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Chapter 18: Problem 42

How sensitive are the untrained noses of students? Exercise \(16.27\) (page 390) gives the lowest levels of dimethyl sulfide (DMS) that 10 students could detect. You want to estimate the mean DMS odor threshold among all students and you would be satisfied to estimate the mean to within \(\pm 0.1\) with \(99 \%\) confidence. The standard deviation of the odor threshold for untrained noses is known to be \(\sigma=7\) micrograms per liter of wine. How large an SRS of untrained students do you need?

Short Answer

Expert verified
You need a sample size of 32,516 students.

Step by step solution

01

Identify Known Quantities

The exercise provides the following details: the target error margin is \(E = 0.1\) micrograms per liter, the confidence level is \(99\%\), and the standard deviation is \(\sigma = 7\) micrograms per liter.
02

Determine the Z-score for Confidence Level

For a confidence level of \(99\%\), we need to find the critical Z-score. The critical Z-score for \(99\%\) confidence can be obtained from Z-tables, which is \(Z = 2.576\).
03

Use the Sample Size Formula

The formula to calculate the required sample size for estimating a mean is given by:\[ n = \left(\frac{Z \times \sigma}{E}\right)^2 \]Substituting the values we have:- \( Z = 2.576 \)- \( \sigma = 7 \)- \( E = 0.1 \)
04

Calculate Sample Size

Substitute the known values into the formula:\[ n = \left(\frac{2.576 \times 7}{0.1}\right)^2 \]Calculate inside the parentheses first:\( 2.576 \times 7 = 18.032 \)Then:\( \frac{18.032}{0.1} = 180.32 \)Finally, square the result:\[ n = (180.32)^2 = 32515.3024 \]
05

Round Up Sample Size

Since we cannot have a fraction of a participant, we round up to the next whole number:\( n = 32516 \). Thus, you need a sample size of 32,516 students.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Level
When we talk about the confidence level, we're expressing how sure we are about the estimate of the population parameter, which in this case is the mean DMS odor threshold. For example, a 99% confidence level means that if we repeated the experiment multiple times, 99% of the calculated confidence intervals would contain the true population mean. This high level of confidence is like a grade that shows how reliable our findings are. To put it into perspective, choosing a 99% confidence level ensures a very high degree of certainty, which usually results in a larger sample size necessary for the same margin of error, compared to a lower confidence level like 95%.
Standard Deviation
The standard deviation, represented by the symbol \( \sigma \), is a measure of how spread out the numbers are in a data set. In our exercise, it's known to be 7 micrograms per liter. It tells us that the odor threshold numbers vary around the average with a spread of 7 units on either side of the mean.Knowing the standard deviation is crucial in estimating how much our sample needs to consider this variability. A higher standard deviation means the data is more spread out, leading to a bigger required sample size to accurately reflect the true population mean within a desired margin of error.
Z-score
The Z-score is a statistical measurement that tells us how many standard deviations away a data point is from the mean. It is crucial when determining the sample size for estimating a mean with a certain confidence level. For our problem, the Z-score needed for a 99% confidence level is 2.576. This number is derived from Z-tables or standard normal distribution charts. The Z-score literally sets the boundary in probability terms, cutting off the extreme 1% of cases in the normal distribution tails, thus providing the desired high level of confidence.
Margin of Error
The margin of error (E) in our scenario ensures how tight the estimate is around the population mean. It's the plus-or-minus range we're comfortable with, expressed in the same units as the data, here being 0.1 micrograms per liter. A smaller margin of error would demand a larger sample size to maintain the same confidence level. In this exercise, wanting a margin of error of just 0.1 with a 99% confidence level influences the need for a surprisingly large sample size of 32,516 students, showcasing how precision in estimates can considerably impact the resources needed to collect the data.

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Most popular questions from this chapter

Suppose that scores on the mathematics part of the National Assessment of Educational Progress (NAEP) test for eighthgrade students follow a Normal distribution with standard deviation \(\sigma=110\). You want to estimate the mean score within \(\pm 10\) with \(90 \%\) confidence. How large an SRS of scores must you choose?

Valium is a common antidepressant and sedative. A study investigated how valium works by comparing its effect on sleep in seven genetically modified mice and eight normal control mice. There was no significant difference between the two groups. The authors say that this lack of significance "is related to the large inter-individual variability that is also reflected in the low power \((20 \%)\) of the test." 12 (a) Explain exactly what power \(20 \%\) against a specific alternative means. (b) Explain in simple language why tests having low power often fail to give evidence against a null hypothesis even when the null hypothesis is really false. (c) What fact about this experiment most likely explains the low power?

How much education children get is strongly associated with the wealth and social status of their parents. In social science jargon, this is socioeconomic status, or SES. But the SES of parents has little influence on whether children who have graduated from college go on to yet more education. One study looked at whether college graduates took the graduate admissions tests for business, law, and other graduate programs. The effects of the parents' SES on taking the LSAT test for law school were "both statistically insignificant and small." (a) What does "statistically insignificant" mean? (b) Why is it important that the effects were small in size as well as insignificant?

The 2013 Youth Risk Behavior Survey found that 349 individuals in its random sample of 1367 Ohio high school students said that they had texted or emailed while driving in the previous 30 days. That's \(25.5 \%\) of the sample. Why is this estimate likely to be biased? Do you think it is biased high or low? Does the margin of error of a \(95 \%\) confidence interval for the proportion of all Ohio high school students who texted or emailed while driving in the previous 30 days allows for this bias?

Example \(16.1\) (page 377) assumed that the body mass index (BMI) of all American young women follows a Normal distribution with standard deviation \(\sigma=7.5\). How large a sample would be needed to estimate the mean BMI \(\mu\) in this population to within \(\pm 1\) with \(95 \%\) confidence?
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