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For a test of \(\mathrm{H}_{0}: p=0.50\), the sample proportion is 0.35 based on a sample size of 100 . a. Show that the test statistic is \(z=-3.0\). b. Find the \(P\) -value for \(H_{a}: p<0.50\). c. Does the \(P\) -value in part b give much evidence against \(\mathrm{H}_{0} ?\) Explain.

Short Answer

Expert verified
The test statistic is \( z = -3.0 \). The \( P \)-value is approximately 0.0013, providing strong evidence against \( H_0 \).

Step by step solution

01

State the Hypotheses

The null hypothesis is \( H_0: p = 0.50 \) and the alternative hypothesis is \( H_a: p < 0.50 \).
02

Identify Parameters

The sample proportion is \( \hat{p} = 0.35 \), the population proportion under the null hypothesis is \( p_0 = 0.50 \), and the sample size is \( n = 100 \).
03

Calculate the Standard Error

The standard error of the sample proportion under the null hypothesis is calculated using the formula: \[ SE = \sqrt{ \frac{p_0(1 - p_0)}{n} } \] Substituting the values, \( SE = \sqrt{ \frac{0.50 \times 0.50}{100} } = 0.05 \).
04

Compute the Test Statistic

The test statistic \( z \) is calculated using the formula: \[ z = \frac{\hat{p} - p_0}{SE} \] Substituting the values, \( z = \frac{0.35 - 0.50}{0.05} = -3.0 \).
05

Calculate the P-Value

Since the alternative hypothesis is \( H_a: p < 0.50 \), we look for the left-tail probability for \( z = -3.0 \). Using a z-table or standard normal distribution calculator, the \( P \)-value is approximately \( 0.0013 \).
06

Evaluate Evidence Against Null Hypothesis

A \( P \)-value of \( 0.0013 \) is very small, which suggests there's strong evidence against the null hypothesis \( H_0 \). Thus, we reject \( H_0 \) at common significance levels (e.g., \( \alpha = 0.05 \)).

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a vital starting point. It represents a default position or statement that there is no effect or no difference, and it is typically denoted as \(H_0\). In our example, the null hypothesis is expressed as \(H_0: p = 0.50\). This indicates the belief that the population proportion is 0.50. Establishing the null hypothesis sets the stage for testing whether the available sample data provide enough evidence to refute this claim.The null hypothesis acts as a baseline or control condition in hypothesis testing. It assumes that any kind of observed effect in the data is due to chance rather than a specific cause. Importantly, it is always stated in such a way to test if it can be rejected or not rejected based on significant contrary evidence.Taking on the null hypothesis as the default allows statisticians to systematically dismantle or uphold it through calculated tests, which provide objectivity in findings and conclusions.
Standard Error
Standard error is a key concept in statistics that helps us understand the variability of a sample mean or proportion by measuring how much these sample estimates are expected to differ from the actual population parameter.In hypothesis testing, calculating the standard error is crucial to understanding how far off our sample proportion might be from the assumed population proportion. The formula used in this context is\[SE = \sqrt{ \frac{p_0(1 - p_0)}{n} }\]where \(p_0\) is the population proportion under the null hypothesis and \(n\) is the sample size.In our example calculation, substituting the known values gives us \(SE = \sqrt{ \frac{0.50 \times 0.50}{100} } = 0.05\). This value tells us that, on average, the sample proportion should be within 0.05 of the actual population proportion given the null hypothesis is true. Standard error provides a measure of spread in the sampling distribution and is a fundamental component of determining the test statistic.
Test Statistic Z-Score
Once the standard error is calculated, we move on to finding the test statistic Z-Score. This statistic helps to determine the position of the sample mean (or proportion) relative to the null hypothesis. The test statistic is calculated using:\[z = \frac{\hat{p} - p_0}{SE}\]where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and \(SE\) is the standard error.In our exercise, with \(\hat{p} = 0.35\), \(p_0 = 0.50\), and a calculated \(SE = 0.05\), the Z-Score becomes \[z = \frac{0.35 - 0.50}{0.05} = -3.0\].This Z-Score tells us how many standard deviations our sample proportion is from the population proportion defined under the null hypothesis. A negative Z-Score indicates that the sample proportion is less than the hypothesized proportion. The magnitude of the score indicates how significantly it differs, with a larger absolute value suggesting a stronger departure from the null hypothesis assumption.
P-Value
The P-Value is a crucial element of hypothesis testing as it provides a probability measure that helps determine the strength of the evidence against the null hypothesis.After calculating the test statistic, in this case, a Z-Score of -3.0, the P-Value represents the probability of observing a test result at least as extreme or more extreme than the one obtained, under the assumption that the null hypothesis is true.For our alternative hypothesis \(H_a: p < 0.50\), we examine the left-tail probability for the Z-Score value. Using a Z-Table or standard normal distribution calculator, we find the P-Value to be approximately 0.0013. This small P-Value means there is a significantly low chance of observing the sample data if the null hypothesis were actually true.When you compare this P-Value with a typical significance level like \(\alpha = 0.05\), it is much smaller, suggesting very strong evidence against the null hypothesis. Therefore, in practice, we would reject \(H_0\), concluding that there is sufficient evidence to support \(H_a\). P-Values enable an objective assessment of whether the observed data align with or contradict the initial assumptions represented by the null hypothesis.

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Most popular questions from this chapter

A recent study \(^{4}\) considered whether dogs could be trained to detect whether a person has lung cancer or breast cancer by smelling the subject's breath. The researchers trained five ordinary household dogs to distinguish, by scent alone, exhaled breath samples of 55 lung and 31 breast cancer patients from those of 83 healthy controls. A dog gave a correct indication of a cancer sample by sitting in front of that sample when it was randomly placed among four control samples. Once trained, the dogs' ability to distinguish cancer patients from controls was tested using breath samples from subjects not previously encountered by the dogs. (The researchers blinded both dog handlers and experimental observers to the identity of breath samples.) Let \(p\) denote the probability a dog correctly detects a cancer sample placed among five samples when the other four are controls. a. Set up the null hypothesis that the dog's predictions correspond to random guessing. b. Set up the alternative hypothesis to test whether the probability of a correct selection differs from random guessing. c. Set up the alternative hypothesis to test whether the probability of a correct selection is greater than with random guessing. d. In one test with 83 Stage I lung cancer samples, the dogs correctly identified the cancer sample 81 times. The test statistic for the alternative hypothesis in part \(\mathrm{c}\) was \(z=17.7 .\) Report the \(\mathrm{P}\) -value to three decimal places and interpret. (The success of dogs in this study made researchers wonder whether dogs can detect cancer at an earlier stage than conventional methods such as MRI scans.

Home pregnancy tests claim an accuracy rate of over \(99 \% .\) A positive test does turn out later to be a false positive. There are several reasons why a woman may obtain a positive result in a home pregnancy test when she is not actually pregnant. a. For the home pregnancy test, explain what a Type I error is and explain the consequence to a woman of this type of error. b. For the home pregnancy test, what is a Type II error? What is the consequence to a woman of this type of error? c. To which diagnostic does the probability of \(99 \%\) refer? d. Can you state that the probability that a woman who receives a positive test result is not actually pregnant is 0.01? Explain your answer.

For a test of \(\mathrm{H}_{0}: p=0.50,\) the \(z\) test statistic equals 1.04 a. Find the P-value for \(\mathrm{H}_{a}: p>0.50\). b. Find the P-value for \(\mathrm{H}_{a}: p \neq 0.50\). c. Find the P-value for \(\mathrm{H}_{a}: p<0.50 .\) (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.

\(z\) test statistic To test \(\mathrm{H}_{0}: p=0.50\) that a population proportion equals 0.50 , the test statistic is a \(z\) -score that measures the number of standard errors between the sample proportion and the \(\mathrm{H}_{0}\) value of \(0.50 .\) If \(z=3.6,\) do the data support the null hypothesis, or do they give strong evidence against it? Explain.

In 2004 , New York Attorney General Eliot Spitzer filed a lawsuit against GlaxoSmithKline pharmaceutical company, claiming that the company failed to publish results of one of its studies that showed that an antidepressant drug (Paxil) may make adolescents more likely to commit suicide. Partly as a consequence, editors of 11 medical journals agreed to a new policy to make researchers and companies register all clinical trials when they begin, so that negative results cannot later be covered up. The International Journal of Medical Journal Editors wrote, "Unfortunately, selective reporting of trials does occur, and it distorts the body of evidence available for clinical decision-making." Explain why this controversy relates to the argument that it is misleading to report results only if they are "statistically significant." (Hint: See the subsection of this chapter on misinterpretations of significance tests.)

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