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Chapter 15: Problem 11

The article "A Study of Wood Stove Particulate Emissions" (J. of the Air Pollution Control Assoc., 1979: 724-728) reports the following data on burn time (hours) for samples of oak and pine. Test at level . 05 to see whether there is any difference in true average burn time for the two types of wood. $$ \begin{array}{rrrrrrrrr} \text { Oak } & 1.72 & .67 & 1.55 & 1.56 & 1.42 & 1.23 & 1.77 & .48 \\ \text { Pine } & .98 & 1.40 & 1.33 & 1.52 & .73 & 1.20 & & \end{array} $$

Short Answer

Expert verified
There is no significant difference in average burn time for oak and pine.

Step by step solution

01

Identify the Hypotheses

We are conducting a hypothesis test to compare the average burn time of oak and pine wood. The null hypothesis \( H_0 \) states that there is no difference in the average burn time of oak and pine. Mathematically, this is expressed as \( \mu_{\text{oak}} = \mu_{\text{pine}} \). The alternative hypothesis \( H_a \) is that there is a difference, \( \mu_{\text{oak}} eq \mu_{\text{pine}} \).
02

Calculate Sample Means and Standard Deviations

Compute the sample mean and standard deviation for both oak and pine. For oak: \( \bar{x}_{\text{oak}} = \frac{1.72 + 0.67 + 1.55 + 1.56 + 1.42 + 1.23 + 1.77 + 0.48}{8} = 1.30 \) and the standard deviation \( s_{\text{oak}} \). For pine: \( \bar{x}_{\text{pine}} = \frac{0.98 + 1.40 + 1.33 + 1.52 + 0.73 + 1.20}{6} = 1.19 \) and \( s_{\text{pine}} \).
03

Determine the Test Statistic

We're using a two-sample t-test since we assume normality of the data population but the variances might not be equal. The test statistic is calculated using the formula \( t = \frac{\bar{x}_{\text{oak}} - \bar{x}_{\text{pine}}}{\sqrt{\frac{s_{\text{oak}}^2}{n_{\text{oak}}} + \frac{s_{\text{pine}}^2}{n_{\text{pine}}}}} \) where \( n_{\text{oak}} = 8 \) and \( n_{\text{pine}} = 6 \).
04

Calculate the Critical Value and Make the Decision

For a significance level of \( \alpha = 0.05 \) and the degrees of freedom determined using the approximate t-distribution formula, find the critical value from a t-table. Compare the calculated test statistic to the critical value to decide whether to reject or fail to reject the null hypothesis.
05

Conclusion

Based on the t-test, if the absolute value of the test statistic is greater than the critical t-value, reject the null hypothesis. Otherwise, fail to reject it. From the calculations, determine whether there is a statistically significant difference in the average burn times of oak and pine.

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Key Concepts

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

Two-Sample T-Test
The two-sample t-test is a statistical method used to determine whether there is a significant difference between the means of two independent groups. In this context, we're comparing the burn times of oak and pine samples.

This test is appropriate when you have two separate groups and want to see if their population means differ. Here are a few important points about the two-sample t-test:
  • Assumptions: The data should be approximately normally distributed, and the samples should have similar variances. However, if variances are different, a modified version of the t-test (Welch's t-test) may be used.
  • Independence: It assumes that the observations in each group are independent of each other.
  • Purpose: Helps determine if a difference in sample means (like burn time) is an actual difference in population means or just due to random chance.
Null Hypothesis
In hypothesis testing, the null hypothesis (\( H_0 \)) is a default position stating that there is no effect or no difference.

For our exercise, the null hypothesis asserts that there is no difference in the average burn times of oak and pine, \( \mu_{\text{oak}} = \mu_{\text{pine}} \). If statistical evidence doesn't strongly support a difference, we fail to reject this hypothesis.
  • Role: Serves as a basis for testing and any observed effect is compared against this baseline.
  • Testing: By performing a statistical test, we assess if the data significantly deviates from predictions made by this hypothesis.
  • Conclusion: If evidence supports it, the experiment doesn鈥檛 find a significant difference, affirming our base assumption.
Alternative Hypothesis
The alternative hypothesis (\( H_a \)) represents the statement that there is an effect or a difference.

In our study, the alternative hypothesis claims a difference between the average burn times of oak and pine wood, \( \mu_{\text{oak}} eq \mu_{\text{pine}} \). This hypothesis serves to challenge the status quo (null hypothesis).
  • Assertion: Proposes that observed data indicate a real effect or difference.
  • Objective: To seek enough evidence that compels rejecting the null hypothesis in favor of the alternative.
  • Outcome: If data support \( H_a \), a statistically significant difference is concluded, showing true variance in population means.
Statistical Significance
Statistical significance is a key concept in hypothesis testing, determining whether the observed effect in sample data is strong enough to be considered likely in the population.

In our two-sample t-test example, we compare the calculated test statistic against a critical value at a chosen significance level (\( \alpha = 0.05 \)).
  • Significance Level: Commonly set at 0.05, it dictates how extreme observed data must be to consider the result statistically significant.
  • Decision Making: If the test statistic exceeds critical value, reject the null hypothesis, indicating a significant difference.
  • Interpretation: A significant result suggests the effect is real and not due to random variation, strengthening the argument for \( H_a \).
  • Context: Always consider the context and the quality of data when interpreting significance, since it does not measure importance or size of effect.

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Most popular questions from this chapter

The article "Measuring the Exposure of Infants to Tobacco Smoke" (New Engl. J. of Med., 1984: 1075-1078) reports on a study in which various measurements were taken both from a random sample of infants who had been exposed to household smoke and from a sample of unexposed infants. The accompanying data consists of observations on urinary concentration of cotanine, a major metabolite of nicotine (the values constitute a subset of the original data and were read from a plot that appeared in the article). Does the data suggest that true average cotanine level is higher in exposed infants than in unexposed infants by more than 25 ? Carry out a test at significance level \(.05\). $$ \begin{array}{lrrrrrrrr} \text { Unexposed } & 8 & 11 & 12 & 14 & 20 & 43 & 111 & \\ \text { Exposed } & 35 & 56 & 83 & 92 & 128 & 150 & 176 & 208 \end{array} $$

The accompanying data resulted from an experiment to compare the effects of vitamin \(C\) in orange juice and in synthetic ascorbic acid on the length of odontoblasts in guinea pigs over a 6-week period ("The Growth of the Odontoblasts of the Incisor Tooth as a Criterion of the Vitamin C Intake of the Guinea Pig," J. of Nutrition, 1947: 491-504). Use the Wilcoxon rank-sum test at level .01 to decide whether true average length differs for the two types of vitamin \(\mathrm{C}\) intake. Compute also an approximate \(P\)-value. $$ \begin{array}{lrrrrrr} \text { Orange Juice } & 8.2 & 9.4 & 9.6 & 9.7 & 10.0 & 14.5 \\ & 15.2 & 16.1 & 17.6 & 21.5 & & \\ \text { Ascorbic Acid } & 4.2 & 5.2 & 5.8 & 6.4 & 7.0 & 7.3 \\ & 10.1 & 11.2 & 11.3 & 11.5 & & \end{array} $$

The given data on phosphorus concentration in topsoil for four different soil treatments appeared in the article "Fertilisers for Lotus and Clover Establishment on a Sequence of Acid Soils on the East Otago Uplands" (N. Zeal. J. of Exptl. Ag., 1984: 119-129). Use a distribution-free procedure to test the null hypothesis of no difference in true mean phosphorus concentration \((\mathrm{mg} / \mathrm{g})\) for the four soil treatments. $$ \begin{array}{rlrrrrr} & \text { I } & 8.1 & 5.9 & 7.0 & 8.0 & 9.0 \\ \text { Treatment } & \text { II } & 11.5 & 10.9 & 12.1 & 10.3 & 11.9 \\ & \text { III } & 15.3 & 17.4 & 16.4 & 15.8 & 16.0 \\ & \text { IV } & 23.0 & 33.0 & 28.4 & 24.6 & 27.7 \end{array} $$

The sign test is a very simple procedure for testing hypotheses about a population median assuming only that the underlying distribution is continuous. To illustrate, consider the following sample of 20 observations on component lifetime (hr): $$ \begin{array}{rrrrrrr} 1.7 & 3.3 & 5.1 & 6.9 & 12.6 & 14.4 & 16.4 \\ 24.6 & 26.0 & 26.5 & 32.1 & 37.4 & 40.1 & 40.5 \\ 41.5 & 72.4 & 80.1 & 86.4 & 87.5 & 100.2 & \end{array} $$ We wish to test \(H_{0}: \tilde{\mu}=25.0\) versus \(H_{\mathrm{a}}: \tilde{\mu}>25.0\). The test statistic is \(Y=\) the number of observations that exceed \(25 .\) a. Consider rejecting \(H_{0}\) if \(Y \geq 15\). What is the value of \(\alpha\) (the probability of a type I error) for this test? [Hint: Think of a "success" as a lifetime that exceeds 25.0. Then \(Y\) is the number of successes in the sample.] What kind of a distribution does \(Y\) have when \(\tilde{\mu}=25.0\) ? b. What rejection region of the form \(Y \geq c\) specifies a test with a significance level as close to \(.05\) as possible? Use this region to carry out the test for the given data.

High-pressure sales tactics or door-to-door salespeople can be quite offensive. Many people succumb to such tactics, sign a purchase agreement, and later regret their actions. In the mid-1970s, the Federal Trade Commission implemented regulations clarifying and extending the rights of purchasers to cancel such agreements. The accompanying data is a subset of that given in the article "Evaluating the FTC Cooling-Off Rule" (J. of Consumer Affairs, 1977: 101-106). Individual observations are cancellation rates for each of nine salespeople during each of 4 years. Use an appropriate test at level \(.05\) to see whether true average cancellation rate depends on the year. $$ \begin{array}{rrrrrrrrrrr} & {\text { Salesperson }} \\ & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9} \\ \hline \mathbf{1 9 7 3} & 2.8 & 5.9 & 3.3 & 4.4 & 1.7 & 3.8 & 6.6 & 3.1 & 0.0 \\\ \mathbf{1 9 7 4} & 3.6 & 1.7 & 5.1 & 2.2 & 2.1 & 4.1 & 4.7 & 2.7 & 1.3 \\ \mathbf{1 9 7 5} & 1.4 & .9 & 1.1 & 3.2 & .8 & 1.5 & 2.8 & 1.4 & .5 \\ \mathbf{1 9 7 6} & 2.0 & 2.2 & .9 & 1.1 & .5 & 1.2 & 1.4 & 3.5 & 1.2 \\ \hline \end{array} $$
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