Question

please show how you got the answer

Find B. Ho: M = 7 us returnel H.M= 7.2 if Seuze $=0.567 x=0.05 n=47, X= 7.112 /

Answer 1

Type II Error is failing to reject the null when it is false

$\beta$ is the probability of type II error.

H:H= 7

H, :μ#7

significance level =

g= 0,05

This is a two-tailed test. We will fail to reject the null (commit a Type II error) if we get a Z statistic greater than -1.96 and less than 1.96.

This -1.96 & 1.96  Z-critical value corresponds to some X critical value ( X critical), such that

$\frac{X_{c_1}-7}{\frac{0.567}{\sqrt{47}}}=1.96$

$X_{c_1}=7.162$

$\frac{X_{c_2}-7}{\frac{0.567}{\sqrt{47}}}=-1.96$

$X_{c_2}=6.838$

So I will incorrectly fail to reject the null as long as a draw a sample mean that greater than 6.838 and less than 7.162.

To complete the problem what I now need to do is compute the probability of drawing a sample mean greater than 6.838 and less than 7.162 given $\mu = 7.2$

Thus, the probability of a Type II error is given by

$P(\frac{6.838-7.2}{\frac{0.567}{\sqrt{47}}}<Z<\frac{7.162-7.2}{\frac{0.567}{\sqrt{47}}})$

$P(-4.377<Z<-0.459)=P(Z<-0.459)-P(Z<-4.377)$

$P(-4.377<Z<-0.459)=P(Z<-0.459)-P(Z<-4.377)=0.3231-0$

the probability of a Type II error is $\beta=0.3231$