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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

To test \(H_{0}: \mu=40\) versus \(H_{1}: \mu>40,\) a random sample of size \(n=25\) is obtained from a population that is known to be normally distributed with \(\sigma=6\) (a) If the sample mean is determined to be \(\bar{x}=42.3\) compute the test statistic. (b) If the researcher decides to test this hypothesis at the \(\alpha=0.1\) level of significance, determine the critical value. (c) Draw a normal curve that depicts the critical region. (d) Will the researcher reject the null hypothesis? Why?

Determine the critical value for a right-tailed test of a population mean with \(\sigma\) unknown at the \(\alpha=0.01\) level of significance with 15 degrees of freedom.

Conduct the appropriate test. Family Size In \(1985,\) the mean for the ideal number of children for a family was considered to be \(2.5 .\) A Gallup Poll of 1006 adults aged 18 or older conducted February \(16-17,2004,\) reported the mean for the ideal number of children to be 2.6 (a) Assuming \(\sigma=1.2,\) is there sufficient evidence to conclude that the mean ideal number of children has changed since 1985 at the \(\alpha=0.1\) level of significance? (b) Do the results have any practical significance? Explain.

Simulate drawing 40 simple random samples of size \(n=20\) from a population that is normally distribution with mean 50 and standard deviation \(10 .\) (a) Test the null hypothesis \(H_{0}: \mu=50\) versus the alternative hypothesis \(H_{1}: \mu \neq 50\) for each of the 40 samples using a \(t\) -test. (b) Suppose we were testing this hypothesis at the \(\alpha=0.05\) level of significance. How many of the 40 samples would you expect to result in a Type I error? (c) Count the number of samples that lead to a rejection of the null hypothesis. Is it close to the expected value determined in part (b)? (d) Describe why we know a rejection of the null hypothesis results in making a Type I error in this situation.

Suppose a chemical company has developed a catalyst that is meant to reduce reaction time in a chemical process. For a certain chemical process, reaction time is known to be 150 seconds. The researchers conducted an experiment with the catalyst 40 times and measured reaction time. The researchers reported that the catalysts reduced reaction time with a \(P\) -value of \(0.02 .\) (a) Identify the null and alternative hypotheses. (b) Explain what this result means. Do you believe that the catalyst is effective?
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