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Magazine Chapter 7: Q8E (page 397)
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鈥揻, 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,
The given alternative hypothesis is two-tailed.
Also,
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:
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