={31960}\ -\ {28040}={3920}\) Width of interval for \(\displaystyle{75}\%\) confidence interval \(\displaystyle={31150.3}\ -\ {28849.7}={2300.6}\) On decreasing Confidence level. Width of confidence level decreases.
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