Answer to Question #258496 in Statistics and Probability for Jac

Question

A random sample of 25 tablets of buffered aspirin contains, on average, 325.05 mg of aspirin per tablet, with a standard deviation of 0.5 mg. Find the 95% tolerance limits that will contain 90% of the tablet contents for this brand of buffered aspirin. Assume that the aspirin content is normally distributed.

Answer 1

Let u' be a soughtful tolerance interval, then:

$u'=(u-k*s;u+k*s)$ , where u - sample mean,s - sample standart deviation and k is known as k-factor. There are some tables with k-factor values for different data, but it seems pretty difficult to find it for 90% coverage, so we should calculate it by ourselves. According to Guenther correction fot Howe's estimation(https://www.itl.nist.gov/div898/handbook/prc/section2/prc263.htm), we have:

$k=T(1-a,n-1)*\sqrt{{\frac {(n-1)(1+{\frac 1 n})} {Xi^{2}(1-b,n-1)}}*(1+{\frac {(n-3)*Xi^{2}(1-b,n-1)} {2(n+1)^{2}}}}$ , where

$T(1-a,n-1)$ - t-statistic(two-sided), n - sample size, a - coverage significance level, $Xi^{2}(1-b, n-1)$ - chi-square, b - confidence significance level

In the given case

n=25, a = 0.1, b = 0.05

$k=2.06*\sqrt{{\frac{24.96} {33.196}}*(1+{\frac {22*33.196} {1352})}}=2.06 * 1.076=2.217$

Then $u - k*s=325.05-2.217*0.5=323.95$ , $u+k*s=326.14$

The tolerance interval: (323.95, 326.14)