Answer to Question #287190 in Statistics and Probability for Pash

Question

Two independent experiments are run in which two different types of paint are compared. Eighteen specimens are painted using type A, and the drying time, hours, is recorded for each. The same is done with type B. The population standard deviations are both known to be 1.0. Assuming that the mean drying time is equal for the two types of paint, if someone did the experiment 10,000 times under such condition, in how many of those 10,000 experiments would there be a difference overline X A - overline X B that was as large as (or larger than) 1.0? Note overline X A and overline X B are average drying times for samples of size = n=18.

Answer 1

From the sampling distribution of $x_a-x_b,$ we know that the distribution is approximately normal with mean

$\mu _{x_a-x_b}=\mu _a-\mu _b=0$, since the mean drying time is equal for the two types of paint.

and variance

$\sigma^2_{x_a-x_b}={\sigma _a^2 \over n_a}+{\sigma_b^2 \over n_b}={1 \over 18}+{1 \over 18}={1 \over 9}$

due to independence of the experiments.

Corresponding to the value $x_a-x_b=1.0,$ we have

$z={1-(\mu_a-\mu _b) \over \sqrt{1/9}}={1-0 \over \sqrt{1/9}}=3.0$

$z={1-(\mu_a-\mu _b) \over \sqrt{1/9}}={1-0 \over \sqrt{1/9}}=3.0$

$z={1-(\mu_a-\mu _b) \over \sqrt{1/9}}={1-0 \over \sqrt{1/9}}=3.0$

We determine,

$P(x_a-x_b>1.0)=P(z>3.0)=1-P(z<3.0)=1-0.998650=0.001350$

if someone did the experiment 10,000 times under such condition, we expect$10000\times0.001350=13.5\approx 14$ experiments with a difference $(overline \space X _A - overline\space X _B)$ that is as large as (or larger than) 1.0