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 t

Rui Baldwin 2021-02-09 Answered
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?
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dieseisB
Answered 2021-02-10 Author has 85 answers
Step 1 Given information Sample size (n)=25 Sample mean x=30,000 km Population standard deviation =5,000 km 3) Significance level (α)=1  0.95=0.05 95% Confidence interval is given by C.I= x ± Zα2 × σn
=30.000 ± 1.9600 × 500025=(28040, 31390)
Z0.052=Z0.025= ± 1.9600 (From Excel = NORM.S.INV(0.025)) Step 2 2) Significance level (α)=1  0.75=0.25
75% Confidence interval is given by C.I= x ± Zα2 × 500025=(28849.7, 31150.3)
Z0.252=Z0.125= ± 1.1503 (From Excel = NORM. S. INV(0.125)) Step 3 3) Width of interval for 95% confidence interval =31960  28040=3920 Width of interval for 75% confidence interval =31150.3  28849.7=2300.6 On decreasing Confidence level. Width of confidence level decreases.
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