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Get P-value from \(z\) For a test of \(\mathrm{H}_{0}: p=0.50,\) the \(z\) test statistic equals 1.04 a. Find the P-value for \(\mathrm{H}_{a} \cdot p>0.50\). b. Find the P-value for \(\mathrm{H}_{a}: p \neq 0.50\). c. Find the P-value for \(\mathrm{H}_{c}: p<0.50 .\) (Hint: The P-values for the two possible one-sided tests must sum to \(1 .)\) d. Do any of the P-values in part a, part b, or part c give strong evidence against \(\mathrm{H}_{0}\) ? Explain.

Short Answer

Expert verified
a. 0.1492 b. 0.2984 c. 0.1492 None provide strong evidence against \( \mathrm{H}_{0} \).

Step by step solution

01

Understand the Set-up

We have the null hypothesis \( \mathrm{H}_{0}: p = 0.50 \) with a test statistic \( z = 1.04 \). We are using the standard normal distribution to find P-values corresponding to alternative hypotheses.
02

P-value for Upper-Tailed Test

For \( \mathrm{H}_{a}: p > 0.50 \), the P-value is the probability of observing a value as extreme or more extreme than \( z = 1.04 \). Therefore, find \( P(Z > 1.04) \) using the standard normal distribution table or a calculator.
03

Evaluate P-value from Z-table

From the standard normal distribution table, \( P(Z < 1.04) \) is approximately 0.8508. Therefore, \( P(Z > 1.04) = 1 - 0.8508 = 0.1492 \). So, the P-value is 0.1492.
04

P-value for Two-Tailed Test

For \( \mathrm{H}_{a}: p eq 0.50 \), the P-value is \( 2 \times P(Z > 1.04) \) because it’s a two-tailed test. Calculate the P-value as \( 2 \times 0.1492 = 0.2984 \).
05

P-value for Lower-Tailed Test

For \( \mathrm{H}_{c}: p < 0.50 \), we have \( P(Z < -1.04) = P(Z > 1.04) = 0.1492 \), from symmetry of the standard normal distribution.
06

Assess Evidence Against Null Hypothesis

A P-value is typically considered significant if it is below 0.05 or 0.01. The smallest P-value obtained here is 0.1492, which is much larger than 0.05. Therefore, none of these P-values provide strong evidence against \( \mathrm{H}_{0} \).

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Key Concepts

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

Z-test
The Z-test is a type of statistical test used to determine if there is a significant difference between sample and population means. It is especially useful when dealing with large sample sizes. The Z-test relies on the Z-statistic, which is calculated to compare the sample mean to the population mean while considering the variance and size of the sample.

When performing a Z-test, you use the standard normal distribution to determine probabilities. This involves calculating a Z-score, which tells you how many standard deviations your test statistic is from the mean of the distribution. This is crucial as the Z-score allows you to find the probability, or P-value, associated with the observed statistic.

**Key Points:**
  • Used when the sample size is large, typically larger than 30.
  • Requires knowledge of the population variance.
  • Useful for hypothesis testing related to means.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement that indicates no effect or no difference in a statistical test. It is the default or initial assumption about a population parameter, such as the mean or proportion. In hypothesis testing, the null hypothesis represents the status quo or the condition that any observed difference in the data is merely due to chance.

The goal of hypothesis testing is to determine whether the observed data provides enough evidence to reject the null hypothesis. Testing \( H_0 \) often involves comparing results to a significance level, commonly set at 0.05, to decide if the evidence is sufficient to question or reject it.

**Characteristics of Null Hypothesis:**
  • Symbolized as \( H_0 \).
  • Assumes no change or effect — a baseline assumption.
  • Tested directly through statistical analysis.
Standard Normal Distribution
The standard normal distribution is a special normal distribution with a mean of 0 and a standard deviation of 1. All normal distributions can be transformed to a standard normal distribution by converting scores into Z-scores, which measures the number of standard deviations an element is from the mean.

This transformation is beneficial because it allows for the use of standard normal distribution tables or calculators to find probabilities and p-values. By standardizing data, any normal distribution question becomes a question about the standard normal distribution, simplifying the process of hypothesis testing.

**Advantages of Standard Normal Distribution:**
  • Simplifies computations by having fixed parameters (mean = 0, sd = 1).
  • Enables easy probability calculations using Z-scores.
  • Facilitates comparisons between different datasets and hypotheses.
Alternative Hypotheses
Alternative hypotheses, denoted as \( H_a \) or \( H_1 \), represent what you aim to support or show evidence for through statistical testing. These hypotheses propose that there is an actual effect or difference compared to the null hypothesis, which assumes no difference or effect.

**Types of Alternative Hypotheses:**
  • **One-tailed test (right-tailed):** Suggests an increase (e.g., \( H_a: p > 0.50 \)).
  • **One-tailed test (left-tailed):** Suggests a decrease (e.g., \( H_a: p < 0.50 \)).
  • **Two-tailed test:** Suggests a difference but not the direction (e.g., \( H_a: p eq 0.50 \)).
Each type of alternative hypothesis has implications on how the test is conducted and which regions of the probability distribution are considered when calculating the P-value. Carefully selecting an alternative hypothesis based on the research question is crucial for accurate and relevant results in hypothesis testing.

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Most popular questions from this chapter

A binomial headache A null hypothesis states that the population proportion \(p\) of headache sufferers who have more pain relief with aspirin than with another pain reliever equals \(0.50 .\) For a crossover study with 10 subjects, all 10 have more relief with aspirin. If the null hypothesis were true, by the binomial distribution the probability of this sample result equals \((0.50)^{10}=0.001\). In fact, this is the small-sample P-value for testing \(\mathrm{H}_{0}: p=0.50\) against \(\mathrm{H}_{a}: p>0.50 .\) Does this P-value give (a) strong evidence in favor of \(\mathrm{H}_{0}\) or \((\mathrm{b})\) strong evidence against \(\mathrm{H}_{0}\) ? Explain why.

\(\mathbf{H}_{0}\) or \(\mathbf{H}_{a}\) ? For parts a and \(\mathrm{b},\) is the statement a null hypothesis, or an alternative hypothesis? a. In Canada, the proportion of adults who favor legalized gambling equals \(0.50 .\) b. The proportion of all Canadian college students who are regular smokers is less than 0.24 , the value it was 10 years ago. c. Introducing notation for a parameter, state the hypotheses in parts a and \(\mathrm{b}\) in terms of the parameter values.

\(\mathbf{H}_{0}\) or \(\mathbf{H}_{a}\) ? For each of the following, is the statement a null hypothesis or an alternative hypothesis? Why? a. The mean IQ of all students at Lake Wobegon High School is larger than 100 . b. The probability of rolling a 6 with a particular die equals \(1 / 6\). c. The proportion of all new business enterprises that remain in business for at least five years is less than 0.50 .

P-value Indicate whether each of the following P-values gives strong evidence or not especially strong evidence against the null hypothesis. a. 0.38 b. 0.001

Fishing for significance A marketing study conducts 60 significance tests about means and proportions for several groups. Of them, 3 tests are statistically significant at the 0.05 level. The study's final report stresses only the tests with significant results, not mentioning the other 57 tests. What is misleading about this?

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