The following graph portrays the decision criterion for a onemean z-test, using the critical-value approach to hypothesis testing. The curve in the graph is the normal curve for the test statistic under the assumption that the null hypothesis is true.

Determine the

a. rejection region.

b. nonrejection region.

c. critical value(s).

d. significance level.

e. Draw a graph that depicts the answers that you obtained in parts (a)-(d).

f. Classify the hypothesis test as two tailed, left tailed, or right tailed.

Short Answer

Expert verified

(a) The rejection region is z1.28

(b) The non- rejection region is z < 1.28

(c) The critical value is 1.28

(d) The significance level is 0.10

Step by step solution

01

Subpart (a) Step 1:

(a)

The rejection zone in this example consists of all z-scores to the right of 1.28. The zone of rejection is:z1.28

02

Subpart (b) step 1:

(b)

The non-rejection zone in this situation is defined as all z-scores to the left of 1.28. The non-rejection zone is defined as:z<1.28

03

Subpart (c) Step 3:

(c)

The rejection and non-rejection areas are separated by the crucial value of 1.28.

04

Subpart (d) Step 1:

(d)

The region to the right of the 1.28 is the significance level. The significance threshold in this example is 0.10.

05

Subpart (e) Step 1:

06

Subpart (f) step 1:

(f)

The provided hypothesis test is a right tailed test since the rejection zone is to the right of the critical value.

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Most popular questions from this chapter

State two reasons why including the P-value is prudent when you are reporting the results of a hypothesis test.

9.128 Two-Tailed Hypothesis Tests and CIs. The following relationship holds between hypothesis tests and confidence intervals for one-mean t-procedures: For a two-tailed hypothesis test at the significance level α, the null hypothesis H0:μ=μ0will be rejected in favor of the alternative hypothesis H2:μ>μ0if and only if μ0lies outside the (1-α)-level confidence interval for μ. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean t-interval (Procedure 8.2 on page 338 ) and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise 9.113
b. Exercise 9.116

The daily charges, in dollars, for a sample of 15 hotels and motels operating in South Carolina are provided on the WeissStats site. The data were found in the report South Carolina Statistical Abstract, sponsored by the South Carolina Budget and Control Board.

Part (a): Use the one-mean z-test to decide, at the 5% significance level, whether the data provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than \(75. Assume a population standard deviation of \)22.40.

Part (b): Obtain a normal probability plot, boxplot, histogram and stem-and-leaf diagram of the data.

Part (c): Remove the outliers (if any) from the data and then repeat part (a).

Part (d): Comment on the advisability of using thez-test here.

The following relationship holds between hypothesis tests and confidence intervals for one-mean t-procedures: For a two-tailed hypothesis test at the significance level α, the null hypothesis H0:μ=μ0will be rejected in favor of the alternative hypothesis Ha:μ>μ0if and only if μ0lies outside the 1-α-level confidence interval for μ. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean t-interval and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.113

Part (b): Exercise 9.116

In Exercise 8.146 on page 345,we introduced one-sided one mean-t-intervals. The following relationship holds between hypothesis test and confidence intervals for one-mean t-procedures: For a right-tailed hypothesis test at the significance level α, the null hypothesis H0:μ=μ0will be rejected in favor of the alternative hypothesis data-custom-editor="chemistry" Ha:μ>μ0if and only if μ0is less than or equal to the 1-α-level lower confidence bound for μ. In each case, illustrate the preceding relationship by obtaining the appropriate lower confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.114 (both parts)

Part (b) Exercise9.115

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