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Chapter 8: Problem 58

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data supports the use of a one-sample \(t\) test. a. What hypotheses should be tested if the investigators believe a priori that the design specification has been satisfied? b. What conclusion is appropriate if the hypotheses of part (a) are tested, \(t=-2.3\), and \(\alpha=.05\) ? c. What conclusion is appropriate if the hypotheses of part (a) are tested, \(t=-1.8\), and \(\alpha=.01\) ? d. What should be concluded if the hypotheses of part (a) are tested and \(t=-3.6 ?\)

Short Answer

Expert verified
Test the hypothesis \( \mu \geq 10 \) vs \( \mu < 10 \). (b) Reject \( H_0 \); (c) Fail to reject \( H_0 \); (d) Reject \( H_0 \).

Step by step solution

01

Define the Hypotheses

For the hypothesis test, we want to determine if the pen's average writing lifetime is less than 10 hours. Therefore, we set up the following hypotheses:- Null Hypothesis \( H_0 \): \( \mu \geq 10 \) hours (The average writing lifetime is at least 10 hours).- Alternative Hypothesis \( H_a \): \( \mu < 10 \) hours (The average writing lifetime is less than 10 hours).
02

Decision Rule for Hypothesis Testing

The decision to reject or not reject the null hypothesis depends on the calculated p-value or the critical value approach. For a one-sample \( t \)-test with a significance level \( \alpha \), compare the test statistic to critical \( t \)-values from the \( t \)-distribution table with \( n-1 \) degrees of freedom (where \( n = 18 \)).
03

Conclusion for Part b - \( t = -2.3 \), \( \alpha = 0.05 \)

Given \( \alpha = 0.05 \), we need to find the critical \( t \)-value for 17 degrees of freedom. The critical \( t \)-value is approximately \( -1.74 \). Since \( t = -2.3 \) is less than \( -1.74 \), we reject the null hypothesis. There is enough evidence to conclude the average lifetime is less than 10 hours.
04

Conclusion for Part c - \( t = -1.8 \), \( \alpha = 0.01 \)

For \( \alpha = 0.01 \) and 17 degrees of freedom, the critical \( t \)-value is approximately \( -2.567 \). Since \( t = -1.8 \) is greater than \( -2.567 \), we fail to reject the null hypothesis. There is not enough evidence to say the average lifetime is less than 10 hours at the 0.01 significance level.
05

Conclusion for Part d - \( t = -3.6 \)

With \( t = -3.6 \), this value is more extreme than any reasonable negative critical \( t \)-value for both \( \alpha = 0.05 \) and \( \alpha = 0.01 \). Therefore, regardless of the significance level, we reject the null hypothesis. There is strong evidence to conclude that the average lifetime is less than 10 hours.

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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 statistical method used to determine if there is enough evidence to reject a null hypothesis, or if we should fail to reject it. In the context of a one-sample t-test, this refers to testing whether a sample mean deviates significantly from a known or hypothesized population mean.

In our example, the pens have a hypothesized average writing lifetime of at least 10 hours under controlled conditions. The null hypothesis (\( H_0 \)) asserts that the true average is greater than or equal to 10 hours (\( \mu \geq 10 \)). The alternative hypothesis (\( H_a \)) contends that the average is actually less than 10 hours (\( \mu < 10 \)).

By drawing a random sample and using the one-sample t-test, we calculate a test statistic that assists us in understanding if the observed sample mean significantly diverges from our hypothesized mean of 10 hours. The result of the test either supports continuing to accept the null hypothesis or suggests sufficient evidence to adopt the alternative hypothesis.
Statistical significance
Statistical significance is a measure of whether the results of a hypothesis test are unlikely to have occurred by random chance. In hypothesis testing, this is typically quantified using a p-value or a critical value.

When the p-value is less than the chosen significance level (\( \alpha \)), we consider the result to be statistically significant, meaning there is strong evidence to suggest the null hypothesis is false. In practice, common significance levels are 0.05 or 0.01. For instance, in our scenario:
  • If \( t = -2.3 \) and \( \alpha = 0.05 \), we find the test is statistically significant because the test statistic exceeds the critical value, prompting us to reject \( H_0 \).
  • If \( t = -1.8 \) and \( \alpha = 0.01 \), we find insufficient statistical significance because the test statistic does not exceed the critical value, so we fail to reject \( H_0 \).
Statistical significance helps determine the likelihood that the observed data could occur under the null hypothesis, guiding our conclusion towards or against it.
Critical value
In hypothesis testing, a critical value is a threshold that the test statistic must exceed for us to reject the null hypothesis. It forms part of the decision rule for hypothesis testing and is derived from the specified significance level (\( \alpha \)) and the degrees of freedom in the dataset.

For a one-sample t-test, the critical value is taken from the t-distribution table, which varies based on the chosen \( \alpha \) level and the number of observations in the sample.

In our pen example, for \( n = 18 \) (hence, 17 degrees of freedom), and \( \alpha = 0.05 \), the critical value was found to be approximately \( -1.74 \). Our test statistic of \( t = -2.3 \) being less than this critical value led us to reject \( H_0 \). By contrast, a test statistic of \( t = -1.8 \) at \( \alpha = 0.01 \) did not exceed the critical value of \( -2.567 \), leading us to fail to reject \( H_0 \). Understanding critical values helps in deciding whether to reject the null hypothesis based on how extreme the test statistic is compared to this benchmark.

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Most popular questions from this chapter

The article "Caffeine Knowledge, Attitudes, and Consumption in Adult Women" ( \(J\). of Nutrition Educ., 1992: 179-184) reports the following summary data on daily caffeine consumption for a sample of adult women: \(n=47, \bar{x}=215 \mathrm{mg}\), \(s=235 \mathrm{mg}\), and range \(=5-1176 .\) a. Does it appear plausible that the population distribution of daily caffeine consumption is normal? Is it necessary to assume a normal population distribution to test hypotheses about the value of the population mean consumption? Explain your reasoning. b. Suppose it had previously been believed that mean consumption was at most \(200 \mathrm{mg}\). Does the given data contradict this prior belief? Test the appropriate hypotheses at significance level .10 and include a \(P\)-value in your analysis.

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Newly purchased tires of a certain type are supposed to be filled to a pressure of \(30 \mathrm{lb} / \mathrm{in}^{2}\). Let \(\mu\) denote the true average pressure. Find the \(P\)-value associated with each given \(z\) statistic value for testing \(H_{0}: \mu=30\) versus \(H_{\mathrm{a}}: \mu \neq 30\). a. \(2.10\) b. \(-1.75\) c. \(-.55\) d. \(1.41\) e. \(-5.3\)

Let \(\mu\) denote true average serum receptor concentration for all pregnant women. The average for all women is known to be 5.63. The article "Serum Transferrin Receptor for the Detection of Iron Deficiency in Pregnancy" (Amer. \(J\). Clinical Nutr., 1991: 1077-1081) reports that \(P\)-value \(>.10\) for a test of \(H_{0}: \mu=5.63\) versus \(H_{\mathrm{a}}: \mu \neq 5.63\) based on \(n=\) 176 pregnant women. Using a significance level of \(.01\), what would you conclude?

A sample of 12 radon detectors of a certain type was selected, and each was exposed to \(100 \mathrm{pCi} / \mathrm{L}\) of radon. The resulting readings were as follows: $$ \begin{array}{rrrrrr} 105.6 & 90.9 & 91.2 & 96.9 & 96.5 & 91.3 \\ 100.1 & 105.0 & 99.6 & 107.7 & 103.3 & 92.4 \end{array} $$ a. Does this data suggest that the population mean reading under these conditions differs from 100 ? State and test the appropriate hypotheses using \(\alpha=.05\). b. Suppose that prior to the experiment, a value of \(\sigma=7.5\) had been assumed. How many determinations would then have been appropriate to obtain \(\beta=.10\) for the alternative \(\mu=95\) ?
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