Question

Really short question! Please help me to solve ONLY part(b) with R code. Thank you!

Problem 4 [26 points] (Section 2.4): Consider a one-sample z-test (known variance) with hypotheses: Ho: μ lo vs H, μ *Ho. The probability of Type II error can be written in the form |ß D(%2_Jnd)-0(-%2_Jnd) where Φ㈠ is the CDF of N(0,1), d Isyo, and δ is the difference between the true mean and the mean under Ho (a) [10 points] Based on the fact that φ(x) [pdf of N(0,1)) is a decreasing function in x when a/2 0. x> 0, show that (i) for specified values of a and n, is a decreasing function of d. (ii) for specified values of α and d, β is a decreasing function of n. (b)[16 points] Suppose the true mean is H1 13.46. Consider the hypotheses Ho H14.0 vs Hi #14.0 (u-0.05) with variance known at ơ2-0.36, what is the minimum sample size to ensure that the probability of Type II error is less than 0.05?

Answer 1

$(b)~Z-statistic=\frac{\sqrt{n}(\bar{X}-13.46)}{\sqrt{0.36}}=\frac{\sqrt{n}(\bar{X}-13.46)}{0.6}\\ We~reject~H_0~at~level~0.05~if\\ \begin{vmatrix} \frac{\sqrt{n}(\bar{X}-13.46)}{0.6} \end{vmatrix}>1.96.\\ Probability~of~Type~II~error=P\left( \begin{vmatrix} \frac{\sqrt{n}(\bar{X}-13.46)}{0.6} \end{vmatrix}\leq 1.96|\mu=14\right )\\ =P\left(-1.96\leq \frac{\sqrt{n}(\bar{X}-14&plus;14-13.46)}{0.6}\leq 1.96|\mu=14 \right )\\ =\Phi\left(1.96-0.9\sqrt{n} \right )-\Phi\left(-1.96-0.9\sqrt{n} \right )<0.05\\$

R code:

n=10:20
p=pnorm(1.96-0.9*sqrt(n))-pnorm(-1.96-0.9*sqrt(n))
k=length(n)
m=1:k
for(i in 1:k)
{
if(p[i]<0.05){
m[i]=n[i]
}
else{
m[i]=0
}
}
N=min(subset(m, m>0))
N

Out put:

> N
[1] 17