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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:μ≠16.7

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

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: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 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

Dementia is the loss of the in actual and social abilities severe enough to interfere with judging behavior and daily functioning. Alzheimer's disease is the most common type of dementia. In the article "Living with Early Onsite dementia: Exploring the Experience and Developing Evidence Guidelines for Practice", P Harris and J Keady explored the experiment struggles of people diagnosed with dementia and their familiar simple random sample \(21\) people with early-onset dementia the following data on age at diagnosis in years.

At the \(1%\) significance level, do the data provide sufficient evidence to conclude that the mean age at diagnosis of all people with early onset dementia is less than \(55\) years old? Assume that the population is standard deviation is \(6.8\) years.

In each part, we have given the value obtained for the test statistic, z, in a one-mean z-test. We have also specified whether the test is two tailed, left tailed, or right tailed. Determine the P-value in each case and decide whether, at the 5%significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.

a. z=-1.25; left-tailed test

b. z=2.36; right-tailed test

c. z=1.83; two-tailed test

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.

Refer to Exercise 9.18. Explain what each of the following would mean.

(a) Type I error.

(b) Type II error.

(c) Correct decision.

Now suppose that the results of carrying out the hypothesis test lead to the rejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean age at diagnosis of all people with early-onset dementia.

(d) is 55 years old.

(e) is less than 55 years old.

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