09_23_09 - STAT 409 Examples for 09/23/2009 Fall 2009...

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Unformatted text preview: STAT 409 Examples for 09/23/2009 Fall 2009 EXCEL: = CHIINV ( α , v ) 2 χ α (v ) for χ 2 distribution with v degrees gives of freedom = CHIDIST ( y , v ) gives the upper tail probability for χ 2 distribution with v degrees of freedom, P ( Y > y ). Recall: If X 1 , X 2 , … , X n are i.i.d. N µ , σ 2 . Then ( n − 1 )⋅ S 2 σ 2 ∑(X i − X ) = σ 2 2 is χ 2 ( n – 1 ). A ( 1 − α ) 100 % confidence interval 2 for the population variance σ (where the population is assumed normal) 2 ( n − 1 )⋅ s , χ 2α 2 2 ( n − 1 )⋅ s χ2 α 1− 2 n – 1 degrees of freedom A ( 1 − α ) 100 % confidence interval for the population standard deviation σ (where the population is assumed normal) ( n − 1 )⋅ s 2 χα 2 2 , ( n − 1 )⋅ s χ 2 1− α 2 2 OR s⋅ ( n −1 ) , s ⋅ χ 2α 2 ( n −1 ) χ2 α 1− 2 n – 1 degrees of freedom 1. A machine makes ½-inch ball bearings. In a random sample of 41 bearings, the sample standard deviation of the diameters of the bearings was 0.02 inch. Assume that the diameters of the bearings are approximately normally distributed. Construct a 90% confidence interval for the standard deviation of the diameters of the bearings. 2. The service time in queues should not have a large variance; otherwise, the queue tends to build up. A bank regularly checks service time by its tellers to determine 2 its variance. A random sample of 22 service times (in minutes) gives s = 8. Assume the service times are normally distributed. a) Find a 95% confidence interval for the overall variance of service time at the bank. b) Find the two one-sided 95% confidence intervals for the overall variance of service time at the bank. c) Find a 95% confidence interval for the overall standard deviation σ of service time at the bank having minimum length. ( “Hint”: Use Table X ( p. 696 ). ) ...
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09_23_09 - STAT 409 Examples for 09/23/2009 Fall 2009...

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