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Chapter 8: Problem 4

Prison Time According to a public service website, $$69.4 \%$$ of white collar criminals get prison time. A randomly selected sample of 165 white collar criminals revealed that 120 were serving or had served prison time. Using $$\alpha=0.05,$$ test the conjecture that the proportion of white collar criminals serving prison time differs from $$69.4 \%$$ in two different ways.

Short Answer

Expert verified
The proportion of white collar criminals serving prison time does not significantly differ from 69.4% at the 0.05 significance level.

Step by step solution

01

Define the Hypothesis

We need to establish the null and alternative hypotheses. The null hypothesis \( H_0 \) states that the proportion of white collar criminals who get prison time is equal to 69.4%, \( p = 0.694 \). The alternative hypothesis \( H_1 \) suggests that the proportion is not equal to 69.4%, which is \( p eq 0.694 \).
02

Gather Sample Data

From the exercise, we have a sample of 165 white collar criminals, with 120 having served prison time. The sample proportion, \( \hat{p} \), can be calculated as \( \hat{p} = \frac{120}{165} \approx 0.727 \).
03

Calculate the Test Statistic

To test our hypothesis, we'll use a z-test for proportion. The test statistic \( z \) is given by the formula:\[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}}\]where \( \hat{p} \) is the sample proportion, \( p_0 \) is the hypothesized population proportion (0.694), and \( n \) is the sample size (165). Plugging in these values:\[z = \frac{0.727 - 0.694}{\sqrt{\frac{0.694 \times 0.306}{165}}} \approx 0.992\]
04

Determine the Critical Value and Decision Rule

For a two-tailed test at \( \alpha = 0.05 \), the critical z-values are approximately \( \pm 1.96 \). This means if our calculated z-value is less than -1.96 or greater than 1.96, we reject the null hypothesis. Otherwise, we fail to reject it.
05

Compare and Conclude

Since our calculated z-value (0.992) lies within the range of -1.96 and 1.96, we do not reject the null hypothesis. Thus, at the 0.05 significance level, there is not enough evidence to conclude that the proportion of white collar criminals serving prison time is different from 69.4%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportion Hypothesis Test
A proportion hypothesis test is a statistical method used to determine if there is enough evidence in a sample to support a claim about a population proportion. This technique is useful when you're dealing with categorical data 鈥� such as whether a white collar criminal gets prison time or not.

For the provided exercise, we start by considering the proportion of white collar criminals that are expected to receive prison time according to the public service website, which is 69.4%. We are interested in testing if our sample data from 165 individuals provides enough evidence to say the true population proportion is different from this stated rate.

The hypothesis test involves taking our sample (120 out of 165 serving time) and computing a sample proportion. We then use this proportion in calculations to decide if the observed sample proportion significantly deviates from the hypothesized population proportion of 69.4%.
Z-Test for Proportions
The z-test for proportions is the statistical approach used in this exercise to test our hypothesis about proportions. It helps determine if the observed sample proportion is significantly different from the population proportion stated in the null hypothesis.

Here's how it works:
  • Calculate the sample proportion, \( \hat{p} \), by dividing the number of successes (120 criminals serving time) by the sample size (165).
  • Use the formula \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}} \] where \( p_0 \) is the expected population proportion (0.694), to find the test statistic, \( z \).
  • Compare the calculated \( z \) value to the critical z-values for a two-tailed test (approximately 卤1.96 for \( \alpha = 0.05 \)).
If your z-value falls outside these critical values, there's significant evidence to suggest the sample proportion is different from the hypothesized proportion.
Null and Alternative Hypotheses
The null and alternative hypotheses are foundational to hypothesis testing. They set up the scenario for testing claims about data.

- The **null hypothesis** \( H_0 \) always presents a statement of no effect or no difference. For our exercise, it claims that the proportion of white collar criminals receiving prison time is 69.4%: \( p = 0.694 \). The goal is to test whether we can reject this statement based on our sample findings.

- The **alternative hypothesis** \( H_1 \) proposes a new claim that we are interested in supporting. It suggests that the true proportion is not equal to 69.4%, represented as \( p eq 0.694 \).

When performing the test, we aim to gather enough evidence to potentially reject the null hypothesis in favor of the alternative. If the evidence is insufficient, we fail to reject the null, which means our sample doesn't provide enough justification to say the proportion is different than expected.

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Most popular questions from this chapter

For Exercises 7 through \(23,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Internet Visits A U.S. Web Usage Snapshot indicated a monthly average of 36 Internet visits a particular website per user from home. A random sample of 24 Internet users yielded a sample mean of 42.1 visits with a standard deviation of 5.3 . At the 0.01 level of significance, can it be concluded that this differs from the national average?

For Exercises I through 25, perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Transmission Service A car dealer recommends that transmissions be serviced at \(30,000\) miles. To see whether her customers are adhering to this recommendation, the dealer selects a random sample of 40 customers and finds that the average mileage of the automobiles serviced is \(30,456 .\) The standard deviation of the population is 1684 miles. By finding the \(P\) -value, determine whether the owners are having their transmissions serviced at \(30,000\) miles. Use \(\alpha=0.10 .\) Do you think the \(\alpha\) value of 0.10 is an appropriate significance level?

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Soda Bottle Content A machine fills 12 -ounce bottles with soda. For the machine to function properly, the standard deviation of the population must be less than or equal to 0.03 ounce. A random sample of 8 bottles is selected, and the number of ounces of soda in each bottle is given. At \(\alpha=0.05,\) can we reject the claim that the machine is functioning properly? Use the \(P\) -value method. $$ \begin{array}{cccc}{12.03} & {12.10} & {12.02} & {11.98} \\ {12.00} & {12.05} & {11.97} & {11.99}\end{array} $$

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Exam Grades A statistics professor is used to having a variance in his class grades of no more than \(100 .\) He feels that his current group of students is different, and so he examines a random sample of midterm grades as shown. At \(\alpha=0.05,\) can it be concluded that the variance in grades exceeds \(100 ?\) $$ \begin{array}{lllll}{92.3} & {89.4} & {76.9} & {65.2} & {49.1} \\ {96.7} & {69.5} & {72.8} & {67.5} & {52.8} \\ {88.5} & {79.2} & {72.9} & {68.7} & {75.8}\end{array} $$

For Exercises I through 25, perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Sick Days A manager states that in his factory, the average number of days per year missed by the employees due to illness is less than the national average of \(10 .\) The following data show the number of days missed by 40 randomly selected employees last year. Is there sufficient evidence to believe the manager's statement at \(\alpha=0.05 ? \sigma=3.63 .\) Use the \(P\) -value method. $$ \begin{array}{cccccccc}{0} & {6} & {12} & {3} & {3} & {5} & {4} & {1} \\ {3} & {9} & {6} & {0} & {7} & {6} & {3} & {4} \\ {7} & {4} & {7} & {1} & {0} & {8} & {12} & {3} \\ {2} & {5} & {10} & {5} & {15} & {3} & {2} & {5} \\ {3} & {11} & {8} & {2} & {2} & {4} & {1} & {9}\end{array} $$
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