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Test the hypothesis using (a) the classical approach and (b) the P-value approach. Be sure to verify the requirements of the test. $$\begin{array}{l}H_{0}: p=0.6 \text { versus } H_{1}: p<0.6 \\\n=250 ; x=124 ; \alpha=0.01\end{array}$$

Short Answer

Expert verified
Reject \( H_0 \) using both approaches.

Step by step solution

01

Define the Hypotheses

The null hypothesis is given as \( H_0: p = 0.6 \). The alternative hypothesis is \( H_1: p < 0.6 \). This sets the framework for our hypothesis test.
02

Verify Test Requirements

Check if the sample size is sufficiently large. The sample size \( n = 250 \) should be large enough if both \( np \) and \( n(1-p) \) are greater than 5. Calculate these:
03

Calculate \(np\) and \(n(1-p)\)

\[ np = 250 \times 0.6 = 150 \]\[ n(1-p) = 250 \times (1-0.6) = 100 \]Since both values are greater than 5, the sample size requirement is satisfied.
04

Test Statistic Calculation

Use the formula for the test statistic for a proportion:\[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \]where \( \hat{p} = \frac{x}{n} = \frac{124}{250} = 0.496 \). Plug in the values:\[ z = \frac{0.496 - 0.6}{\sqrt{\frac{0.6 \times 0.4}{250}}} = \frac{-0.104}{0.03098} \approx -3.36 \]
05

Step 5a: Classical Approach

Find the critical value for \( \alpha = 0.01 \) in a left-tailed test. From the z-table, the critical value is approximately \( -2.33 \). Compare the test statistic with the critical value:\[ z = -3.36 < -2.33 \]Since the test statistic is less than the critical value, reject the null hypothesis.
06

Step 5b: P-value Approach

Find the P-value corresponding to the test statistic \( z = -3.36 \) from the z-table. The P-value is approximately 0.0004. Compare this with \( \alpha = 0.01 \):Since the P-value is less than \( \alpha \), reject the null hypothesis.

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Key Concepts

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

Classical Approach
In hypothesis testing, the classical approach involves comparing the test statistic with a critical value to decide whether to reject the null hypothesis. This method begins by defining the null and alternative hypotheses. Next, a significance level \( \alpha \) is chosen; in this example, it is 0.01. A test statistic is calculated, which often follows a standard distribution such as the normal distribution. By comparing the test statistic against the critical value, we decide if we can reject the null hypothesis. If the test statistic is in the critical region (the area beyond the critical value), we reject the null hypothesis. This approach is straightforward because it relies on pre-calculated thresholds.
P-value Approach
The P-value approach is another method to conduct hypothesis testing. Instead of comparing the test statistic to a critical value, we calculate the P-value, which represents the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value under the assumption that the null hypothesis is true. We then compare this P-value to the significance level \( \alpha \). If the P-value is less than \( \alpha \), we reject the null hypothesis. For instance, in the example given, the calculated P-value is approximately 0.0004. Because this is less than 0.01, we reject the null hypothesis. This approach provides the exact probability and can be more intuitive for interpreting the strength of evidence against the null hypothesis.
Test Statistic
The test statistic is a standardized value that measures the degree of agreement between a sample data set and the null hypothesis. It essentially quantifies how far our sample statistic is from the hypothesized population parameter in units of standard error. For a proportion test, the test statistic use the formula \[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \], where \( \hat{p} \) is the sample proportion, \( p \) is the population proportion under the null hypothesis, and \( n \) is the sample size. In the example, this calculation yielded a test statistic of -3.36, which tells us that our sample proportion is 3.36 standard errors below the hypothesized proportion of 0.6.
Sample Size
The sample size \( n \) is a crucial component in hypothesis testing. It influences the accuracy and reliability of the results. In the context of proportion tests, the sample size should be large enough to satisfy the conditions: both \( np \) and \( n(1-p) \) should be greater than 5. This ensures that the sampling distribution of the sample proportion is approximately normal, enabling the use of z-tests. For the given example, with a sample size of 250, calculations showed that both \( 250 \times 0.6 = 150 \) and \( 250 \times (1-0.6) = 100 \) are greater than 5, thus meeting the requirement and validating the test results.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement that there is no effect or no difference, serving as a baseline or default position. In hypothesis testing, we seek evidence against this baseline. For the given example, the null hypothesis is \( H_0: p = 0.6 \), implying that the population proportion is 0.6. The goal of the test is to gather enough evidence to reject \( H_0 \) in favor of the alternative hypothesis. A strong test, indicated by a low P-value or a test statistic in the critical region, suggests that the data are inconsistent with \( H_0 \).
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_1 \) or \( H_a \), represents what we aim to support. It is the hypothesis that there is an effect or a difference. In the given problem, the alternative hypothesis is \( H_1: p < 0.6 \), suggesting that the population proportion is less than 0.6. The alternative hypothesis is typically based on the question of interest or the research question. By rejecting the null hypothesis, we provide support for the alternative hypothesis, indicating that our sample data provide sufficient evidence to infer that the population proportion is less than 0.6.

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Most popular questions from this chapter

According to the Centers for Disease Control, \(15.2 \%\) of American adults experience migraine headaches. Stress is a major contributor to the frequency and intensity of headaches. A massage therapist feels that she has a technique that can reduce the frequency and intensity of migraine headaches. (a) Determine the null and alternative hypotheses that would be used to test the effectiveness of the massage therapist's techniques. (b) A sample of 500 American adults who participated in the massage therapist's program results in data that indicate that the null hypothesis should be rejected. Provide a statement that supports the massage therapist's program. (c) Suppose, in fact, that the percentage of patients in the program who experience migraine headaches is \(15.3 \%\). Was a Type I or Type II error committed?

To test \(H_{0}: p=0.40\) versus \(H_{1}: p>0.40,\) a simple random sample of \(n=200\) individuals is obtained and \(x=84\) successes are observed. (a) What does it mean to make a Type II error for this test? (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, compute the probability of making a Type II error if the true population proportion is 0.44. What is the power of the test? (c) Redo part (b) if the true population proportion is 0.47

If we do not reject the null hypothesis when the statement in the alternative hypothesis is true, we have made a Type ____ error.

From Super Bowl I (1967) through Super Bowl XXXI (1997), the stock market increased if an NFL team won the Super Bowl and decreased if an AFL team won. This condition held 28 out of 31 years. (a) Suppose the likelihood of predicting the direction of the stock market (increasing or decreasing) in any given year is \(0.50 .\) Decide on the appropriate null and alternative hypotheses to test whether the outcome of the Super Bowl can be used to predict the direction of the stock market. (b) Use the binomial probability distribution to determine the \(P\) -value for the hypothesis test from part (a). (c) Comment on the dangers of using the outcome of the hypothesis test to judge investments. Be sure your comment includes a discussion of circumstances in which associations have a causal relationship.

A manufacturer of high-strength, lowalloy steel beams requires that the standard deviation of yield strength not exceed 7000 pounds per square inch (psi). The quality-control manager selected a sample of 20 steel beams and measured their yield strength. The standard deviation of the sample was 7500 psi. Assume that yield strengths are normally distributed. Does the evidence suggest that the standard deviation of yield strength exceeds 7000 psi at the \(\alpha=0.01\) level of significance?

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