91Ó°ÊÓ

Examples \(1,3,\) and 5 referred to a study about astrology. Another part of the study used the following experiment: Professional astrologers prepared horoscopes for 83 adults. Each adult was shown three horoscopes, one of which was the one an astrologer prepared for him or her and the other two were randomly chosen from ones prepared for other subjects in the study. Each adult had to guess which of the three was his or hers. Of the 83 subjects, 28 guessed correctly. a. Define the parameter of interest and set up the hypotheses to test that the probability of a correct prediction is \(1 / 3\) against the astrologers' claim that it exceeds \(1 / 3 .\) b. Show that the sample proportion \(=0.337,\) the standard error of the sample proportion for the test is \(0.052,\) and the test statistic is \(z=0.08\). c. Find the P-value. Would you conclude that people are more likely to select their horoscope than if they were randomly guessing, or are results consistent with random guessing?

Step by step solution

01

Define the Parameter and Hypotheses

The parameter of interest is the true proportion \( p \) of individuals who can correctly identify their horoscope. The null hypothesis \( H_0 \) is \( p = \frac{1}{3} \), meaning the probability of a correct guess is consistent with random guessing. The alternative hypothesis \( H_a \) is \( p > \frac{1}{3} \), suggesting that individuals can choose their horoscope better than random guessing.

02

Calculate the Sample Proportion

The sample proportion \( \hat{p} \) is calculated as the number of correct guesses (28) divided by the total number of subjects (83). Thus, \( \hat{p} = \frac{28}{83} \approx 0.337 \).

03

Compute the Standard Error

The standard error of the sample proportion is given by \( SE = \sqrt{\frac{p(1-p)}{n}} \), where \( p = \frac{1}{3} \) and \( n = 83 \). Calculating this, \( SE = \sqrt{\frac{\frac{1}{3}(1-\frac{1}{3})}{83}} \approx 0.052 \).

04

Calculate the Test Statistic

The test statistic \( z \) is calculated using the formula \( z = \frac{\hat{p} - p}{SE} \). Substituting the known values, \( z = \frac{0.337 - \frac{1}{3}}{0.052} \approx 0.08 \).

05

Determine the P-value

Using a standard normal distribution table or calculator, find the P-value for \( z = 0.08 \). This P-value corresponds to the area to the right of \( z = 0.08 \), which is approximately 0.468. This is larger than any common significance level (e.g., \( \alpha = 0.05 \)).

06

Conclusion Interpretation

Since the P-value is greater than common significance levels, we fail to reject the null hypothesis. This implies that individuals are not more likely to select their horoscope than if they were guessing randomly.

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Key Concepts

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

Sample Proportion

When dealing with hypothesis testing, understanding the concept of "sample proportion" is crucial. A sample proportion is a statistic that estimates the proportion of individuals in a population who possess a certain attribute. In this experiment about horoscopes, the sample proportion is the fraction of adults who guessed their own horoscope correctly. To calculate this, divide the number of successful guesses (28) by the total number of subjects (83).
This yields a sample proportion \( \hat{p} = \frac{28}{83} \approx 0.337 \).
This value represents the estimated probability of a correct guess based on the data from the sample. By comparing this sample proportion to a theoretical probability, analysts can make inferences about the population from which the sample was taken.

• If the sample proportion is significantly different from the hypothesized population proportion, there might be an effect or characteristic worth exploring further.

Standard Error

The concept of "standard error" is another key element in hypothesis testing. It reflects how much variability there is in the sample proportion from the true proportion of the population. More simply, it measures how much the sample proportion might differ from the true population proportion if the experiment were repeated multiple times.
The formula for standard error of a sample proportion is given by: \[ SE = \sqrt{\frac{p(1-p)}{n}} \]
where \( p \) is the expected proportion (in this case, \( \frac{1}{3} \)) and \( n \) is the sample size (83). In the example, the standard error is calculated as \( SE = \sqrt{\frac{\frac{1}{3}(1-\frac{1}{3})}{83}} \approx 0.052 \).
This helps researchers determine how precisely the sample proportion estimates the true population proportion and is essential for calculating the test statistic.

Z-Test

The "z-test" is a type of hypothesis test used when determining whether there is a significant difference between the observed sample proportion and a theoretical expectation. It's especially handy when the data is assumed to follow a normal distribution, which is a typical assumption for large samples.
The test statistic for a z-test is calculated as follows: \[ z = \frac{\hat{p} - p}{SE} \]
where \( \hat{p} \) is the sample proportion, \( p \) is the expected proportion, and \( SE \) is the standard error of the sample proportion.For the horoscope experiment, substituting the known values produces: \[ z = \frac{0.337 - \frac{1}{3}}{0.052} \approx 0.08 \].
This small z-value suggests that there isn't a strong deviation of the sample proportion from what's expected under the null hypothesis, hinting that random guessing might actually be at play.

P-value

Last but not least is the "P-value," a cornerstone of hypothesis testing that helps assess the significance of the results. It represents the probability of obtaining test results at least as extreme as the ones observed during the study, under the assumption that the null hypothesis is true. In practical terms, a low P-value (typically less than 0.05) would suggest that the observed results appear not to occur randomly, indicating a potential effect or association. However, a higher P-value means the data does not provide strong evidence against the null hypothesis. For the horoscope study, the z-value of 0.08 leads us to a P-value of approximately 0.468.

• This P-value is greater than the usual significance levels (like 0.05), meaning there's not enough evidence to reject the null hypothesis.
• Thus, the results suggest that the accuracy of guessing one's horoscope is consistent with random choice.