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

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 \(H_{a}: p<0.5\) Sample data: \(\hat{p}=38 / 100=0.38\) with \(n=100\)

Short Answer

Expert verified
The null and alternative hypotheses are \(H_{0}: p = 0.5\) and \(H_{a}: p < 0.5\) respectively. The sample data provides a sample proportion of \(\hat{p}=38/100 = 0.38\) based on a sample size \(n=100\). This data can be used on statistical software like StatKey to generate a random distribution and compute the p-value.

Step by step solution

01

Hypothesis Formulation

The null hypothesis \(H_{0}\) posits that the population proportion \(p = 0.5\). The alternative hypothesis \(H_{a}\) suggests that the population proportion \(p < 0.5\). It's essential to remember that the null hypothesis often includes an equals sign.
02

Sample Data Interpretation

Our sample proportion is given by \(\hat{p}=38/100 = 0.38\). This is based on a sample size \(n=100\). The sample implies that in 100 trials, the predicted scenario occurred 38 times.
03

Use of Statistical Software

With our hypotheses and sample data, we can now use StatKey or compatible software to conduct our tests. Select 'Test for a Single Proportion' then 'Edit Data' where you insert your sample data.
04

Calculation of p-value

Using the software, generate a random distribution. The p-value can be subsequently calculated. The p-value is the chance of obtaining our observed sample data (or more extreme) assuming that the null hypothesis is true. It measures the strength of evidence in support of a null hypothesis.

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.

Null and Alternative Hypotheses
Understanding null and alternative hypotheses is a cornerstone of hypothesis testing. In any experimental setup, a null hypothesis (\(H_{0}\)) is a statement that there is no effect or no difference, and it specifies a population proportion that we assume to be true before collecting any data. In contrast, the alternative hypothesis (\(H_{a}\) or \(H_{1}\) ) is a statement that indicates what we suspect might be true instead.

Using the example from the exercise, the null hypothesis is \(H_{0}: p=0.5\) suggesting no difference from the expected value. The alternative hypothesis, \(H_{a}: p<0.5\) in this case, suggests that the actual population proportion is less than what we initially believed. Accurately formulating these hypotheses is essential for analyzing the direction and the robustness of the evidence provided by the sample data.
Population Proportion
The term population proportion, denoted by \(p\) indicates the likelihood of a particular outcome within a population. In hypothesis testing, we use the sample proportion (\(\hat{p}\) as an estimate of the true population proportion based on the sample data. The sample proportion is calculated by dividing the number of times the event of interest occurs by the total number of trials.

In the context of the textbook exercise, the sample proportion was calculated to be \(\hat{p}=38/100=0.38\) from a sample size of 100. This estimate plays a pivotal role in testing our hypotheses and determining the statistical significance of our findings.
P-value Calculation
P-value calculation is a crucial step in hypothesis testing that quantifies the evidence against the null hypothesis. The p-value is the probability of observing a sample statistic as extreme or more extreme than the one calculated from the sample data, given that the null hypothesis is true.

The smaller the p-value, the stronger the evidence against the null hypothesis. A commonly used threshold for significance is \(p < 0.05\) which indicates that if the null hypothesis were true, there would be less than a 5% chance of observing a result as extreme as the sample data. Calculating the p-value allows researchers to make an informed decision on whether to reject the null hypothesis in favor of the alternative.
StatKey Software
To streamline the process of hypothesis testing, tools like StatKey software are invaluable. StatKey is designed to conduct statistical tests, including hypothesis testing for a single proportion, and create the relevant distributions for analysis. The user-friendly interface allows for easy entry of sample data, and it automates the randomization process and the calculation of the p-value.

Following the steps in the provided exercise, StatKey simplifies these processes by allowing the user to input their sample information directly and obtain results that guide decision-making in statistical testing. By promptly generating a randomization distribution and the associated p-value, the software aids in substantiating the statistical reasoning.
Randomization Distribution
A randomization distribution is a representation of what the distribution of a sample statistic (like a sample proportion) would look like, assuming the null hypothesis were true. This distribution is generated by randomly reassigning outcomes among the observations in the sample and calculating the statistic for many randomized samples.

The central purpose of observing a randomization distribution is to determine how unusual the observed statistic is. The extremity of the sample data is assessed in the context of the distribution, which culminates in the calculation of the p-value. By comparing the sample statistic to this distribution, we essentially determine the likelihood of obtaining our sample or more extreme results, should the null hypothesis hold true. The randomization test is a non-parametric method of hypothesis testing which does not rely on underlying population distributions, making it versatile and robust for a range of data.

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

We are conducting many hypothesis tests to test a claim. In every case, assume that the null hypothesis is true. Approximately how many of the tests will incorrectly find significance? 100 tests conducted using a significance level of \(5 \%\).

An article noted that it may be possible to accurately predict which way a penalty-shot kicker in soccer will direct his shot. \({ }^{27}\) The study finds that certain types of body language by a soccer player \(-\) called "tells"-can be accurately read to predict whether the ball will go left or right. For a given body movement leading up to the kick, the question is whether there is strong evidence that the proportion of kicks that go right is significantly different from one-half. (a) What are the null and alternative hypotheses in this situation? (b) If sample results for one type of body movement give a p-value of 0.3184 , what is the conclusion of the test? Should a goalie learn to distinguish this movement? (c) If sample results for a different type of body movement give a p-value of \(0.0006,\) what is the conclusion of the test? Should a goalie learn to distinguish this movement?

The data in Hurricanes contains the number of hurricanes that made landfall on the eastern coast of the United States over the 101 years from 1914 to 2014 . Suppose we are interested in testing whether the number of hurricanes is increasing over time. (a) State the null and alternative hypotheses for testing whether the correlation between year and number of hurricanes is positive, which would indicate the number of hurricanes is increasing. (b) Describe in detail how you would create a randomization distribution to test this claim (if you had many more hours to do this exercise and no access to technology).

Mating Choice and Offspring Fitness: MiniExperiments Exercise 4.153 explores the question of whether mate choice improves offspring fitness in fruit flies, and describes two seemingly identical experiments yielding conflicting results (one significant, one insignificant). In fact, the second source was actually a series of three different experiments, and each full experiment was comprised of 50 different mini-experiments (runs), 10 each on five different days. (a) Suppose each of the 50 mini-experiments from the first study were analyzed individually. If mating choice has no impact on offspring fitness, about how many of these \(50 \mathrm{p}\) -values would you expect to yield significant results at \(\alpha=0.05 ?\) (b) The 50 p-values, testing the alternative \(H_{a}\) : \(p_{C}>p_{N C}\) (proportion of flies surviving is higher in the mate choice group) are given below: $$ \begin{array}{lllllllllll} \text { Day 1: } & 0.96 & 0.85 & 0.14 & 0.54 & 0.76 & 0.98 & 0.33 & 0.84 & 0.21 & 0.89 \\ \text { Day 2: } & 0.89 & 0.66 & 0.67 & 0.88 & 1.00 & 0.01 & 1.00 & 0.77 & 0.95 & 0.27 \\ \text { Day 3: } & 0.58 & 0.11 & 0.02 & 0.00 & 0.62 & 0.01 & 0.79 & 0.08 & 0.96 & 0.00 \\ \text { Day 4: } & 0.89 & 0.13 & 0.34 & 0.18 & 0.11 & 0.66 & 0.01 & 0.31 & 0.69 & 0.19 \\ \text { Day 5: } & 0.42 & 0.06 & 0.31 & 0.24 & 0.24 & 0.16 & 0.17 & 0.03 & 0.02 & 0.11 \end{array} $$ How many are actually significant using \(\alpha=0.05 ?\) (c) You may notice that two p-values (the fourth and last run on day 3 ) are 0.00 when rounded to two decimal places. The second of these is actually 0.0001 if we report more decimal places. This is very significant! Would it be appropriate and/or ethical to just report this one run, yielding highly statistically significant evidence that mate choice improves offspring fitness? Explain. (d) You may also notice that two of the p-values on day 2 are 1 (rounded to two decimal places). If we had been testing the opposite alternative, \(H_{a}:\) \(p_{C}

It is believed that sunlight offers some protection against multiple sclerosis (MS) since the disease is rare near the equator and more prevalent at high latitudes. What is it about sunlight that offers this protection? To find out, researchers \({ }^{23}\) injected mice with proteins that induce a condition in mice comparable to MS in humans. The control mice got only the injection, while a second group of mice were exposed to UV light before and after the injection, and a third group of mice received vitamin D supplements before and after the injection. In the test comparing UV light to the control group, evidence was found that the mice exposed to UV suppressed the MS-like disease significantly better than the control mice. In the test comparing mice getting vitamin D supplements to the control group, the mice given the vitamin D did not fare significantly better than the control group. If the p-values for the two tests are 0.472 and 0.002 , which p-value goes with which test?

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