Question

An experimenter fed different rations, \(A\) and \(B\), to two groups of 100 chicks each. Assume that all factors other than rations are the same for both groups. Of the chicks fed ration \(\mathrm{A}, 13\) died, and of the chicks fed ration \(\mathrm{B}, 6\) died. a. Construct a \(98 \%\) confidence interval for the true difference in mortality rates for the two rations. b. Can you conclude that there is a difference in the mortality rates for the two rations?

Short answer

Answer: Yes, there is a significant difference in the mortality rates for chicks fed rations A and B at the 98% confidence level. Ration A has a significantly higher mortality rate than ration B.

Step-by-step solution

Calculate the proportions of mortality rates

Let p_A and p_B represent the proportions of mortality rates in each group. To calculate the proportions, we need to divide the number of chicks that died by the total number of chicks in each group: $$ p_A = \frac{13}{100} = 0.13 \\ p_B = \frac{6}{100} = 0.06 $$

Calculate the difference in proportions

Now we'll find the observed difference in proportions: $$ d = p_A - p_B = 0.13 - 0.06 = 0.07 $$

Calculate the standard error

In order to calculate the standard error of the difference in proportions, we'll use the formula: $$ SE = \sqrt{\frac{p_A(1-p_A)}{n_A} + \frac{p_B(1-p_B)}{n_B}} \\ SE = \sqrt{\frac{0.13(1-0.13)}{100} + \frac{0.06(1-0.06)}{100}} \\ SE \approx 0.0395 $$

Determine the critical value

To calculate the 98% confidence interval, we'll use the standard normal distribution. We need to find the critical value, which is also called Z-score, corresponding to the 98% confidence level. For a 98% confidence interval, the critical value (Z-score) is approximately 2.33 (you can read this value from a standard normal table (Z-table), or you can use statistical software).

Calculate the confidence interval

Now, we'll compute the 98% confidence interval for the true difference in mortality rates using the formula: $$ CI = (d - Z \cdot SE, d + Z \cdot SE) \\ CI = (0.07 - 2.33 \cdot 0.0395, 0.07 + 2.33 \cdot 0.0395) $$ After calculating, we obtain the confidence interval: $$ CI \approx (0.0061, 0.1339) $$

Interpret the confidence interval and draw conclusions

From the 98% confidence interval, we can see that the true difference in mortality rates between the two rations lies between 0.61% and 13.39%. Since zero is not included in the confidence interval, we can conclude that there is a significant difference in the mortality rates for the two rations at the 98% confidence level. This means that ration A has a significantly higher mortality rate than ration B.

Key concepts

Mortality Rate Comparison

Comparing mortality rates between different treatments or conditions is a vital procedure in experiments studying the efficacy or safety of an intervention. In the scenario of the experiment with ration A and B given to chicks, the mortality rate is simply the proportion of chicks that died during the study period. What's critical is the ability to accurately assess whether a difference in mortality rates is statistically significant or could reasonably be attributed to random chance.

In this case, mortality rate comparison is not just about subtracting one proportion from another. It involves constructing a confidence interval around the difference to gauge the precision of the estimate. This process helps in determining whether a seemingly obvious difference in mortality rates (in this case, between two dietary rations for chicks) is a true difference with practical implications, or if it might simply be due to natural variation within the samples being compared.

Standard Error Calculation

Understanding the calculation of standard error (SE) is essential for measuring the accuracy of an estimate. In hypothesis testing and confidence interval construction, the standard error of the difference in proportions provides a gauge for the variability in the estimate. To put it simply, SE measures how much sample means are spread out around the true population mean.

The formula mentioned, \(SE = \sqrt{\frac{p_A(1-p_A)}{n_A} + \frac{p_B(1-p_B)}{n_B}}\), is grounded in the principles of probability and enables us to estimate how much the computed difference between mortality rates could fluctuate if different random samples were taken. It's the variability within each group (intrinsic to any biological or experimental data) that contributes to the overall uncertainty in the comparison of the proportions. A larger SE would indicate greater variability, hence a less precise estimate of the true difference, while a smaller SE suggests a more precise estimate.

Hypothesis Testing Concepts

When we reference hypothesis testing concepts in statistics, we're diving into a systematic way to test claims or hypotheses about a parameter. The core idea is to determine whether there is enough evidence in a sample of data to infer about the entire population. In the context of our example where we compare mortality rates, hypothesis testing enables us to challenge the assumption that the two diets result in the same mortality rates, essentially testing the hypothesis that the true difference in proportions is zero.

In the steps provided for solving the exercise, hypothesis testing wasn't explicitly described, but it's wrapped up in the construction of the confidence interval. When zero is not included in a 98% confidence interval for the difference in proportions, we implicitly reject the null hypothesis which claims there's no difference between the mortality rates of rations A and B. The key takeaway is that hypothesis testing in the form of confidence intervals can help us make inferences about population parameters and guide decisions in an experimental or real-world context.