Answer to Question #258497 in Statistics and Probability for Jac

Question

The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint:

3.4 2.5 4.8 2.9 3.6

2.8 3.3 5.6 3.7 2.8

4.4 4.0 5.2 3.0 4.8

Assuming that the measurements represent a random sample from a normal population, find a 95% prediction interval for the drying time for the next trial of the paint.

Answer 1

Lets mark u' as prediction interval

$u' = (u-Cr*s, u+Cr*s)$ , where u - sample mean, Cr - testing critical value, s - sample standart deviation

$u = {\frac {3.4+...+4.8} {15}}={\frac {56.8} {15}}=3.79$

$s^2={\frac {(3.4-3.79)^{2}+...+(4.8-3.79)^{2}} {14}}=0.986\to s=0.993$

Since the sample population has normal distribution, it is appropriate to calculate critical value using t-statistic with 15 - 1 = 14 degrees of freedom

$P(t(14) > Cr) = {\frac a 2}=0.025$ , where a - level of significance. We use ${\frac a 2}$ cause we need a two-sided test

Now the value of the Cr can be found from tables of t-statistic

Сr = 2.145

So, $u-Cr*s=3.79-2.145*0.93=1.66$ , $u+Cr*s=5.92$

The 95% prediction interval is (1.66, 5.92)