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Exercises 4.117 to 4.122 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: \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) Sample data: \(\hat{p}=28 / 40=0.70\) with \(n=40\)

Short Answer

Expert verified
Given the hypotheses and the sample data, calculate the sample proportion, generate a randomization distribution using a statistical tool and compute the p-value. The p-value then aids in determining whether to reject or fail to reject the null hypothesis.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis is \(H_{0}: p=0.5\) and the alternative hypothesis is \(H_{a}: p \neq 0.5\) . Given that, under \(H_{0}\), the population proportion \(p\) is equal to 0.5.
02

Compute the Sample Proportion

Based on the sample data, the sample proportion \(\hat{p}\) is calculated as 28 / 40 = 0.70. This represents the proportion of success in the sample.
03

Generate Randomization Distribution and Calculate p-value

Utilize technology like StatKey to create a randomization distribution based on the null hypothesis and calculate the corresponding p-value. The p-value is the probability of observing a stat as extreme as, or more extreme than, the actual observed statistic, assuming that the null hypothesis is true.
04

Interpret the p-value

If the p-value is less than the significance level (commonly 0.05), then that is strong evidence against the null hypothesis, so it's rejected in favour of the alternative hypothesis. If it's greater, there's insufficient evidence to reject the null hypothesis.

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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 an initial statement or assumption made during hypothesis testing. In this context, the null hypothesis postulates that there is no effect or difference, and any observed variation is due to chance or sampling error. The purpose of the null hypothesis is to provide a statement that can be rigorously tested and potentially refuted using statistical methods.

Typically, the null hypothesis involves a statement of equality. For example, in the provided exercise, it is expressed as \(H_0: p = 0.5\). This means that the initial claim is that the population proportion \(p\) is exactly 0.5.

To conduct the hypothesis test, we begin by assuming \(H_0\) is true. This involves calculating probabilities related to our observed data under the assumption that the population parameter equals the specific value stated in the null hypothesis.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_a\) or sometimes \(H_1\), is set up as the counterpart to the null hypothesis and suggests that there is a statistically significant effect or difference. It is what researchers typically want to prove. Unlike the null hypothesis, the alternative hypothesis encompasses all other possible outcomes apart from the one proposed by the null.

In many tests, the alternative hypothesis can take different forms depending on the nature of the test:
  • Two-tailed: Indicates that the parameter is not equal to a certain value (e.g., \(H_a: p eq 0.5\)).
  • One-tailed: Indicates that the parameter is either less than or greater than a particular value.
In the given exercise, for example, the alternative hypothesis is \(H_a: p eq 0.5\). This suggests that the researcher expects the population proportion to be different from 0.5, in either direction. Determining which hypothesis to accept depends on the evidence provided by the data in conjunction with statistical calculations.
P-Value Calculation
The p-value is a fundamental measure in hypothesis testing, providing a way to quantify the probability of observing data as extreme as the sample data, under the assumption that the null hypothesis is true. In simple terms, a smaller p-value indicates stronger evidence against the null hypothesis.

Here is how it works: after establishing a randomization distribution, technology like StatKey can simulate what the sampling distribution of the test statistic would look like if the null hypothesis were true. By using this simulation, the p-value is calculated as the proportion of samples that are at least as extreme as the observed sample.

The p-value helps decide whether to reject or fail to reject the null hypothesis. Typically, if the p-value is less than a pre-determined significance level (commonly set at 0.05), it suggests there is sufficient evidence to reject the null hypothesis in favor of the alternative. If the p-value is greater, then the data isn't enough to reject the null hypothesis. This inferential approach helps to make data-driven decisions based on probability.

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Most popular questions from this chapter

Exercises 4.117 to 4.122 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 \(n=200\)

Exercise Hours Introductory statistics students fill out a survey on the first day of class. One of the questions asked is "How many hours of exercise do you typically get each week?" Responses for a sample of 50 students are introduced in Example 3.25 on page 207 and stored in the file ExerciseHours. The summary statistics are shown in the computer output. The mean hours of exercise for the combined sample of 50 students is 10.6 hours per week and the standard deviation is 8.04 . We are interested in whether these sample data provide evidence that the mean number of hours of exercise per week is different between male and female statistics students. Variable Gender N Mean StDev Minimum Maximum \(\begin{array}{lllllll}\text { Exercise } & \text { F } 30 & 9.40 & 7.41 & 0.00 & 34.00\end{array}\) \(\begin{array}{llll}20 & 12.40 & 8.80 & 2,00\end{array}\) Discuss whether or not the methods described below would be appropriate ways to generate randomization samples that are consistent with \(H_{0}: \mu_{F}=\mu_{M}\) vs \(H_{a}: \mu_{F} \neq \mu_{M} .\) Explain your reasoning in each case. (a) Randomly label 30 of the actual exercise values with " \(\mathrm{F}^{\prime \prime}\) for the female group and the remaining 20 exercise values with " \(\mathrm{M} "\) for the males. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\) (b) Add 1.2 to every female exercise value to give a new mean of 10.6 and subtract 1.8 from each male exercise value to move their mean to 10.6 (and match the females). Sample 30 values (with replacement) from the shifted female values and 20 values (with replacement) from the shifted male values. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\) - (c) Combine all 50 sample values into one set of data having a mean amount of 10.6 hours. Select 30 values (with replacement) to represent a sample of female exercise hours and 20 values (also with replacement) for a sample of male exercise values. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\)

Divorce Opinions and Gender In Data 4.4 on page \(227,\) we introduce the results of a May 2010 Gallup poll of 1029 U.S. adults. When asked if they view divorce as "morally acceptable," \(71 \%\) of the men and \(67 \%\) of the women in the sample responded yes. In the test for a difference in proportions, a randomization distribution gives a p-value of \(0.165 .\) Does this indicate a significant difference between men and women in how they view divorce?

Exercises 4.117 to 4.122 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: \(H_{0}: p=0.6\) vs \(H_{a}: p>0.6\) Sample data: \(\hat{p}=52 / 80=0.65\) with \(n=80\)

In Exercises 4.146 to \(4.149,\) hypotheses for a statistical test are given, followed by several possible confidence intervals for different samples. In each case, use the confidence interval to state a conclusion of the test for that sample and give the significance level used. Hypotheses: \(H_{0}: \mu=15\) vs \(H_{a}: \mu \neq 15\) (a) 95\% confidence interval for \(\mu: \quad 13.9\) to 16.2 (b) 95\% confidence interval for \(\mu: \quad 12.7\) to 14.8 (c) 90\% confidence interval for \(\mu: \quad 13.5\) to 16.5

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