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Chapter 10: Problem 10

To test \(H_{0}: \mu=4.5\) versus \(H_{1}: \mu>4.5,\) a simple random sample of size \(n=13\) is obtained from a population that is known to be normally distributed. (a) If \(\bar{x}=4.9\) and \(s=1.3,\) compute the test statistic. (b) Draw a \(t\) -distribution with the area that represents the \(P\) -value shaded. (c) Approximate and interpret the \(P\) -value. (d) If the researcher decides to test this hypothesis at the \(\alpha=0.1\) level of significance, will the researcher reject the null hypothesis? Why?

Short Answer

Expert verified
Test statistic is 1.1081. P-value is approximately 0.1456. Do not reject the null hypothesis at the 0.1 significance level.

Step by step solution

01

Compute the test statistic

To compute the test statistic, use the following formula for the t-score: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] Substitute the given values: \( \bar{x} = 4.9, \mu = 4.5, s = 1.3 \) and \( n = 13 \). \[ t = \frac{4.9 - 4.5}{1.3/\sqrt{13}} \] Calculate the standard error: \[ 1.3/\sqrt{13} \approx 0.3609 \] Now, compute the t-score: \[ t = \frac{4.9 - 4.5}{0.3609} \approx 1.1081 \]
02

Draw a t-distribution for the P-value

Draw a t-distribution curve with degrees of freedom: \[ df = n - 1 = 13 - 1 = 12 \] Shaded the area to the right of \( t = 1.1081 \). This represents the P-value.
03

Approximate and interpret the P-value

To find the P-value, use a t-distribution table or a calculator. For \( t = 1.1081 \) with 12 degrees of freedom, the P-value can be found: \[ P(t > 1.1081) \approx 0.1456 \] Interpretation: The P-value of approximately 0.1456 indicates that there is a 14.56% probability of observing a test statistic as extreme as 1.1081, assuming the null hypothesis is true.
04

Decision on null hypothesis at \( \alpha=0.1 \) level

Compare the P-value \( 0.1456 \) with the significance level \( \alpha = 0.1 \). Since the P-value is greater than \( \alpha \), we fail to reject the null hypothesis. Interpretation: There is not enough evidence to conclude that the population mean \( \mu \) is greater than 4.5 at the 0.1 significance level.

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Key Concepts

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

t-test
A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It's especially useful when the sample size is small, and the population standard deviation is unknown. The t-test calculates a t-score, which tells us how far our sample mean is from the population mean in units of standard error. There are several types of t-tests, but in this context, we are using a one-sample t-test to compare the sample mean to a known value.
P-value
The P-value is a crucial part of hypothesis testing. It indicates the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. In simpler terms, it tells us how likely it is to get our results by random chance. A small P-value (typically less than 0.05) suggests that the observed data is unlikely under the null hypothesis, leading us to reject the null. Conversely, a large P-value indicates that the data is consistent with the null hypothesis.
null hypothesis
The null hypothesis (denoted as \(H_0\)) is a statement that there is no effect or no difference, and it serves as the starting assumption for statistical testing. In our exercise, the null hypothesis is \(H_0: \mu = 4.5\). It suggests that the population mean is equal to 4.5. The goal of hypothesis testing is to assess whether there is enough evidence to reject this statement in favor of the alternative hypothesis (\(H_1\)), which in our case is \(H_1: \mu > 4.5\).
significance level
The significance level, denoted as \(\alpha\), is the threshold used to determine whether the P-value is low enough to reject the null hypothesis. Common choices for the significance level are 0.05, 0.01, and 0.1. In our exercise, we use \(\alpha = 0.1\). This means we are willing to accept a 10% risk of rejecting the null hypothesis when it is actually true. If the P-value is less than \(\alpha\), we reject the null hypothesis; otherwise, we fail to reject it. The choice of \(\alpha\) should be based on the context and the consequences of making a Type I error (rejecting a true null hypothesis).

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Most popular questions from this chapter

Carl Reinhold August Wunderlich said that the mean temperature of humans is \(98.6^{\circ} \mathrm{F}\). Researchers Philip Mackowiak, Steven Wasserman, \(\begin{array}{lllllll}\text { and } & \text { Myron } & \text { Levine } & {[\mathrm{JAMA},} & \text { Sept. } & 23-30 & 1992\end{array}\) \(268(12): 1578-80]\) thought that the mean temperature of humans is less than \(98.6^{\circ}\) F. They measured the temperature of 148 healthy adults 1 to 4 times daily for 3 days, obtaining 700 measurements. The sample data resulted in a sample mean of \(98.2^{\circ} \mathrm{F}\) and a sample standard deviation of \(0.7^{\circ} \mathrm{F}\) (a) Test whether the mean temperature of humans is less than \(98.6^{\circ} \mathrm{F}\) at the \(\alpha=0.01\) level of significance using the classical approach. (b) Determine and interpret the \(P\) -value.

In Problems \(3-8,\) test the hypothesis, using (a) the classical approach and then (b) the P-value approach. Be sure to verify the requirements of the test. $$\begin{aligned}&H_{0}: p=0.3 \text { versus } H_{1}: p>0.3\\\&n=200 ; x=75 ; \alpha=0.05\end{aligned}$$

Suppose the consequences of making a Type I error are severe. Would you choose the level of significance, \(\alpha,\) to equal \(0.01,0.05,\) or \(0.10 ?\) Why?

Confidence in Schools In \(1995,40 \%\) of adults aged 18 years or older reported that they had "a great deal" of confidence in the public schools. On June \(1,2005,\) the Gallup Organization released results of a poll in which 372 of 1004 adults aged 18 years or older stated that they had "a great deal" of confidence in the public schools. Does the evidence suggest at the \(\alpha=0.05\) level of significance that the proportion of adults aged 18 years or older having "a great deal" of confidence in the public schools is significantly lower in 2005 than the 1995 proportion?

Inmates In \(2002,\) the mean age of an inmate on death row was 40.7 years, according to data obtained from the U.S. Department of Justice. A sociologist wondered whether the mean age of a death-row inmate has changed since then. She randomly selects 32 deathrow inmates and finds that their mean age is \(38.9,\) with a standard deviation of \(9.6 .\) (a) Do you believe the mean age has changed? Use the \(\alpha=0.05\) level of significance. (b) Construct a \(95 \%\) confidence interval about the mean age. What does the interval imply?
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