/*! 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 1 Suppose that eight sets of hypot... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that eight sets of hypotheses about a population proportion of the form $$H_{0}: p=0.3 \quad H_{1}: p>0.3$$ have been tested and that the \(P\) -values for these tests are 0.15 , \(0.83,0.103,0.024,0.03,0.07,0.09,\) and 0.13. Use Fisher's procedure to combine all of these \(P\) -values. Is there sufficient evidence to conclude that the population proportion exceeds \(0.30 ?\)

Short Answer

Expert verified
Combine using Fisher's method, qnd chi-square with 16 df to check significance.

Step by step solution

01

Understanding Fisher's Method

Fisher's method is used to combine multiple \(P\)-values from independent tests to determine the overall significance. If \(P_1, P_2, ..., P_m\) are the \(P\)-values from \(m\) independent tests, Fisher's method combines them using the test statistic \(X^2 = -2 \sum_{i=1}^{m} \ln(P_i)\), which follows a chi-squared distribution with \(2m\) degrees of freedom under the null hypothesis.
02

Calculate Chi-Square Statistic

First, calculate \(-2 \ln(P_i)\) for each \(P\)-value given: \(0.15, 0.83, 0.103, 0.024, 0.03, 0.07, 0.09, 0.13\). Then sum these values to find \(X^2\).
03

Compute Individual Contributions

Calculate contributions: \(-2 \ln(0.15) \approx 3.91\)\(-2 \ln(0.83) \approx 0.38\)\(-2 \ln(0.103) \approx 4.54\)\(-2 \ln(0.024) \approx 7.42\)\(-2 \ln(0.03) \approx 7.00\)\(-2 \ln(0.07) \approx 5.41\)\(-2 \ln(0.09) \approx 4.79\)\(-2 \ln(0.13) \approx 4.33\).
04

Sum Contributions for Total Statistic

Add the contributions from each \(P\)-value: \[3.91 + 0.38 + 4.54 + 7.42 + 7.00 + 5.41 + 4.79 + 4.33 = 37.78\].

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.

Hypothesis Testing
Hypothesis testing is a crucial statistical tool used to evaluate two competing statements about a population parameter. It starts with setting up two opposing hypotheses:
  • Null hypothesis (denoted as \(H_0\)): A statement of no effect or no difference, usually involving equality.
  • Alternative hypothesis (denoted as \(H_1\)): A statement indicating the presence of an effect or difference, often suggesting greater or lesser values.
In the scenario of testing population proportions, one might want to assess whether a proportion exceeds a particular value. For example, we could have \(H_0: p = 0.3\) and \(H_1: p > 0.3\), where \(p\) represents the population proportion.
The hypothesis test involves calculating a test statistic and comparing it with a critical value to make a decision. The decision could lead to either rejecting the null hypothesis (suggesting the alternative might be true) or failing to reject it if the evidence isn't strong enough.
Chi-Squared Distribution
The chi-squared distribution is a significant concept in statistical testing, particularly tied to tests like Fisher's method. It is a distribution that arises when we sum the squares of independent standard normal random variables. The shape of the chi-squared distribution is determined by the degrees of freedom, which can affect its variance and skewness.
In Fisher’s method for combining \(P\)-values, the resulting test statistic follows a chi-squared distribution. Specifically, if you have \(m\) independent tests, the test statistic calculated (\(X^2 = -2 \sum \ln(P_i)\)) follows a chi-squared distribution with \(2m\) degrees of freedom. This allows one to find the overall significance from multiple tests by referencing chi-squared tables to find critical values.
Understanding this distribution's properties is essential for correctly interpreting test results and assessing statistical significance.
Population Proportion
The population proportion is a parameter central to many statistical studies and represents the fraction of the population that holds a particular characteristic. For example, if you're studying a population where 30% are left-handed, \(p = 0.3\) would be the population proportion.
In hypothesis testing, we often test claims about a population proportion. This involves setting up hypotheses around a claimed proportion and utilizing sample data to assess these claims. For the hypotheses \(H_0: p = 0.3\) vs \(H_1: p > 0.3\), you're assessing whether there is enough statistical evidence from your sample to conclude that the true population proportion is greater than 0.3.
Calculating population proportions may involve estimations from sample data, and these estimates are subjected to statistical tests to reach conclusions about the entire population.
P-values
P-values are fundamental to hypothesis testing, quantifying the strength of evidence against the null hypothesis. A \(P\)-value indicates the probability of observing the test results, or more extreme results, given that the null hypothesis is true.
  • If the \(P\)-value is very low, it suggests that the observed result is not very likely under the null hypothesis, thereby providing evidence against it.
  • Conversely, a high \(P\)-value suggests that the null hypothesis could be true, or there's not enough evidence to refute it.
In the context of multiple testing, combining \(P\)-values from several independent tests, like with Fisher’s method, provides a way to understand the collective evidence from all tests together. Each \(P\)-value contributes to a chi-squared test statistic, which can be evaluated to determine overall significance.
Statistical Significance
Statistical significance is a measure to determine whether the results observed in your study are likely to have occurred through random chance alone. When a test result is statistically significant, it means there's strong evidence against the null hypothesis and in favor of the alternative hypothesis.
In hypothesis testing, one often sets a predetermined threshold (alpha level, commonly 0.05 or 0.01). Results below this threshold indicate statistical significance, leading to rejecting the null hypothesis.
Fisher’s method combines multiple related tests, so statistical significance achieved through this method reflects the combined strength of evidence across all tests. Calculating and interpreting statistical significance empowers researchers to make informed decisions about their hypotheses and understand the broader implications of their 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

For the hypothesis test \(H_{0}: \mu=7\) against \(H_{1}: \mu \neq 7\) with variance unknown and \(n=20,\) approximate the \(P\) -value for each of the following test statistics. a. \(t_{0}=2.05\) b. \(t_{0}=-1.84\) c. \(t_{0}=0.4\)

The proportion of residents in Phoenix favoring the building of toll roads to complete the freeway system is believed to be \(p=0.3 .\) If a random sample of 10 residents shows that one or fewer favor this proposal, we will conclude that \(p<0.3\). a. Find the probability of type I error if the true proportion is \(p=0.3\). b. Find the probability of committing a type II error with this procedure if \(p=0.2\). c. What is the power of this procedure if the true proportion is \(p=0.2 ?\)

The mean bond strength of a cement product must be at least 10000 psi. The process by which this material is manufactured must show equivalence to this standard. If the process can manufacture cement for which the mean bond strength is at least 9750 psi, it will be considered equivalent to the standard. A random sample of six observations is available, and the sample mean and standard deviation of bond strength are 9360 psi and 42.6 psi, respectively. a. State the appropriate hypotheses that must be tested to demonstrate equivalence. b. What are your conclusions using \(\alpha=0.05 ?\)

Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once it is implanted in the patient's body, but the battery pack needs to be recharged about every 4 hours. A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume that battery life is normally distributed with standard deviation \(\sigma=0.2\) hour. a. Is there evidence to support the claim that mean battery life exceeds 4 hours? Use \(\alpha=0.05 .\) b. What is the \(P\) -value for the test in part (a)? c. Compute the power of the test if the true mean battery life is 4.5 hours. d. What sample size would be required to detect a true mean battery life of 4.5 hours if you wanted the power of the test to be at least \(0.9 ?\) e. Explain how the question in part (a) could be answered by constructing a one-sided confidence bound on the mean life.

In a little over a month, from June 5,1879 , to July 2 , 1879 , Albert Michelson measured the velocity of light in air 100 times (Stigler, Annals of Statistics, 1977 ). Today we know that the true value is \(299,734.5 \mathrm{~km} / \mathrm{sec} .\) Michelson's data have a mean of \(299,852.4 \mathrm{~km} / \mathrm{sec}\) with a standard deviation of \(79.01 .\) a. Find a two-sided \(95 \%\) confidence interval for the true mean (the true value of the speed of light). b. What does the confidence interval say about the accuracy of Michelson's measurements?

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.