The diameter of a random sample of roll pins was measured. The average diameter

Question

The diameter of a random sample of roll pins was measured. The average diameter of the twenty pins was found to be 24.8246 nun and the standard deviation of the samples taken was found to be 0.9469 mm. A new random sample of measurements were taken from the population of pins. The average diameter of the sixty pins was found to be 24.8778 mm and the standard deviation of the samples taken was found to be 0.8953 nun. Report, all answers to the fourth decimal place. Homework should be typed, including equations. Your submission should include excel spreadsheets or mat lab code. However, simply printing excel files or matlab code will not be accepted as your submission. Calculate the 95% confidence intervals for both data sets: Calculate the following considering only the large data set: The probability of a measurement being within 1 standard deviation of the mean. The probability of a measurement being between 24.5 and 24.8 nun.

Explanation / Answer

1)for first sample

std error=std deviation/(n)1/2 =0.2117

and from excel at (n-1=19) degrees of freedom and 95% CI, t=2.093 ; excel formula : tinv(0.05,19)

hence 95% confidence interval for first data =mean +/- *std error =24.3814 ; 25.2678

second sample

std error=std deviation/(n)1/2 =0.1222

and from excel at (n-1=59) degrees of freedom and 95% CI, t=2.001   excel formula : tinv(0.05,59)

hence 95% confidence interval for first data =mean +/- *std error =24.6332 ; 25.1224 ;

2 from larger sample probabilty of being with in 1 standard deviation from mean =P(-1<t<1) =0.6786

hence 67.86% values from excel formula :(1-tdist(1,59,2)

probability of measurement b/w 24.5 and 24.8

P(24.5<t<24.8) =P((24.5-24.8778)/0.1224<t<(24.8-24.8778)/0.124) =P(-3.0905<t<-0.6364) =0.2620

hence 26.20% values

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