# 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
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?

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.

### Relevant Questions

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