Statistical Measurements and Graph
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This problem has to do with graphical plotting and calculation for certain distribution. Please outline a detailed response to the questions listed with appropriate graphical and statistical data representation(s).
NB. The 90 percent confidence interval is for both the mean and for the variance too.
1. The following are the spark plug gaps, in mm, of eight spark plugs in a package (drawn from a normally distributed population): 2.50, 2.52, 2.48, 2.53, 2.47, 2.51, 2.50 and 2.55
Find a 90% percent confidence interval for all spark plug gaps (use chi-square distribution).
2. A six-cylinder, 4000-hp marine diesel engine has six fuel injectors. Because heavy fuel oil is used, the injectors get dirty after approximately 1000 hours. Consequently, pressure adjustment is lost, resulting in poor combustion and possible component failure. Assume that the chance of an injector failing is 1 in 200 after the 1000 hours of operation. Assume that the occurrences of failure of the injectors are independent of each other and that the engine can run if more than three injectors keep operating. What is the probability of the engine failing after 1000 hours of operation (use binomial distribution)?
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1. The following are the spark plug gaps, in mm, of eight spark plugs in a package (drawn from a normally distributed population): 2.50, 2.52, 2.48, 2.53, 2.47, 2.51, 2.50 and 2.55
Find a 90% percent confidence interval for all spark plug gaps (use chi-square distribution).
Solution. Denote the sample mean by and denote the sample standard deviation by s. Denote the population mean by and denote the sample standard deviation by . Note that the sample size is n=8. Use Excel, we get the following table.
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Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
Recent Feedback
- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
- "Thank you"
- "Thank you very much for your valuable time and assistance!"
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