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In each of Exercises 9.41-9.46 ,determine the critical values for a one-mean z-test. For each exercise, draw a graph that illustrates your answer

A left-tailed test withα=0.05

Short Answer

Expert verified

A left-tailed test with α=0.05.

In the left tail test that α=0.05, from normal local tables values are significant
z0=-1.645.

Step by step solution

01

Step 1. Given 

A left-tailed test withα=0.05

02

Step 2. Graph 

A left-tailed test with α=0.05.

in the left tail test that α=0.05, from normal local tables values are significant z0=-1.645.

The graph is shown below :

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Most popular questions from this chapter

Apparel and Services. According to the document Consumer Expenditures, a publication of the Bureau of Labor Statistics, the average consumer unit spent \( 1736 on apparel and services in 2012 That same year, 25 consumer units in the Northeast had the following annual expenditures, in dollars, on apparel and services.

12791457202016821273
22232233219216111734
26882029216618602444
18441765226715222012
19901751211322021712

At the 5 % significance level, do the data provide sufficient evidence to conclude that the 2012 mean annual expenditure on apparel and services for consumer units in the Northeast differed from the national mean of \)1736? (Note: The sample mean and sample standard deviation of the data are \(1922.76 and \)350.90, respectively.)

Cadmium, a heavy metal, is toxic to animals. Mushrooms, however are able to absorb and accumulate cadmium at high concentrations. The Czech and Slovak governments have set a safety limit for cadmium in dry vegetables at \(0.5\) part per million M. Melgar et al. measured the cadmium levels in a random sample of the edible mushroom Boletus pinicola and published the results in the paper "Influence of Some Factors in Toxicity and Accumulation of Cd from Edible Wild Macrofungi in NW Spain. A hypothesis test is to be performed to decide whether the mean cadmium level in Boletus pinicola mushrooms is greater than the government's recommended limit.

a. determine the null hypothesis

b. determine the alternative hypothesis

c. classify the hypothesis test as two tailed, left tailed or right tailed.

Ankle Brachial Index. The ankle brachial index (ABI) compares the blood pressure of a patient's arm to the blood pressure of the patient's leg. The ABI can be an indicator of different diseases, including arterial diseases. A healthy (or normal) ABI is 0.9 or greater. In a study by M. McDermott et al. titled "Sex Differences in Peripheral Arterial Disease: Leg Symptoms and Physical Functioning" (Journal of the American Geriatrics Society, Vol. 51, No. 2, Pp. 222-228), the researchers obtained the ABI of 187 women with peripheral arterial disease. The results were a mean ABI of 0.64 with a standard deviation of 0.15 At the 1 % significance level, do the data provide sufficient evidence to conclude that, on average, women with peripheral arterial disease have an unhealthy ABI?

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

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.

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