A random sample of n observations is selected from a normal population to test the null hypothesis that muμequals=10. Specify the rejection region for each of the following combinations of HSubscript aa, alphaα, and n. a. HSubscript aa: muμnot equals≠10; alphaαequals=0.010.01; nequals=1313 b. HSubscript aa: muμgreater than>10; alphaαequals=0.100.10; nequals=2323 c. HSubscript aa: muμgreater than>10; alphaαequals=0.050.05; nequals=99 d. HSubscript aa: muμless than<10; alphaαequals=0.100.10; nequals=1111 e. HSubscript aa: muμnot equals≠10; alphaα equals=0.050.05; nequals=2020 f. HSubscript aa: muμless than<10; alphaαequals=0.010.01; nequals=77 a. Select the correct choice below and fill in the answer box within your choice.

Step-by-step explanation:

a) H0: μ = 10

Ha: μ ≠ 10

This is a two tailed test

n = 13

Since α = 0.01, the critical value is determined from the t distribution table. Recall that this is a two tailed test. Therefore, we would find the critical value corresponding to 1 - α/2 and reject the null hypothesis if the absolute value of the test statistic is greater than the value of t 1 - α/2 from the table.

1 - α/2 = 1 - 0.01/2 = 1 - 0.005 = 0.995

The critical value is 3.012

The rejection region is area > 3.012

b) Ha: μ > 10

This is a right tailed test

n = 23

α = 0.1

We would reject the null hypothesis if the test statistic is greater than the table value of 1 - α

1 - α = 1 - 0.1 = 0.9

The critical value is 1.319

The rejection region is area > 1.319

c) Ha: μ > 10

This is a right tailed test

n = 99

α = 0.05

We would reject the null hypothesis if the test statistic is greater than the table value of 1 - α

1 - α = 1 - 0.05 = 0.95

The critical value is 1.66

The rejection region is area > 1.66

d) Ha: μ < 10

This is a left tailed test

n = 11

α = 0.1

We would reject the null hypothesis if the test statistic is lesser than the table value of 1 - α

1 - α = 1 - 0.1 = 0.9

The critical value is 1.363

The rejection region is area < 1.363

e) H0: μ = 10

Ha: μ ≠ 10

This is a two tailed test

n = 20

Since α = 0.05, we would find the critical value corresponding to 1 - α/2 and reject the null hypothesis if the absolute value of the test statistic is greater than the value of t 1 - α/2 from the table.

1 - α/2 = 1 - 0.05/2 = 1 - 0.025 = 0.975

The critical value is 2.086

The rejection region is area > 2.086

f) Ha: μ < 10

This is a left tailed test

n = 77

α = 0.01

We would reject the null hypothesis if the test statistic is lesser than the table value of 1 - α

1 - α = 1 - 0.01 = 0.99

The critical value is 2.376

The rejection region is area < 2.376

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb