/*! 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 158 Hypotheses for a statistical tes... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 158

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: \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) (a) 95\% confidence interval for \(p: \quad 0.53\) to 0.57 (b) \(95 \%\) confidence interval for \(p: \quad 0.41\) to 0.52 (c) 99\% confidence interval for \(p: \quad 0.35\) to 0.55

Short Answer

Expert verified
For part (a), the null hypothesis \(H_{0}: p=0.5\) is rejected at a 95% confidence level because the interval does not include the hypothesis value. For parts (b) and (c), we do not reject the null hypothesis at a 95% and 99% confidence levels respectively because the confidence intervals include the value of the null hypothesis.

Step by step solution

01

Analyzing Confidence Interval (a)

The given 95% confidence interval for \(p\) is from 0.53 to 0.57. The value for \(H_{0}\) which is 0.5 is not lying in this range. Therefore, the null hypothesis of \(p = 0.5\) is rejected, concluding \(H_{a}: p \neq 0.5\).
02

Analyzing Confidence Interval (b)

The given 95% confidence interval for \(p\) is from 0.41 to 0.52. Here, the hypothesis value \(p = 0.5\) lies within the interval. Therefore, we do not reject the null hypothesis, implying there's not enough evidence to conclude \(H_{a}: p \neq 0.5\).
03

Analyzing Confidence Interval (c)

The given 99% confidence interval for \(p\) is from 0.35 to 0.55. The value \(p = 0.5\) lies within this interval. Therefore, we do not reject the null hypothesis in this case as well. Again, we cannot conclude \(H_{a}: p \neq 0.5\).

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 91影视!

Key Concepts

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

Confidence Intervals
Confidence intervals are crucial in statistical hypothesis testing. They provide a range of values, derived from a sample, likely to contain the population parameter. The main purpose of confidence intervals is to give an estimate of uncertainty. For instance, if we have a 95% confidence interval, it suggests that if we were to take many samples and build a confidence interval from each one, about 95% of these intervals would contain the true parameter.
In this exercise, several confidence intervals are analyzed:
  • Interval (a) is from 0.53 to 0.57. Since the hypothesized value 0.5 does not lie within, we reject the null hypothesis.
  • Interval (b) is from 0.41 to 0.52. Here, 0.5 is inside, so we fail to reject the null hypothesis.
  • Interval (c) expands even more, from 0.35 to 0.55, and also contains 0.5, leading to the conclusion we cannot reject the null hypothesis.
Different confidence intervals provide distinct insights about the data, based on sample size and variability.
Null Hypothesis
In statistical hypothesis testing, the null hypothesis (denoted as \(H_{0}\)) is a statement we aim to test. It usually represents a default position that there is no effect or difference, and it suggests any observation is due to chance.
In our scenario, the null hypothesis \(H_{0}: p=0.5\) proposes that the probability \(p\) of a certain outcome is 0.5. This is a neutral claim suggesting no deviation from an expected standard. The goal of testing the null hypothesis is to determine if there is enough statistical evidence to reject it in favor of an alternative hypothesis (\(H_{a}\)), which proposes a change or effect. Here, \(H_{a}: p eq 0.5\) implies that the probability is either greater or less than 0.5.
Assessing whether to reject \(H_{0}\) requires examining if the hypothesized value falls within the given confidence interval. If it does fall within, as it did in examples (b) and (c), it means there is not enough evidence against \(H_{0}\).
Significance Level
The significance level is a threshold that defines the probability of rejecting the null hypothesis when it is actually true. Often denoted as \(\alpha\), this value is typically set before conducting the analysis. Common significance levels include 0.05, 0.01, etc. These correspond to 5% and 1% chances, respectively, of making a Type I error, which is the mistake of rejecting a true null hypothesis.
In the context of our exercise:
  • Using a 95% confidence interval translates to a significance level of 0.05. This means we accept a 5% risk of claiming a difference when there is none.
  • For the 99% confidence interval, the significance level is 0.01, indicating just a 1% risk of error.
The choice of significance level can influence the decision to reject or fail to reject the null hypothesis. A smaller \(\alpha\), like in the 99% confidence interval example, implies a stricter standard for rejecting \(H_{0}\). As shown, 0.5 was not outside the interval in cases (b) and (c), meaning we do not have sufficient evidence to reject \(H_{0}\) at the given significance levels.

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

By some accounts, the first formal hypothesis test to use statistics involved the claim of a lady tasting tea. \({ }^{11}\) In the 1920 's Muriel Bristol- Roach, a British biological scientist, was at a tea party where she claimed to be able to tell whether milk was poured into a cup before or after the tea. R.A. Fisher, an eminent statistician, was also attending the party. As a natural skeptic, Fisher assumed that Muriel had no ability to distinguish whether the milk or tea was poured first, and decided to test her claim. An experiment was designed in which Muriel would be presented with some cups of tea with the milk poured first, and some cups with the tea poured first. (a) In plain English (no symbols), describe the null and alternative hypotheses for this scenario. (b) Let \(p\) be the true proportion of times Muriel can guess correctly. State the null and alternative hypothesis in terms of \(p\).

Numerous studies have shown that a high fat diet can have a negative effect on a child's health. A new study \(^{22}\) suggests that a high fat diet early in life might also have a significant effect on memory and spatial ability. In the double-blind study, young rats were randomly assigned to either a high-fat diet group or to a control group. After 12 weeks on the diets, the rats were given tests of their spatial memory. The article states that "spatial memory was significantly impaired" for the high-fat diet rats, and also tells us that "there were no significant differences in amount of time exploring objects" between the two groups. The p-values for the two tests are 0.0001 and 0.7 . (a) Which p-value goes with the test of spatial memory? Which p-value goes with the test of time exploring objects? (b) The title of the article describing the study states "A high-fat diet causes impairment" in spatial memory. Is the wording in the title justified (for rats)? Why or why not?

In this exercise, we see that it is possible to use counts instead of proportions in testing a categorical variable. Data 4.7 describes an experiment to investigate the effectiveness of the two drugs desipramine and lithium in the treatment of cocaine addiction. The results of the study are summarized in Table 4.14 on page \(323 .\) The comparison of lithium to the placebo is the subject of Example 4.34 . In this exercise, we test the success of desipramine against a placebo using a different statistic than that used in Example 4.34. Let \(p_{d}\) and \(p_{c}\) be the proportion of patients who relapse in the desipramine group and the control group, respectively. We are testing whether desipramine has a lower relapse rate then a placebo. (a) What are the null and alternative hypotheses? (b) From Table 4.14 we see that 20 of the 24 placebo patients relapsed, while 10 of the 24 desipramine patients relapsed. The observed difference in relapses for our sample is $$\begin{aligned}D &=\text { desipramine relapses }-\text { placebo relapses } \\\&=10-20=-10\end{aligned}$$ If we use this difference in number of relapses as our sample statistic, where will the randomization distribution be centered? Why? (c) If the null hypothesis is true (and desipramine has no effect beyond a placebo), we imagine that the 48 patients have the same relapse behavior regardless of which group they are in. We create the randomization distribution by simulating lots of random assignments of patients to the two groups and computing the difference in number of desipramine minus placebo relapses for each assignment. Describe how you could use index cards to create one simulated sample. How many cards do you need? What will you put on them? What will you do with them?

Do iPads Help Kindergartners Learn: A Series of Tests Exercise 4.147 introduces a study in which half of the kindergarten classes in a school district are randomly assigned to receive iPads. We learn that the results are significant at the \(5 \%\) level (the mean for the iPad group is significantly higher than for the control group) for the results on the HRSIW subtest. In fact, the HRSIW subtest was one of 10 subtests and the results were not significant for the other 9 tests. Explain, using the problem of multiple tests, why we might want to hesitate before we run out to buy iPads for all kindergartners based on the results of this study.

4.151 Does Massage Really Help Reduce Inflammation in Muscles? In Exercise 4.112 on page \(301,\) we learn that massage helps reduce levels of the inflammatory cytokine interleukin-6 in muscles when muscle tissue is tested 2.5 hours after massage. The results were significant at the \(5 \%\) level. However, the authors of the study actually performed 42 different tests: They tested for significance with 21 different compounds in muscles and at two different times (right after the massage and 2.5 hours after). (a) Given this new information, should we have less confidence in the one result described in the earlier exercise? Why? (b) Sixteen of the tests done by the authors involved measuring the effects of massage on muscle metabolites. None of these tests were significant. Do you think massage affects muscle metabolites? (c) Eight of the tests done by the authors (including the one described in the earlier exercise) involved measuring the effects of massage on inflammation in the muscle. Four of these tests were significant. Do you think it is safe to conclude that massage really does reduce inflammation?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.