4) A certain company manufactures steel dowels to be transmissions. The diameter
ID: 3065213 • Letter: 4
Question
4) A certain company manufactures steel dowels to be transmissions. The diameter of each dowel is supposed to be 6 mm. Yo the diameters of the dowels have a Normal Distribution. The worried if the manufactured dowels have a mean diameter that is not 6 used as pins in automobile u may assume that company executives will be mm or if their standard deviation is greater than 0.25 mm. A quality control employee collects a random sample of 61 dowels. The sample mean is 6.09 mm and the sample standard deviation is 0.15 mm. Construct a 98% confidence interval for the population mean. Show the work. Make sure you use the correct distribution when computing the margin of error. a) b) Construct a 98% confidence interval for the population standard deviation. Show the work. Based on the above two confidence intervals, will the executives be worried? Give a good explanation with your answer. Be specific. c)Explanation / Answer
X : Diameter of each dowel
X ~ N( mu =6 , sigma2)
From the information
n = xbar = 6.09 mm, sample standard deviation = s = 0.15 mm
a) Since population standard deviation is not known. use t-disribution to find the confidence interval for population mean.
The 98 % confidence interval for population mean is
( xbar - margin of error , xbar + margin of error)
where margin of error = s/ sqrt(n) * t n-1, alpha/2
t n-1, alpha/2 = t 60, 0.01 = 2.3901
s/sqrt(n) = 0.15 / sqrt(61) = 0.0192
Margin of error = 2.3901 * 0.0192 = 0.0458
Hence 98 % confidence interval is
( 6.09 - 0.0458, 6.09 + 0.0458)
= ( 6.0442, 6.1358)
b) 98% Confidence interval for population standard deviation is
( sqrt ( ( n-1) s2 / Chi-square n-1,alpha/2) , sqrt( ( n-1) s2 / Chi-square n-1,1-alpha/2) )
from the Chi-square table
chi-square n-1,alpha/2 = chi-square 60, 0.01 = 88.3794
chi-square n-1,1-alpha/2 = chi-square 60, 0.99 = 37.4848
Hence confidence interval for population standard deviation is
( sqrt ( 60 * 0.152 /88.3794), sqrt ( 60 * 0.152 /37.4848))
= ( 0.1235, 0.1897)
c) Since sample mean is always unbiased estimator population mean and 98% of values of standard deviation lie between ( 0.1235 , 0.1897 ) the maximum value of estimate value of standard deviation is 0.1897 mm which is less than 0.25 .
Hence by the condition ( the standard deviation is greater than 0.25 mm) the executives will not be worried.
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