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Chapter 9: Problem 29

Practice mechanics of a \(t\) test A study has a random sample of 20 subjects. The test statistic for testing \(\mathrm{H}_{0}: \mu=100\) is \(t=2.40 .\) Find the approximate P-value for the alternative, (a) \(\mathrm{H}_{a}: \mu \neq 100\) (b) \(\mathrm{H}_{a}: \mu>100,\) and \((\mathrm{c}) \mathrm{H}_{a}: \mu<100\). (Among others, you can use the \(t\) Distribution web app to find the answer.)

Short Answer

Expert verified
(a) P-value 鈮� 0.027, (b) P-value 鈮� 0.0135, (c) P-value 鈮� 0.9865.

Step by step solution

01

Identify the Degrees of Freedom

First, determine the degrees of freedom for the sample. The degrees of freedom (df) for a t-test with a sample size of 20 is calculated as follows: \( df = n - 1 \), where \( n \) is the sample size. Therefore, \( df = 20 - 1 = 19 \).
02

Find the P-value for (a) Two-tailed Test

For the alternative hypothesis \( \mathrm{H}_{a}: \mu eq 100 \), you need to determine the P-value for a two-tailed test. Using a t-distribution table or a statistical software/app, find the probability \( P \) such that \( \left|t\right| > 2.40 \) with \( df = 19 \). The two-tailed P-value is \( 2 \times P(t > 2.40) \). Check the table or tool; you'll find the P-value is approximately 0.027.
03

Find the P-value for (b) Right-tailed Test

For the alternative hypothesis \( \mathrm{H}_{a}: \mu > 100 \), calculate the P-value for the right-tailed test. You only need the probability \( P(t > 2.40) \) with \( df = 19 \). Using the table or app, you will find this P-value is approximately 0.0135.
04

Find the P-value for (c) Left-tailed Test

For the alternative hypothesis \( \mathrm{H}_{a}: \mu < 100 \), calculate the P-value for the left-tailed test. This is the probability \( P(t < 2.40) \), which is \( 1 - P(t > 2.40) = 1 - 0.0135 \), resulting in approximately 0.9865.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics used to make inferences about populations based on sample data. It begins with forming a null hypothesis (\(H_0\)) which represents a default claim that is tested against an alternative hypothesis (\(H_a\)). This alternative hypothesis suggests there is a significant difference or effect.To conduct a hypothesis test:
  • Start by stating both the null and the alternative hypotheses.
  • Determine the significance level (commonly written as \(\alpha\), often set at 0.05), which defines the probability of rejecting the null hypothesis when it is actually true.
  • Select the appropriate statistical test, in this case, a t-test, which is particularly useful when sample sizes are small or when population standard deviation is unknown.
  • Computing the test statistic allows us to determine how far the sample result deviates from the null hypothesis.
A decision is made by comparing the P-value and the significance level. If the P-value is less than \(\alpha\), the null hypothesis is rejected in favor of the alternative hypothesis, indicating that the result is statistically significant.
Degrees of Freedom
Degrees of freedom (\(df\)) is a crucial concept in statistical calculations, particularly in the context of t-tests. It is essentially the number of values in the final calculation of a statistic that are free to vary.For a sample of size \(n\), the formula to determine degrees of freedom in a t-test is:
  • \(df = n - 1\)
For example, in the exercise provided with a sample size of 20, the degrees of freedom would be \(19 (20 - 1)\).Degrees of freedom are essential as they influence the shape of the t-distribution which is used to determine the critical values for the test. Effectively, it tailors the breadth of possible variability, influencing how data is interpreted in tests, thereby making accurate statistical conclusions possible.
P-value
The P-value is a measure used in hypothesis testing to help decide whether to reject the null hypothesis. It quantifies the probability of observing the test results under the null hypothesis assumption.A lower P-value suggests that the observation is less likely under the null hypothesis, supporting the alternative hypothesis. Here's how to interpret P-values:
  • If \(P\)-value \(< 0.05\), reject the null hypothesis (result is statistically significant)
  • If \(P\)-value \(> 0.05\), do not reject the null hypothesis (result is not statistically significant)
In hypothesis testing scenarios:
  • For a two-tailed test, check if \(\left|t\right|\) is more extreme than the critical value.
  • For a right-tailed test, look at whether \(t\) is greater than the critical value.
  • For a left-tailed test, consider if \(t\) is less than the critical value.
Using these principles allows you to compute the P-value and make data-driven decisions effectively. In essence, the P-value helps determine the strength of the evidence against the null hypothesis.

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Most popular questions from this chapter

Electricity prices According to the U.S. Energy Information Administration, the average monthly household electricity bill in 2014 was \(\$ 114\) before taxes and fees. A consumer association plans to investigate if the average amount has changed this year. Define the population parameter of interest and state the null and alternative hypotheses for this investigation.

In the webcomic on the link http://xkcd.com/882/, a girl claims that jelly beans cause acne. Scientists investigate and find no link between the two \((p>0.05)\). They are asked to check if jelly beans of a particular color cause acne. They test 20 different colors each at a significance level of \(5 \%\) and find a link between green jelly beans and acne. This leads to a newspaper headline, "Green Jellybeans Cause Acne" where the \(5 \%\) chance of the link is mentioned as \(95 \%\) confidence. When the scientists repeat the same experiment, they are unable to find any link between acne and color of jelly beans. They conclude that the earlier result might be coincidental. Using this example, explain why you need to have some skepticism when research suggests that some therapy or drug has an impact in treating a disease.

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?

\(\begin{array}{ll}\ & \mathbf{H}_{0} \text { or } \mathbf{H}_{a} \text { ? For parts a and } \mathrm{b} \text { , is the statement a null }\end{array}\) 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 b in terms of the parameter values.

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.
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