Question

Two independent experiments are run in which two different types of paints are compared .eighteen specimens are painted using type a and the drying time in houses is recorded for each .the same is done with tyoe 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 point , find p(xa-xb)>1.0), where xa and xb are average drying times for sample size na=nb=18

Answer 1

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

$$\mu _{x_a-x_b}=\mu _a-\mu _b=0$$

and variance

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

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

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

So

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

$$=1-0.998650=0.001350$$