Question

Problem Statement In lecture we saw a strategy for constructing 95% confidence intervals for the mean of a normally distributed population. We did this by first selecting values U and V so that we could write a probability interval statement of the form: Pr( V < W ) 0.95 Then we did some algebraic manipulation on the inequalities that define the interval to eventually obtain values for L and U such that Pr(L* μ U*) 0.95 = Now let's generalize this procedure. Part (a) We'll start by constructing a 90% confidence interval for the mean of a normally distributed population. First, determine values for V and W so that s 0.90 Part (b) Use the probability interval in Part (a) to determine values for L and U such that: Part (c) Now we'll consider the general problem of constructing a 1- a confidence interval for the mean of a normally distributed population. Let Qz() denote the quantile function for the standard normal distribution, so that for instance: Q2(0.025) -1.96 = Qz(0.975)+1.96 Use the Qz) function to express values for V and W so that

Answer 1

a) $Q_z(0.05)=-1.645=V\ \and\ Q_z(0.95)=1.645=W$

b)

$L^*=V*\frac{\sigma}{\sqrt{n}}-\bar{X}=-1.645\frac{\sigma}{\sqrt{n}}-\bar{X}\\ U^*=W*\frac{\sigma}{\sqrt{n}}-\bar{X}=1.645\frac{\sigma}{\sqrt{n}}-\bar{X}$

c)

$V=Q_z(\alpha/2)\ and\ W=Q_z(1-\alpha/2)$

d)

$L^*=Q_z(\alpha/2)*\frac{\sigma}{\sqrt{n}}-\bar{X}\\ U^*=Q_z(1- \alpha/2)*\frac{\sigma}{\sqrt{n}}-\bar{X}$