Question

An entomologist conducted an experiment to see if wounding a tomato plant would induce changes that improve its defense against insect attack. She grew larvae of the tobacco hornworm (Manduca sexta) on wounded plants or control plants. The accompanying table shows the weights (mg) of the larvae after 7 days of growth. ${}^{41}$ (Assume that the data are normally distributed.) How strongly do the data support the researcher's expectation? Use a $t$ test at the $5\mathrm{\%}$ significance level. Let $H_A$ be that wounding the plant tends to diminish larval growth. [Note: Formula (6.7.1) yields 31.8 df.] $$\begin{array}{|lcc|}\hline & \text{ Wounded }& \text{ Control }\\ n& 16& 18\\ \overline{y}& 28.66& 37.96\\ s& 9.02& 11.14\\ \hline\end{array}$$

Short answer

Calculate the test statistic and compare it to the critical value. If the test statistic is less than the negative of the critical value, the data supports the researcher's expectation, otherwise it does not.

Step-by-step solution

State the Hypotheses

Specify the null hypothesis ($H_0$) and the alternative hypothesis ($H_A$). For this test: $H_0:\text{The mean weight of larvae from wounded plants is equal to the mean weight of larvae from control plants.}$ $H_A:\text{The mean weight of larvae from wounded plants is less than the mean weight of larvae from control plants.}$

Calculate the Test Statistic

Use the data to compute the pooled standard deviation and then the $t$-test statistic using the formula $t=\frac{{\overline{y}}_1-{\overline{y}}_2}{\text{sp}\times \text{sqrt}\frac{1}{n_1}+\frac{1}{n_2}}$, where ${\overline{y}}_1$ and ${\overline{y}}_2$ are the sample means, $n_1$ and $n_2$ are the sample sizes, and $\text{sp}$ is the pooled standard deviation calculated as $\text{sp}=\text{sqrt}\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$. Insert the given values to obtain the test statistic.

Calculate the Degrees of Freedom

The degrees of freedom (df) should be calculated according to the procedure for a $t$-test with two independent samples. As given in the exercise, use 31.8 for the df.

Determine the Critical Value

Find the critical $t$-value from the $t$-distribution table at the $5\mathrm{\%}$ significance level with the given degrees of freedom (31.8). Since we're conducting a one-tailed test, look for the critical $t$-value that corresponds to the $95\mathrm{\%}$ confidence level.

Compare the Test Statistic and Critical Value

If the test statistic calculated in Step 2 is less than the negative of the critical $t$-value from Step 4, then we reject the null hypothesis $H_0$. Otherwise, we fail to reject the null hypothesis.

Conclusion

Based on the comparison from Step 5, draw a conclusion about whether or not the data supports the researcher's expectation that wounding the plant diminishes larval growth.

Key concepts

Hypothesis Testing

Hypothesis testing plays a pivotal role in biology, particularly when researchers need to determine whether their experimental findings are significant or just due to random chance. It starts by proposing two opposing statements: the null hypothesis ((H_0)), which suggests that there is no effect or difference, and the alternative hypothesis ((H_A)), which represents what the researcher is trying to prove. In our entomologist's study, (H_0) indicates there's no difference in larval weight between wounded and control plants, while (H_A) suggests wounded plants have a deterrent effect on larval growth.

To perform hypothesis testing, we need to calculate a test statistic that assesses how much the observed data deviates from (H_0). If this deviation is large enough, as determined by a threshold called the significance level (in this case 5%), we reject (H_0). A common error students make is forgetting that rejecting (H_0) doesn't prove (H_A) is true, only that the data is unlikely under (H_0). It's like saying, 'It's not impossible for it to rain if the clouds are absent, but it's highly unlikely.'

Pooled Standard Deviation

The pooled standard deviation is a weighted average of standard deviations from two or more independent samples. It's used when we assume that the two populations have the same standard deviation but differ in sizes. In our example with the tomato plants, the pooled standard deviation combines variability information from the wounded and control plant groups to gain a more comprehensive measure of dispersion.

The calculation involves squaring the standard deviations, multiplying by their respective sample sizes minus one, summing these values, and finally dividing by the total number of observations from all groups minus the number of groups. The formula is complex, but crucial, as students sometimes overlook the importance of using pooled values rather than individual sample standard deviations. It is key to ensure accurate results when comparing groups with different sample sizes or variances.

Degrees of Freedom

Degrees of freedom (df) is a somewhat abstract yet essential concept in statistics, reflecting the number of independent values that can vary in an analysis without breaking any constraints. It's like the number of 'choices' left after making certain decisions. For the entomologist's t-test, the degrees of freedom inform us which distribution to use when determining the critical value of t. It's calculated as the total number of observations minus the number of groups.

Students often mistake the degrees of freedom as a mere subtraction exercise rather than recognizing its role in shaping the distribution for our t-test. The right df ensures that the critical value for determining statistical significance is correctly identified. Miscalculating df could lead to incorrect conclusions about the research hypothesis. Think about it as the difference between using a map designed for city navigation versus off-road: both might be maps, but they serve different terrains and guide you differently.

Statistical Significance

Statistical significance is a term that quantifies the probability of observing an effect given that the null hypothesis is true. It's typically associated with a p-value or, in this specific biology research case, a predetermined level of significance such as 5%. If the test statistic exceeds a critical value at this level, the result is statistically significant, meaning it's unlikely the observed effect is due to random chance. Students should remember that statistical significance does not mean practical significance or that the results are biologically meaningful; it merely indicates the results are not random.

It's a stamp of 'this is worth a second look', not 'we've found definitive proof'. Many students confuse statistical significance with the importance of the outcome. An outcome can be statistically significant (e.g., technically non-random) but still be of no real-world importance if the effect size is very small or if it lacks meaningful context in the biological study.