Question

Ecological researchers measured the concentration of red cells in the blood of 27 field-caught lizards (Sceloporus occidetitalis). In addition, they examined each lizard for infection by the malarial parasite Plasmodium. The red cell counts \(\left(10^{-3} \times\right.\) cells per \(\left.\mathrm{mm}^{3}\right)\) were as reported in the table. \({ }^{37}\) $$ \begin{array}{|lcc|} \hline & \begin{array}{c} \text { Infected } \\ \text { animals } \end{array} & \begin{array}{c} \text { Noninfected } \\ \text { animals } \end{array} \\ \hline n & 12 & 15 \\ \bar{y} & 972.1 & 843.4 \\ s & 245.1 & 251.2 \\ \hline \end{array} $$ One might expect that malaria would reduce the red cell count, and in fact previous research with another lizard species had shown such an effect. Do the data support this expectation? Assume that the data are normally distributed. Test the null hypothesis of no difference against the alternative that the infected population has a lower red cell count. Use a \(t\) test at (a) \(\alpha=0.05\) (b) \(\alpha=0.10\) [Note: Formula (6.7.1) yields 24 df.]

Short answer

First, identify the null and alternative hypotheses. Then calculate the test statistic using the mean red cell counts, pooled standard deviation, and the sample sizes for both groups. Compare the calculated t-statistic to the critical t values for \(\alpha = 0.05\) and \(\alpha = 0.10\) to determine if the null hypothesis should be rejected at each alpha level.

Step-by-step solution

Identify the Hypotheses

Formulate the null hypothesis (\(H_0\right)\) stating that there is no difference in red cell counts between infected and noninfected animals, i.e., \(H_0: \bar{y}_{infected} - \bar{y}_{noninfected} = 0\). The alternative hypothesis (\(H_a\right)\) is that infected animals have a lower red cell count, i.e., \(H_a: \bar{y}_{infected} < \bar{y}_{noninfected}\).

Calculate the test statistic

Use the formula for the test statistic in a two-sample t-test, which is \[t = \frac{\bar{y}_{infected} - \bar{y}_{noninfected}}{\text{s}_{\text{pooled}} \times \root{2}{\frac{1}{n_{infected}} + \frac{1}{n_{noninfected}}}}\], where \text{s}_{\text{pooled}}\ is the pooled standard deviation. Based on the sample sizes and standard deviations given, find \[\text{s}_{\text{pooled}} = \root{1}{\frac{\frac{(12-1)(245.1)^2 + (15-1)(251.2)^2}{12 + 15 - 2}}\].

Compute the Pooled Standard Deviation

Use the formula \[\text{s}_{\text{pooled}} = \root{1}{\frac{\frac{(n_{infected}-1)\times s_{infected}^2 + (n_{noninfected}-1)\times s_{noninfected}^2}{df}}}\], and substitute the given standard deviations and the degrees of freedom (24) to find \text{s}_{\text{pooled}}\.

Determine the Degrees of Freedom

The degrees of freedom (df) for this test is the total number of observations minus the number of groups, which is \[df = n_{infected} + n_{noninfected} - 2 = 12 + 15 - 2 = 25\]. Note: The exercise has given 24 df, use 24 df for further calculations.

Calculate the Test Statistic

Substitute the means, standard deviation, and sample sizes for infected and noninfected groups into the formula for the t-statistic.

Compare to Critical t Value

Using a t-distribution table or software, find the critical t value for the given alpha levels and degrees of freedom. Then compare the calculated t-statistic to the critical t value to decide whether to reject the null hypothesis.

Make a Conclusion

If the absolute value of the calculated t-statistic is greater than the critical t value, the null hypothesis can be rejected, indicating the data supports that infected lizards have a lower red cell count. If not, the null hypothesis cannot be rejected.

Key concepts

Null Hypothesis

At the heart of hypothesis testing is the concept of the null hypothesis, symbolized as \(H_0\). In essence, it is a statement suggesting that there is no effect or no difference present in the context of the research question. In the given exercise, the null hypothesis is that there is no difference in red cell counts between infected and noninfected lizards. Mathematically, it's expressed as \(H_0: \bar{y}_{infected} - \bar{y}_{noninfected} = 0\).

The null hypothesis serves as a benchmark or standard against which the research findings will be measured. If the data we collect provide enough evidence, we may reject the null hypothesis, paving the way for the acceptance of the alternative hypothesis. However, failure to find strong evidence will lead us to hold onto the null hypothesis, indicating that our sample data does not show a significant difference or effect.

Alternative Hypothesis

Complementary to the null hypothesis is the alternative hypothesis, denoted as \(H_a\) or \(H_1\). This hypothesis states that there is an effect or a difference, and it's what researchers are typically trying to find evidence for. In our example problem, the alternative hypothesis posits that infected lizards have a lower red cell count than noninfected ones, represented as \(H_a: \bar{y}_{infected} < \bar{y}_{noninfected}\).

Researchers hope to reject the null hypothesis in favor of the alternative hypothesis, but this can only be justified if the data significantly deviate from what would be expected under the null hypothesis. The type of alternative hypothesis dictates the nature of the statistical test to be used – in this case, a one-tailed t-test, since we are specifically looking for a decrease rather than any difference.

Degrees of Freedom

Degrees of freedom (df) are a critical component in the realm of hypothesis testing as they influence the shape and the spread of the t-distribution, which is used to estimate population parameters. In a statistical test, degrees of freedom refer to the number of independent values or quantities that can vary in an analysis without breaking any constraints. For a two-sample t-test, \(df\) is computed as the sum of the number of observations in both groups minus two (since we have two group means).

Our exercise provided a slight discrepancy, suggesting to use 24 degrees of freedom while the calculation results in 25. For accuracy, it is important to use the correct \(df\) which is critical in determining the appropriate critical t value from t-distribution tables or using statistical software. The number of degrees of freedom impacts the critical value we compare our test statistic against, which in turn affects the conclusion whether to reject or not reject the null hypothesis.

Pooled Standard Deviation

When comparing two independent samples, as we are in our lizard red cell count study, we need a measure to estimate the variability within the groups, which we call the pooled standard deviation. This metric is a weighted average of the two groups' variances that accounts for the different sample sizes. The formula to calculate the pooled standard deviation involves the standard deviations of each group and their respective sample sizes.

In this two-sample t-test, we apply the formula \(s_{pooled} = \sqrt{\frac{(n_{infected}-1)\times s_{infected}^2 + (n_{noninfected}-1)\times s_{noninfected}^2}{n_{infected} + n_{noninfected} - 2}}\), using the sample data provided. It is crucial to use the pooled standard deviation in the test statistic formula because it gives us a more accurate measure of the overall variability, taking into account the variances of both samples.

Test Statistic

The test statistic is the calculated value that results from the formula employed in a statistical test, which we then use to make a decision about the null hypothesis. In a two-sample t-test, the test statistic determines whether the difference in means observed between two groups is statistically significant, given the variability and size of the samples. The formula for the test statistic in our lizard example has been provided in Step 2 of the problem's solution.

We calculate the test statistic by taking the difference between the group means and dividing it by the estimated standard error of the difference in means. The result is a t-value that we compare to a critical value from the t-distribution given the degrees of freedom \(df\) and the chosen significance level \(\alpha\). If the absolute value of our test statistic exceeds the critical value, the difference between the group means is deemed statistically significant, and we reject the null hypothesis, favoring the alternative hypothesis.