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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 two-tailed test withα=0.05

Short Answer

Expert verified

Here we set the α=0.05value level of the given test that describes z with a two-tailed view. For a single z-definition test, under the null hypothesis, z-test statistics have a fairly common distribution.

z-check. In each test Two double-tailed tail tests, we have a drop point on both sides of the regular curve.

Therefore the value of two z points divides the area below the normal curve into the central area of 0.95 and the two areas outside of 0.025

That means key values ±z0.025

Used for areas under the standard curve table, ±z0.025=±1.96.

Step by step solution

01

Step 1. Given 

A two -tailed test withα=0.05

02

Step 2. Graph 

Here we set the α=0.05value level of the given test that describes z with a two-tailed view. For a single z-definition test, under the null hypothesis, z-test statistics have a fairly common distribution.

z-check. In each test Two double-tailed tail tests, we have a drop point on both sides of the regular curve.

Therefore the value of two z points divides the area below the normal curve into the central area of 0.95 and the two areas outside of 0.025

That means key values ±z0.025

Used for areas under the standard curve table, ±z0.025=±1.96

The rejection position is therefore found as in the following graph

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Most popular questions from this chapter

The normal probability curve and stem-to-leaf diagram of the data are shown in figure; σis unknown.

Perform Hypothesis test for mean of the population from which data is obtained and decide whether to use z-test, t-test or neither. Explain your answer.

In the article "Business Employment Dynamics: New data on Gross Job Gains and Losses", J. Spletzer et al. examined gross job gains and losses as a percentage of the average of previous and current employment figures. A simple random sample of 20quarters provided the net percentage gains for jobs as presented on the WeissStats site. Use the technology of your choice to do the following.

Part (a): Decide whether, on average, the net percentage gain for jobs exceeds 0.2. Assume a population standard deviation of 0.42. Apply the one-mean z-test with a 5%significance level.

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

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

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

As we mentioned on page 378, the following relationship holds between hypothesis test and confidence intervals for one-mean z-procedures: For a two-tailed hypothesis test at the significance level α, the null hypothesis role="math" localid="1653038937481" H0:μ=μ0will be rejected in favor of the alternative hypothesis Ha:μμ0if 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 z-interval and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.84

Part (b): Exercise9.87

Serving Time. According to the Bureau of Crime Statistics and Research of Australia, as reported on Lawlink, the mean length of imprisonment for motor-vehicle theft offenders in Australia is 16.7 months. One hundred randomly selected motor-vehicle-theft offenders in Sydney, Australia, had a mean length of imprisonment of 17.8 months. At the5% significance level, do the data provide sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle theft offenders in Sydney differs from the national mean in Australia? Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in Sydney is 6.0 months.

Purse Snatching. The Federal Bureau of Investigation (FBI) compiles information on robbery and property crimes by type and selected characteristic and publishes its findings in Uniform Crime Reports. According to that document, the mean value lost to purse snatching was 468in 2012. For last year, 12randomly selected purse-snatching offenses yielded the following values lost, to the nearest dollar.

Use a t-test to decide, at the5%significance level, whether last year's mean value lost to purse snatching has decreased from the 2012mean. The mean and standard deviation of the data are 455.0and 86.8, respectively.

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