Question

without using the table of Tolerance factors for normal distributions, find a tolerance interval that gives two sided 95% bounds on 90% of the distribution of packages of 95% lean beef. Assume the data came from an approximately normal distribution. where n=30, sample mean=96.2%, and standard deviation= 0.8%.

Answer 1

If you don't want to use table for tolerance factors, you need to use Standard Normal Probability table and Chi Square distribution Table

The formula for tolerance interval is xbar ± k₂ s

xbar is mean 96.2

s is Sample standard deviation 0.8

To calculate k₂ we will use the following formula

where is the critical value of the chi-square distribution with degrees of freedom, N - 1, that is exceeded with probability and is the critical value of the normal distribution which is exceeded with probability (1-p)/2.

We have proportion p = 0.90

Therefore = (1-p)/2 = (1-.90)/2 = 0.05

From the Normal distribution table get value i.e. which is 1.645 for = 0.05

And from Chi Square table get the value for under the column labelled 0.95 in the row labelled degrees of freedom = 29 which is 17.708

Putting the value in the above equation k₂ =

_{= 2.14}

Hence the tolerance interval is 96 .2 ± (2.14)(0.8)

The lower and upper bounds are 94.5 and 97 .9. Hence we can say that, we are 95% confident that the above range covers the central 90% of the distribution of 95% lean beef.