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A lawyer believes that a certain judge imposes prison sentences for property crimes that are longer than the state average 11.7 months. He randomly selects 36 of the judge's sentences and obtains mean 13.8 and standard deviation 3.9 months. a. Perform the test at the \(1 \%\) level of significance using the critical value approach. b. Compute the observed significance of the test. c. Perform the test at the \(1 \%\) level of significance using the \(p\) -value approach. You need not repeat the first three steps, already done in part (a).

Short Answer

Expert verified
Reject the null hypothesis; the judge imposes significantly longer sentences.

Step by step solution

01

State the Null and Alternative Hypotheses

For hypothesis testing, we need to define our null hypothesis (\(H_0\)) and alternative hypothesis (\(H_a\)). \[H_0: \mu = 11.7\text{ months} \] \( \text{(The judge's average sentence is equal to the state average)} \) \[H_a: \mu > 11.7 \text{ months} \] \( \text{(The judge's average sentence is longer than the state average)} \)
02

Find the Critical Value

Since we're conducting a one-tailed test at the \(1\%\) level of significance, we find the critical value for a \(z\)-test using statistical z-tables. The critical value \(z_c\) for a one-tailed test at \(\alpha = 0.01\) is approximately \(2.33\).
03

Compute the Test Statistic

Use the formula for the \(z\)-statistic:\[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] where:- \(\bar{x} = 13.8\) (sample mean),- \(\mu = 11.7\) (population mean),- \(s = 3.9\) (sample standard deviation),- \(n = 36\) (sample size).Substitute the values to get:\[ z = \frac{13.8 - 11.7}{\frac{3.9}{\sqrt{36}}} = \frac{2.1}{0.65} \approx 3.23\]
04

Make a Decision (Critical Value Approach)

The calculated test statistic \(z = 3.23\) is greater than the critical value \(z_c = 2.33\). Since the test statistic is in the rejection region, we reject the null hypothesis, \(H_0\).
05

Compute the p-value (Observed Significance)

To find the \(p\)-value, look up \(z = 3.23\) in the \(z\)-table or use statistical software. The \(p\)-value is around \(0.0006\), which is the probability that a standard normal variable is greater than \(3.23\).
06

Make a Decision (p-value Approach)

Since the \(p\)-value (\(0.0006\)) is less than the significance level \(\alpha = 0.01\), we reject the null hypothesis. This suggests the judge's sentences are significantly longer than the state average.

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Key Concepts

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

Null Hypothesis
The Null Hypothesis, often denoted as \(H_0\), is a fundamental concept in hypothesis testing. It represents a statement of no effect or no difference. In simpler terms, it is a hypothesis that suggests that there is nothing unusual happening, and any observed effect is really just due to random chance. In the original exercise, the Null Hypothesis was posed to test whether a judge's average prison sentence for property crimes is equal to the state average. Thus, we write \(H_0: \mu = 11.7\text{ months}\), meaning the judge's sentences do not differ from the norm.

When performing hypothesis testing, we strive to provide evidence against the null hypothesis in order to justify an alternative hypothesis. Rejecting the null hypothesis would imply that there is non-random, significant evidence suggesting a difference from the state average. It's important to note that failing to reject the null does not prove it true, it simply means there is insufficient evidence against it.
Alternative Hypothesis
The Alternative Hypothesis, represented as \(H_a\), is what you might call the rival hypothesis. It's the statement you propose when you think there actually is an effect or a significant difference. In the given context, it reflects the belief that the judge's average sentence is longer than the state average. We express this as \(H_a: \mu > 11.7 \text{ months}\).

The alternative hypothesis is directional in this scenario, specifying not just a difference but a greater average sentence compared to the state standard. This specificity requires a one-tailed test, where we only look for the evidence in one direction. When we predefine our alternative hypothesis, it helps us tailor our statistical test to only consider outcomes that would show an increase rather than any difference, which adds precision to our hypothesis testing.
p-value
A p-value is a measure that helps determine the significance of your results in hypothesis testing. It gives us the probability of observing the test results, or something more extreme, assuming the null hypothesis is true. In other words, it tells us how likely the sample data is if there truly is no effect. For this exercise, the calculated p-value was approximately \(0.0006\).

A small p-value indicates strong evidence against the null hypothesis, which means we have a low probability of observing such a sample mean if the null hypothesis were true. Therefore, a very low p-value, as found here, suggests that the judge's sentences are indeed significantly longer than the state average. Remember, the p-value helps us understand how extreme the test results are, given the null hypothesis.
significance level
The significance level, often represented by \(\alpha\), is a threshold you set to decide whether to reject the null hypothesis. It represents the level of risk you are willing to take to incorrectly reject the null hypothesis, also known as a Type I error. For many tests, a standard significance level might be \(0.05\), but in this exercise, we used a stricter \(1\%\) or \(0.01\) level.

This means that only if there is a less than \(1\%\) chance of the observed data occurring under the null hypothesis, we will reject it. Using a low significance level like \(0.01\) implies that we are demanding stronger evidence to support a claim against the null hypothesis. In our example, since the p-value of \(0.0006\) is less than the \(0.01\) significance level, we conclude that there is significant evidence to reject the null hypothesis, suggesting the judge's sentencing is indeed different from the norm.

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Most popular questions from this chapter

Five years ago \(3.9 \%\) of children in a certain region lived with someone other than a parent. A sociologist wishes to test whether the current proportion is different. Perform the relevant test at the \(5 \%\) level of significance using the following data: in a random sample of 2,759 children, 119 lived with someone other than a parent.

Compute the value of the test statistic for the indicated test, based on the information given. a. Testing \(H_{0}: \mu=342\) vs. \(H a: \mu<342, \sigma=11.2, n=40, x-=339, s=10.3\) b. Testing \(H_{0}: \mu=105\) vs. Ha: \(\mu>105, \sigma=5.3, n=80, x-=107, s=5.1\) c. Testing \(H_{0}: \mu=-13.5\) vs. Ha: \(\mu \neq-13.5, \sigma\) unknown, \(n=32, x-=-13.8, s=1.5\) d. Testing \(H_{0}: \mu=28\) vs. \(H a: \mu \neq 28, \sigma\) unknown, \(n=68, x-27.8, s=1.3\)

Two years ago \(72 \%\) of household in a certain county regularly participated in recycling household waste. The county government wishes to investigate whether that proportion has increased after an intensive campaign promoting recycling. In a survey of 900 households, 674 regularly participate in recycling. Perform the relevant test at the \(10 \%\) level of significance.

Find the rejection region (for the standardized test statistic) for each hypothesis test. Identify the test as left-tailed, right-tailed, or two- tailed. a. \(\quad H 0: \mu=141\) VS. \(H a: \mu<141\) \(@ \alpha=0.20 .\) b. \(\quad H 0: \mu=-54\) VS. \(H a: \mu<-54\) @ \(\alpha=0.05 .\) C. \(\quad H 0: \mu=98.6\) VS. \(H a: \mu \neq 98.6\) \(@ \alpha=0.05 .\) d. \(\quad H 0: \mu=3.8\) VS. \(H a: \mu>3.8\) @ \(\alpha=0.001\)

Ice cream is legally required to contain at least \(10 \%\) milk fat by weight. The manufacturer of an economy ice cream wishes to be close to the legal limit, hence produces its ice cream with a target proportion of 0.106 milk fat. A sample of five containers yielded a mean proportion of 0.094 milk fat with standard deviation 0.002. Test the null hypothesis that the mean proportion of milk fat in all containers is 0.106 against the alternative that it is less than \(0.106,\) at the \(10 \%\) level of significance. Assume that the proportion of milk fat in containers is normally distributed. Hint: This exercise could have been presented in an earlier section.

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