Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A reporter on cnn.com stated in July 2010 that \(95 \%\) of all court cases that go to trial result in a guilty verdict. To test the accuracy of this claim, we collect a random sample of 2000 court cases that went to trial and record the proportion that resulted in a guilty verdict. (a) What is/are the relevant parameter(s)? What sample statistic(s) is/are used to conduct the test? (b) State the null and alternative hypotheses. (c) We assess evidence by considering how likely our sample results are when \(H_{0}\) is true. What does that mean in this case?

Short Answer

Expert verified
(a) The relevant parameter is the population proportion \(p\) of guilty verdicts in all court cases, with sample proportion \(\hat{p}\) for the assessment. (b) The null hypothesis is \(H_{0}\): \(p = 0.95\) and the alternative hypothesis is \(H_{a}\): \(p \neq 0.95\). (c) The assessment involves determining how likely the observed sample result would be if \(H_{0}\) (95% of court cases result in a guilty verdict) is true.

Step by step solution

01

Identifying relevant parameters and statistics

The relevant parameter in this scenario is the population proportion of all court cases that result in a guilty verdict. Let's denote this as \(p\). The sample statistic used to conduct the test is the sample proportion of court cases sampled that resulted in a guilty verdict, denoted as \(\hat{p}\).
02

Formulating the null and alternative hypotheses

The null hypothesis, \(H_{0}\), is a statement that the parameter, \(p\), is equal to a specific value. The alternative hypothesis, \(H_{a}\), states that \(p\) takes on a value that is different, larger, or smaller than the value specified in \(H_{0}\). Here, the null hypothesis \(H_{0}\) : \(p = 0.95\) (95% of all cases are declared guilty, according to the reporter). The alternate hypothesis could assume that the proportion is different from 0.95, so \(H_{a}\): \(p \neq 0.95\).
03

Understanding the Assessment

The sentence 'We assess evidence by considering how likely our sample results are when \(H_{0}\) is true' means that we would think about how probable it would be to obtain our sample data if indeed 95% of all court cases resulted in a guilty verdict. If the sample data is very unusual assuming the null hypothesis is true, we would reject \(H_{0}\) in favor of \(H_{a}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Population Proportion
In the context of hypothesis testing, the population proportion refers to a specific characteristic of an entire population, such as the percentage of court cases that result in a guilty verdict. It's denoted as \( p \) and is a fixed value, although unknown. When we seek to estimate the population proportion, we are trying to find what percentage of all possible cases, not just our sample, would end in a similar outcome. If the population is the pool of all court cases that go to trial, then the population proportion is the overall percentage of these cases that lead to a guilty verdict.
Sample Statistic
A sample statistic is a numerical summary about a sample, which serves as an estimate for the unknown population parameter. In our court case scenario, the relevant sample statistic is the sample proportion, \( \hat{p} \) , which is the percentage of cases in our selected sample that resulted in a guilty verdict. By comparing this \( \hat{p} \) with the claimed population proportion, we can conduct a hypothesis test to assess the claim's accuracy. Sample statistics vary from sample to sample due to variability in the selection process, which is a crucial concept in statistics known as sampling variability.
Null Hypothesis
The null hypothesis, symbolized as \( H_{0} \) , is a statement for a statistical hypothesis test that assumes no effect or no difference. It represents the skeptical perspective, providing a benchmark against which we measure the evidence provided by our sample statistic. In our case, the null hypothesis states that the true population proportion of guilty verdicts in court cases (\( p \) ) is equal to 95%, \( H_{0} : p = 0.95 \) . If the data collected from our sample significantly deviates from this assumption, we might reject the null hypothesis in favor of the alternative.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_{a} \) or \( H_{1} \) , is a statement that proposes a new effect, relationship, or difference, contrasting the skeptical stance of the null hypothesis. For our exercise, the alternative hypothesis presents the possibility that the population proportion (\( p \) ) is not 95%. Formally, we write \( H_{a}: p eq 0.95 \) . It reflects the belief that what was stated by the reporter might be incorrect and allows us to challenge the status quo presented by \( H_{0} \) with the sample evidence.
Evidence Assessment in Statistics
In evidence assessment, we use our collected sample data to evaluate how consistent it is with the null hypothesis. This process involves calculating the probability of observing our sample statistic, like the sample proportion of guilty verdicts, given that the null hypothesis is true. If this probability, known as a p-value, is very low, it suggests that our sample result is unlikely under the null hypothesis, and thus, we may have enough evidence to reject \( H_{0} \) in favor of \( H_{a} \) . In our example, if the likelihood of obtaining our sample's guilty verdict proportion is very small under the assumption that the actual population proportion is 95%, this would cast doubt on the accuracy of the reporter's claim.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Figure 4.25 shows a scatterplot of the acidity (pH) for a sample of \(n=53\) Florida lakes vs the average mercury level (ppm) found in fish taken from each lake. The full dataset is introduced in Data 2.4 on page 71 and is available in FloridaLakes. There appears to be a negative trend in the scatterplot, and we wish to test whether there is significant evidence of a negative association between \(\mathrm{pH}\) and mercury levels. (a) What are the null and alternative hypotheses? (b) For these data, a statistical software package produces the following output: $$ r=-0.575 \quad p \text { -value }=0.000017 $$ Use the p-value to give the conclusion of the test. Include an assessment of the strength of the evidence and state your result in terms of rejecting or failing to reject \(H_{0}\) and in terms of \(\mathrm{pH}\) and mercury. (c) Is this convincing evidence that low \(\mathrm{pH}\) causes the average mercury level in fish to increase? Why or why not?

In Exercise 3.89 on page \(239,\) we found a \(95 \%\) confidence interval for the difference in proportion of rats showing compassion, using the proportion of female rats minus the proportion of male rats, to be 0.104 to \(0.480 .\) In testing whether there is a difference in these two proportions: (a) What are the null and alternative hypotheses? (b) Using the confidence interval, what is the conclusion of the test? Include an indication of the significance level. (c) Based on this study would you say that female rats or male rats are more likely to show compassion (or are the results inconclusive)?

Give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.7\) vs \(H_{a}: p<0.7\) Sample data: \(\hat{p}=125 / 200=0.625\) with \(n=200\)

Mating Choice and Offspring Fitness Does the ability to choose a mate improve offspring fitness in fruit flies? Researchers have studied this by taking female fruit flies and randomly dividing them into two groups; one group is put into a cage with a large number of males and able to freely choose who to mate with, while flies in the other group are each put into individual vials, each with only one male, giving no choice in who to mate with. Females are then put into egg laying chambers, and a certain number of larvae collected. Do the larvae from the mate choice group exhibit higher survival rates? A study \(^{44}\) published in Nature found that mate choice does increase offspring fitness in fruit flies (with p-value \(<0.02\) ), yet this result went against conventional wisdom in genetics and was quite controversial. Researchers attempted to replicate this result with a series of related experiments, \({ }^{45}\) with data provided in MateChoice. (a) In the first replication experiment, using the same species of fruit fly as the original Nature study, 6067 of the 10000 larvae from the mate choice group survived and 5976 of the 10000 larvae from the no mate choice group survived. Calculate the p-value. (b) Using a significance level of \(\alpha=0.05\) and \(\mathrm{p}\) -value from (a), state the conclusion in context. (c) Actually, the 10,000 larvae in each group came from a series of 50 different runs of the experiment, with 200 larvae in each group for each run. The researchers believe that conditions dif- fer from run to run, and thus it makes sense to treat each \(\mathrm{run}\) as a case (rather than each fly). In this analysis, we are looking at paired data, and the response variable would be the difference in the number of larvae surviving between the choice group and the no choice group, for each of the 50 runs. The counts (Choice and NoChoice and difference (Choice \(-\) NoChoice) in number of surviving larva are stored in MateChoice. Using the single variable of differences, calculate the p-value for testing whether the average difference is greater than \(0 .\) (Hint: this is a single quantitative variable, so the corresponding test would be for a single mean.) (d) Using a significance level of \(\alpha=0.05\) and the p-value from (c), state the conclusion in context. (e) The experiment being tested in parts (a)-(d) was designed to mimic the experiment from the original study, yet the original study yielded significant results while this study did not. If mate choice really does improve offspring fitness in fruit flies, did the follow-up study being analyzed in parts (a)-(d) make a Type I, Type II, or no error? (f) If mate choice really does not improve offspring fitness in fruit flies, did the original Nature study make a Type I, Type II, or no error?

A study \(^{20}\) conducted in June 2015 examines ownership of tablet computers by US adults. A random sample of 959 people were surveyed, and we are told that 197 of the 455 men own a tablet and 235 of the 504 women own a tablet. We want to test whether the survey results provide evidence of a difference in the proportion owning a tablet between men and women. (a) State the null and alternative hypotheses, and define the parameters. (b) Give the notation and value of the sample statistic. In the sample, which group has higher tablet ownership: men or women? (c) Use StatKey or other technology to find the pvalue.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free