Question

Considered the situation of case study on problem (2) with a larger sample of metal pieces. The diameters are as follows: 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 1.01, 1.03, 0.99, 1.00, 1.00, 0.99, 0.98, 1.01, 1.02, 0.99 centimeters. Once again the normality assumption may be made. Do the following and compare your results to those of the case study. Discuss how they are different and why.

a. Compute a 99% CI on the mean diameter
b. Compute a 99% PI on the next diameter to be measured.
c. Compute a 99% tolerance interval for coverage of the central 95% of the distribution of diameters.

Answer 1

a

`0.9876 <  \mu < 1.0174`

b

`0.941 <  x_o <  1.064`

c

`0.933 , 1.072`

Explanation:

From the question we are told that

The data is

1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 1.01, 1.03, 0.99, 1.00, 1.00, 0.99, 0.98, 1.01, 1.02, 0.99

Generally sample mean is mathematically represented as

`\= x = (\sum x_i)/(n)`

=>
`\= x = (1.01 +  0.97 +  1.03 \cdots +1.02 +  0.99)/(16)`

=>
`\= x =1.0025`

Generally the standard deviation is mathematically represented as

`\sigma =  \sqrt{(\sum (x - \= x) ^2)/(n) }`

=>
`\sigma =  \sqrt{( (1.01 - 1.0025 ) ^2 + (0.97 - 1.0025 ) ^2+ \cdots  + (0.99 - 1.0025 ) ^2 )/(16) }`

=>
`\sigma =  0.0202`

Given that the confidence level is 99% then the level of significance is

`\alpha  =  (100 - 99 ) \%`

=>
`\alpha  = 0.01`

From the t distribution table the critical value for
`(\alpha )/(2)  = (0.05)/(2)  =  0.005` is
`t_{(\alpha )/(2) } =  2.947`

Generally the margin of error is mathematically represented as

`E =  t_{(\alpha )/(2) } *  (\sigma )/(√(n) )`

=>
`E = 2.947 *  (0.0202 )/(√(16) )`

=>
`E = 0.01488`

Generally the 99% confidence interval is

`\= x  - E <  \mu < \= x  + E`

=>
`1.0025  - 0.01488<  \mu < 1.0025  + 0.01488`

=>
`0.9876 <  \mu < 1.0174`

Generally the 99% Prediction interval is mathematically represented as

`\= x  -[t_{(\alpha )/(2) } *  \sigma  *  \sqrt{1 + (1)/(n) } ] <  x_o <\= x  +[t_{(\alpha )/(2) } *  \sigma  *  \sqrt{1 + (1)/(n) } ]`

So

=>
`1.0025  -[2.947 *  0.0202  *  \sqrt{1 + (1)/(16) } ] <  x_o <1.0025  +[2.947 *  0.0202  *  \sqrt{1 + (1)/(16) } ] </p><p>=>  [tex]0.941 <  x_o <  1.064`

Generally the two sided limit tolerance factor at sample size of n = 16 , Prediction level of 99% and confidence level of 95 is
`r =  3.421` , this value is obtained from the statistical table

Generally the 99% tolerance interval for coverage of the central 95% of the distribution of diameters is mathematically represented as

`\= x - (r * \sigma ) , \= x + (r * \sigma )`

=>
`1.0025 - ( 3.421 * 0.0202) , 1.0025 + ( 3.421 * 0.0202)`

=>
`0.933 , 1.072`