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Chapter 9: Problem 19

Consider \(H_{0}: \mu=72\) versus \(H_{1}: \mu>72\). A random sample of 16 observations taken from this population produced a sample mean of \(75.2 .\) The population is normally distributed with \(\sigma=6 .\) a. Calculate the \(p\) -value. b. Considering the \(p\) -value of part a, would you reject the null hypothesis if the test were made at the significance level of \(.01\) ? c. Considering the \(p\) -value of part a, would you reject the null hypothesis if the test were made at the significance level of \(.025\) ?

Short Answer

Expert verified
a. The p-value is approximately 0.016.\nb. At the significance level of 0.01, we would not reject the null hypothesis as p-value > 0.01.\nc. At the significance level of 0.025, we would reject the null hypothesis as p-value < 0.025.

Step by step solution

01

Calculate the Standard Error

The standard error (SE) is calculated by dividing the standard deviation \(\sigma\) by the square root of the sample size \(n\). Using the given values, \(\sigma=6\) and \(n=16\), the calculation becomes: SE = \(\sigma/\sqrt{n}\) = \(6/\sqrt{16}\) = 1.5
02

Calculate the Test Statistic

The test statistic (Z) is found by subtracting the hypothesized population mean \(\mu\) from the sample mean \(x虆\) and then dividing by the standard error. Using the given values, \(\mu=72\) and \(x虆=75.2\), and the calculated SE = 1.5, the calculation becomes:Z = \((x虆 - \mu)/SE\) = \((75.2 - 72)/1.5\) = 2.13
03

Calculate the p-value

The p-value is the probability of getting a result at least as extreme as the one found, assuming the null hypothesis is true. As the alternative hypothesis \(H_{1}: \mu>72\) suggests a one-tailed test, the p-value is found by looking up the Z value of 2.13 in the Z table or using a statistical calculator. The result is a p-value of 0.016.
04

Test the Hypothesis at Significance Level of 0.01

The p-value is compared to the significance level to decide whether to reject the null hypothesis. If the p-value is less than or equal to the significance level, we reject the null hypothesis. Using a significance level of 0.01, since p-value = 0.016 > 0.01, we would not reject the null hypothesis at this level of significance.
05

Test the Hypothesis at Significance Level of 0.025

Using a significance level of 0.025, since p-value = 0.016 < 0.025, we would reject the null hypothesis at this level of significance.

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Key Concepts

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

Understanding the Z-Test
The Z-test is a statistical method used to determine if there is a significant difference between a sample mean and a known population mean. It is particularly useful when the population standard deviation is known and the sample size is large. In this exercise, we're dealing with a scenario where the population mean is hypothesized to be 72, and the sample mean is 75.2.
  • This test calculates a Z-score, which tells us how many standard deviations the sample mean is from the population mean.
  • The Z-score is then used to determine the probability of observing a sample mean as extreme or more extreme than the actual sample mean, assuming the null hypothesis is true.
The Z-test assumes that the population is normally distributed, just like it mentions in our problem.
Calculating the P-Value
The p-value plays a crucial role in hypothesis testing. It represents the probability of obtaining results as extreme as the observed results, assuming the null hypothesis is true. In our exercise, the p-value is calculated using the Z-score derived from the sample data. Since we're dealing with a one-tailed test (more on this later), we look up the corresponding probability of our calculated Z-score, which is 2.13.
  • With a Z-score of 2.13, consulting a standard normal distribution table or using statistical software gives us a p-value of 0.016.
  • A lower p-value indicates stronger evidence against the null hypothesis.
Therefore, the p-value helps us decide whether we should reject or accept the null hypothesis.
Understanding Significance Levels
A significance level is a threshold set by the researcher which determines how much evidence is needed to reject the null hypothesis. It is often denoted as alpha (伪) and commonly set at 0.05, 0.01, or other values based on context. This exercise includes significance levels of 0.01 and 0.025, which helps in decision-making after calculating the p-value.
  • At a significance level of 0.01, if the p-value is less than or equal to 0.01, it indicates strong evidence against the null hypothesis, leading to rejection.
  • For 0.025, a p-value less than or equal to this indicates sufficient evidence to reject the null hypothesis.
The chosen significance level impacts whether the null hypothesis is rejected or not.
The Role of Standard Error
Standard Error (SE) is a statistical term that measures the accuracy with which a sample mean represents a population mean. Essentially, it gives you an idea of how much sample means vary from each other. In the context of hypothesis testing, SE is crucial in calculating the Z-score.
  • It is computed as the standard deviation divided by the square root of the sample size (i.e., SE = \( \sigma/\sqrt{n} \)).
  • For our example, the standard deviation is 6, and the sample size is 16, resulting in an SE of 1.5.
A smaller SE indicates that the sample mean is a more precise estimate of the population mean.
One-Tailed Test Design
A one-tailed test in hypothesis testing determines whether there is a statistically significant difference in one particular direction. In other words, it focuses on whether the parameter of interest is greater than or less than a specified value, as opposed to "different from" it. In this exercise, the alternative hypothesis is one-sided: \(H_1: \mu > 72\).
  • Use a one-tailed test when you have a specific hypothesis about the direction of the difference.
  • This approach considers only one tail of the distribution while calculating the p-value, providing a clearer answer to directional research questions.
One-tailed tests allow for more focused testing but require that the hypothesized direction of the test be justified beforehand.

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Most popular questions from this chapter

In an observational study at Turner Field in Atlanta, Georgia, \(43 \%\) of the men were observed not washing their hands after going to the bathroom (see Exercise \(7.80\) ). Suppose that in a random sample of 95 men who used the bathroom at Camden Yards in Baltimore, Maryland, 26 did not wash their hands. a. Using the critical-value approach and \(\alpha=.10\), test whether the percentage of all men at Camden Yards who use the bathroom and do not wash their hands is less than \(43 \%\). b. How do you explain the Type I error in part a? What is the probability of making this error in part a? c. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.10 ?\)

Thirty percent of all people who are inoculated with the current vaccine used to prevent a disease contract the disease within a year. The developer of a new vaccine that is intended to prevent this disease wishes to test for significant evidence that the new vaccine is more effective. a. Determine the appropriate null and alternative hypotheses. b. The developer decides to study 100 randomly selected people by inoculating them with the new vaccine. If 84 or more of them do not contract the disease within a year, the developer will conclude that the new vaccine is superior to the old one. What significance level is the developer using for the test? c. Suppose 20 people inoculated with the new vaccine are studied and the new vaccine is concluded to be better than the old one if fewer than 3 people contract the disease within a year. What is the significance level of the test?

A telephone company claims that the mean duration of all long-distance phone calls made by its residential customers is 10 minutes. A random sample of 100 long-distance calls made by its residential customers taken from the records of this company showed that the mean duration of calls for this sample is \(9.20\) minutes. The population standard deviation is known to be \(3.80\) minutes. a. Find the \(p\) -value for the test that the mean duration of all long- distance calls made by residential customers is different from 10 minutes. If \(\alpha=.02\), based on this \(p\) -value, would you reject the null hypothesis? Explain. What if \(\alpha=.05\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02\). Does your conclusion change if \(\alpha=.05 ?\)

More and more people are abandoning national brand products and buying store brand products to save money. The president of a company that produces national brand coffee claims that \(40 \%\) of the people prefer to buy national brand coffee. A random sample of 700 people who buy coffee showed that 259 of them buy national brand coffee. Using \(\alpha=.01\), can you conclude that the percentage of people who buy national brand coffee is different from \(40 \%\) ? Use both approaches to make the test.

For each of the following examples of tests of hypothesis about \(\mu\), show the rejection and nonrejection regions on the \(t\) distribution curve. a. A two-tailed test with \(\alpha=.02\) and \(n=20\) b. A left-tailed test with \(\alpha=.01\) and \(n=16\) c. A right-tailed test with \(\alpha=.05\) and \(n=18\)
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