/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 Give the null and alternative hy... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Give the null and alternative hypotheses for each test, and state whether a one-proportion z-test or a two-proportion z-test would be appropriate. a. You test a person to see whether he can tell tap water from bottled water. You give him 20 sips selected randomly (half from tap water and half from bottled water) and record the proportion he gets correct to test the hypothesis. b. You test a random sample of students at your college who stand on one foot with their eyes closed and determine who can stand for at least 10 seconds, comparing athletes and nonathletes.

Short Answer

Expert verified
a. The null hypothesis (H0) is 'The person can't differentiate between tap water and bottled water', and the alternative hypothesis (Ha) is 'The person can differentiate between tap water and bottled water'. This is a one-proportion z-test scenario. b. The null hypothesis (H0) is 'There is no difference between athletes and non-athletes ability to stand on one foot for at least 10 seconds' and the alternative hypothesis (Ha) is 'There is a difference between the athletes' and non-athletes' ability to stand on one foot for at least 10 seconds'. This is a two-proportion z-test scenario.

Step by step solution

01

Task A: Determine the Hypotheses

We are testing one person's ability to differentiate between bottled and tap water. The null hypothesis, \(H_0\), could be: 'The person can't differentiate between tap water and bottled water', and the alternative hypothesis, \(H_a\), 'The person can differentiate between tap water and bottled water'. Since we're dealing with a single proportion - the success rate of the person being tested - this is a one-proportion z-test.
02

Task B: Determine the Hypotheses

We are comparing the abilities of two groups of students - athletes and non-athletes - to maintain their balance. The null hypothesis, \(H_0\), is: 'There is no difference between athletes and non-athletes ability to stand on one foot for at least 10 seconds'. The alternative hypothesis, \(H_a\), 'There is a difference between the athletes' and non-athletes' ability to stand on one foot for at least 10 seconds'. As we are comparing two proportions - the proportion of athletes who can balance for 10 seconds and the proportion of nonathletes who can do the same - this is a two-proportion z-test.

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 91Ó°ÊÓ!

Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis, denoted as \(H_0\), serves as the starting assumption that there is no effect or no difference in the context of the test. It's a statement that any kind of difference or significance you observe in your data is due to chance. This hypothesis is crucial as it sets a baseline for testing, and it's what we aim to test against.

For example, in the exercise where the person is tested to differentiate between tap and bottled water, the null hypothesis is: "The person can't differentiate between tap water and bottled water." This means we assume initially that the person's ability to distinguish the two types of water is no better than random guessing.

The null hypothesis is often associated with equality such as \(H_0: p = p_0\), where \(p_0\) is a specified value, demonstrating no effect or no difference.
Alternative Hypothesis
The alternative hypothesis, represented as \(H_a\), is what you want to prove. It's the statement that indicates the presence of an effect or a difference. If the statistical test shows evidence against the null hypothesis, we might consider the alternative hypothesis as more plausible.

In our bottled versus tap water example, the alternative hypothesis is stated as: "The person can differentiate between tap water and bottled water." This suggests that the individual has some proficiency that is better than merely guessing.

The alternative hypothesis is formulated as being different from what is stated in the null. It can be either one-sided or two-sided, such as \(H_a: p eq p_0\), \(H_a: p > p_0\), or \(H_a: p < p_0\), depending on what you aim to test.
One-Proportion Z-Test
A one-proportion z-test is used when you want to test hypothesis about a single sample proportion. This test is appropriate when you have one group or sample, and you're testing whether the proportion of a certain characteristic in this sample aligns with a known value or different from a hypothesized value.

In our water taste test scenario, we are dealing with one person and we want to evaluate their success in distinguishing between the two types of water. The success rate, in this case, would be compared to an expected value if decisions were made randomly. The objective is to determine whether their success rate is significantly different from this random expectation.

A critical assumption here is that the sample size should be large enough for the normal approximation to hold. The test statistic in a one-proportion z-test is given by:\[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]where \(\hat{p}\) is the sample proportion, \(p_0\) is the population proportion under null hypothesis and \(n\) is the sample size.
Two-Proportion Z-Test
When you need to compare proportions from two independent groups or samples, a two-proportion z-test is the right tool for assessing differences. It's designed to help you decide if there are significant differences between the two groups through their sample proportions.

In the exercise involving athletes and non-athletes trying to stand on one foot, we are comparing two distinct groups. We're interested in understanding if the ability to balance for a set time differs significantly between athletes and non-athletes. To find this difference, you need to calculate and compare their proportions:
  • The proportion of athletes who can stand for 10 seconds.
  • The proportion of non-athletes who can successfully stand for the same duration.
The test statistic for the two-proportion z-test is calculated as follows:\[z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}}\]where \(\hat{p}_1\) and \(\hat{p}_2\) are the sample proportions from the first and second group, \(\hat{p}\) is the pooled sample proportion, and \(n_1\) and \(n_2\) are the sample sizes respectively. This test helps ascertain if the observed differences in proportions are statistically significant.

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

In a 2018 study reported in The Lancet, Molina et al. reported on a study for treatment of patients with HIV-1. The study was a randomized, controlled, double-blind study that compared the effectiveness of ritonavir-boosted darunavir (rbd), the drug currently used to treat HIV-1, with dorovirine, a newly developed drug. Of the 382 subjects taking ritonavir-boosted darunavir, 306 achieved a positive result. Of the 382 subjects taking dorovirine, 321 achieved a positive outcome. See page 430 for guidance. a. Find the sample percentage of subjects who achieved a positive outcome in each group. b. Perform a hypothesis test to test whether the proportion of patients who achieve a positive outcome with the current treatment (ritonavir-boosted darunavir) is different from the proportion of patients who achieve a positive outcome with the new treatment (dorovirine). Use a significance level of \(0.01\). Based on this study, do you think dorovirine might be a more effective treatment option for HIV-1 than ritonavir-boosted darunavir? Why or why not?

Choose one of the answers in each case. In statistical inference, measurements are made on a ______ (sample or population), and generalizations are made to a _____ (sample or population).

Votes for Independents Judging on the basis of experience, a politician claims that \(50 \%\) of voters in Pennsylvania have voted for an independent candidate in past elections. Suppose you surveyed 20 randomly selected people in Pennsylvania, and 12 of them reported having voted for an independent candidate. The null hypothesis is that the overall proportion of voters in Pennsylvania that have voted for an independent candidate is \(50 \%\). What value of the test statistic should you report?

A hospital readmission is an episode when a patient who has been discharged from a hospital is readmitted again within a certain period. Nationally the readmission rate for patients with pneumonia is \(17 \% .\) A hospital was interested in knowing whether their readmission rate for pneumonia was less than the national percentage. They found 11 patients out of 70 treated for pneumonia in a two-month period were readmitted. a. What is \(\hat{p}\), the sample proportion of readmission? b. Write the null and alternative hypotheses. c. Find the value of the test statistic and explain it in context. d. The p-value associated with this test statistic is \(0.39 .\) Explain the meaning of the \(\mathrm{p}\) -value in this context. Based on this result, does the \(\mathrm{p}\) -value indicate the null hypothesis should be doubted?

If we do not reject the null hypothesis, is it valid to say that we accept the null hypothesis? Why or why not?

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.