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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.

Short Answer

Expert verified

Hypothesis Testing is a type of statistical analysis in which you put your assumptions about a population parameter to the test. It is used to estimate the relationship between two statistical variables.

Step by step solution

01

Introduction

  • The Null Hypothesis states that the event will not take place. Unless it is rejected, the null hypothesis has no influence on the study's outcome.
  • Its sign is H0, which is pronounced H-naught.
  • The Null Hypothesis is the logical polar opposite of the Alternate Hypothesis.
  • The alternative hypothesis is accepted once the null hypothesis is rejected. The symbol for it is H1.
02

 Determine the critical value for level of significance 

Given information:

The standard deviation is 6,the mean is 17.8,the sample size is 100 and the level of significance is 0.05.

Calculation:

The hypothesis are,

H0:μ=16.7

Ha:μ16.7

The test statistics is,

z=x¯-μσn

Substitute the given values in above equation.

z=17.8-16.76100

=1.83

The critical value for level of significance of 0.05is ±1.96.

03

Sketch the graph

The graph is shown below.

Thus, the null hypothesis is not rejected and it is not reasonable to believe that the mean length of imprisonment for motor vehicle theft offenders in Sydney differ from the national mean in Australia.

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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.)

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

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

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.

Job Gains and Losses. In the article "Business Employment Dynamics: New Data on Gross Job Gains and Losses" (Monthly Labor Review, Vol. 127, Issue 4, pp. 29-42), 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 20 quarters provided the net percentage gains (losses are negative gains) for jobs as presented on the WeissStats site. Use the technology of your choice to do the following.

a. Decide whether, on average, the net percentage gain for jobs exceeds 0.2. Assume a population standard deviation of 0.42. Apply

which we provide on the WeissStats site. Use the technology of your choice to do the following.

a. Obtain a normal probability plot, boxplot, histogram, and stemand-leaf diagram of the data.

b. Based on your results from part (a), can you reasonably apply the one-mean z-test to the data? Explain your reasoning.

c. At the 1%significance level, do the data provide sufficient evidence to conclude that the mean body temperature of healthy humans differs from 98.6°F? Assume that σ=0.63°F.

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