Question

For each of the following rejection regions, sketch the sampling distribution for z and indicate the location of the rejection region.

a. \({H_0}:\mu \le {\mu _0}\) and \({H_a}:\mu > {\mu _0};\alpha = 0.1\)

b. \({H_0}:\mu \le {\mu _0}\) and \({H_a}:\mu > {\mu _0};\alpha = 0.05\)

c. \({H_0}:\mu \ge {\mu _0}\) and \({H_a}:\mu < {\mu _0};\alpha = 0.01\)

d. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.05\)

e. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.1\)

f. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.01\)

g. For each rejection region specified in parts a–f, state the probability notation in z and its respective Type I error value.

Short answer

a)

b)

c)

d)

e)

f)

g)

Step-by-step solution

Given information

For each part, the null and the alternative hypotheses are given with a specific significance level

(a) Sketching the sampling distribution of z for H0:μ⩽μ0 and Ha:μ>μ0;α=0.1 

Consider,

\({H_0}:\mu \le {\mu _0}\)

\({H_a}:\mu > {\mu _0}\)

The given alternative hypothesis is right-tailed.

Also,

\(\alpha = 0.1\)

Therefore, from the table of Standard Normal Distribution z score valueis 1.282.

The sampling distribution of Z and the rejection region's location is shown in the following diagram.

(b) Sketching the sampling distribution of z for H0:μ⩽μ0and Ha:μ>μ0;α=0.05 

Consider,

\({H_0}:\mu \le {\mu _0}\)

\({H_a}:\mu > {\mu _0}\)

The given alternative hypothesis is right-tailed.

Also,

\(\alpha = 0.05\)

Therefore, from the table of Standard Normal Distribution z score value is 1.96.

The sampling distribution of Z and the rejection region's location is shown in the following diagram.

(c) Sketching the sampling distribution of z for H0:μ⩾μ0 and Ha:μ<μ0;α=0.01 

Consider,

\({H_0}:\mu \ge {\mu _0}\)

\({H_a}:\mu < {\mu _0}\)

Given alternative hypothesis is left-tailed.

Also,

\(\alpha = 0.01\)

Therefore, from the table of Standard Normal Distribution z score value is -2.326.

The sampling distribution of Z and the location of the rejection region at \(Z < - 2.326\) is shown in the following diagram.

(d) Sketching the sampling distribution of z for H0:μ=μ0 and Ha:μ≠μ0;α=0.05    

Consider,

\({H_0}:\mu = {\mu _0}\)

\({H_a}:\mu \ne {\mu _0}\)

The given alternative hypothesis is two-tailed.

Also,

\(\alpha = 0.05\)

Therefore, from the table of Standard Normal Distribution z score value is 1.96.

The sampling distribution of Z and the rejection region's location is shown in the following diagram.

(e) Sketching the sampling distribution of z for H0:μ=μ0 and Ha:μ≠μ0;α=0.1

Consider,

\({H_0}:\mu = {\mu _0}\)

\({H_a}:\mu \ne {\mu _0}\)

The given alternative hypothesis is two-tailed.

Also,

\(\alpha = 0.1\)

Therefore, from the Standard Normal distribution table, z score values are -1.282 and 1.282.

The sampling distribution of Z and the rejection region's location is shown in the following diagram.

Consider,

\({H_0}:\mu = {\mu _0}\)

\({H_a}:\mu \ne {\mu _0}\)

The given alternative hypothesis is two-tailed.

Also,

\(\alpha = 0.1\)

Therefore, from the Standard Normal distribution table, z score values are -1.282 and 1.282.

(f) Sketching the sampling distribution of z for H0:μ=μ0 and Ha:μ≠μ0;α=0.01

Consider,

$$\begin{gathered} H_0:\mu ={\mu }_0 \\ {H}_a:\mu \ne {\mu }_0 \end{gathered}$$

The given alternative hypothesis is two-tailed.

Also,

$\alpha =0.01$

Therefore, from the Standard Normal distribution table, z score values are -2.326 and 2.326.

The sampling distribution of Z and the rejection region's location is shown in the following diagram.

(g) Stating the probabilities of Type I errors.

Since Type I error is the probability of rejecting the null hypothesis when it is true.

Now, the required table is as follows: