Question

\(\mathbf{6 . 2 2 0}\) Diet Cola and Calcium Exercise B.3 on page 349 introduces a study examining the effect of diet cola consumption on calcium levels in women. A sample of 16 healthy women aged 18 to 40 were randomly assigned to drink 24 ounces of either diet cola or water. Their urine was collected for three hours after ingestion of the beverage and calcium excretion (in \(\mathrm{mg}\) ) was measured. The summary statistics for diet cola are \(\bar{x}_{C}=56.0\) with \(s_{C}=4.93\) and \(n_{C}=8\) and the summary statistics for water are \(\bar{x}_{W}=49.1\) with \(s_{W}=3.64\) and \(n_{W}=8 .\) Figure 6.20 shows dotplots of the data values. Test whether there is evidence that diet cola leaches calcium out of the system, which would increase the amount of calcium in the urine for diet cola drinkers. In Exercise \(\mathrm{B} .3\), we used a randomization distribution to conduct this test. Use a t-distribution here, after first checking that the conditions are met and explaining your reasoning. The data are stored in ColaCalcium.

Short answer

The solution process involves stating the hypotheses, checking the conditions for using a t-distribution, calculating the t-score using the provided statistics, finding the p-value using the t-score and degrees of freedom, and making a conclusion based on the p-value. A t-test could reveal if there is a statistically significant difference in calcium excretion between the diet cola and water consumers.

Step-by-step solution

State the hypotheses

Firstly, state the null hypothesis \(H_0\): 'the means of the two populations are equal', so there is no difference in calcium excretion between the cola and water consumers. Conversely, the alternative hypothesis \(H_a\) is 'the means of the two populations are not equal', implying that there is a difference in calcium excretion between the two groups.

Check the conditions

To utilize the t-distribution, several conditions must be satisfied. Firstly, the samples need to be random, which is satisfied according to the exercise. Secondly, the populations the samples are taken from are approximately normally distributed. This information is not provided, but we will proceed on the assumption that it's plausible. Lastly, the populations have equal standard deviations. Although the standard deviations provided (4.93 for cola and 3.64 for water) differ slightly, they are close enough to be reasonable for this assumption.

Calculate the t-score

The t-score is a measure of the degree to which our groups differ standardized by the variability of our measurements. We use the following formula to calculate it: \( t = \frac{{\bar{x}_C - \bar{x}_W}}{{s_p \sqrt{\frac{1}{n_C} + \frac{1}{n_W}}}} \) where \( s_p = \sqrt{\frac{{(n_C-1)s^2_C + (n_W-1)s^2_W}}{{n_C + n_W - 2}}} \) is the pooled standard deviation.

Find the p-value

Next, use the calculated t-score and the degrees of freedom \( df = n_C + n_W - 2 \) to find the p-value from a two-tailed t-distribution table. The p-value tells us the probability of finding a test statistic as extreme as, or more so, than what was observed, assuming the null hypothesis is true.

Draw the conclusion

If the p-value is less than the chosen significance level, typically 0.05, we reject the null hypothesis. Otherwise, we cannot reject the null hypothesis. According to the p-value, determine which conclusion to make. Note that a significant result does not tell us about the magnitude or the practical importance of the difference, just that we can reasonably rule out chance as an explanation for the observed effect.