Caffeine in Soft Drinks Listed below are measured amounts of c...

Question

Caffeine in Soft Drinks Listed below are measured amounts of caffeine (mg per 12 oz of drink) obtained in one can from each of 20 brands (7UP, A&W Root Beer, Cherry Coke, . . . , TaB).

b. Given that Exercise 20 in Section 7-2 used the same data for a 99% confidence interval based on use of the t distribution, and given that the data do not appear to be from a normally distributed population, which confidence interval is likely to be better: The confidence interval from part (a) or the confidence interval found in Exercise 20 in Section 7-2?

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says a food testing agency is evaluating the amount of iron in milligrams per serving in breakfast cereals. Two independent random samples are tested sample A with N equal to 15, produced a 95% confidence interval for the mean iron content, which is equal to. The interval of 8.45 to 9.88 mg, while sample B with equal to 12 had a 99% confidence interval for the mean iron content of the interval of 7.62 to 10.14 mg. Given that the data is not from a normally distributed population, which of the two confidence intervals is likely to provide a better estimate of the population mean? And we're given 4 possible choices as our answers. For choice A, confidence interval from sample A, for choice B, we have confidence interval from sample B. For choice C, both confidence confidence intervals are equally valid, and for choice D could not be determined. Now if we look at both of our samples, we have relatively small sample sizes with N equal to 15 for sample A and N equal to 12 for sample B. And if we look at sample A, we're told that we have a 90.5% confidence interval, and the interval goes from 8.45 to 9.88 mg. And for sample B, it has a 99% confidence level, with the interval going from 7.62. 2 10.14 mg. So comparing our two samples, we see that sample A has a narrower interval, but has a lower confidence level, whereas sample B has the wider interval, but a higher confidence level. Now, if we look at our data, we see that it is not from a normally distributed population. So recall when data is not from a normally distributed population and sample sizes are small, which is what we have in this case, confidence intervals based on the T distribution may be unreliable. However, higher confidence levels offer more conservative estimates, which is preferred in uncertain conditions. And also recall that the T distribution is used for small samples and assumes normality or approximate symmetry, which we do not technically have in this case. Now, since the data is not from a normal population, we want to choose the 99% confidence interval from sample B as being more appropriate, even though the interval is actually wider, it better accounts for um uncertainty in the estimate. So, based on that information, if we look at our answer choices, we see that the correct answer choice corresponds to answer choice B, which is the confidence interval for sample B, since it had the higher confidence level. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Confidence Interval

t Distribution

Normality Assumption

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