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Chapter 6: Problem 109

In Exercises 6.109 to 6.111 , we examine the effect of different inputs on determining the sample size needed. Find the sample size needed to give, with \(95 \%\) confidence, a margin of error within 卤10 . Within 卤5 . Within 卤1 . Assume that we use \(\tilde{\sigma}=30\) as our estimate of the standard deviation in each case. Comment on the relationship between the sample size and the margin of error.

Short Answer

Expert verified
The margin of error being 卤10 results in a sample size of roughly 35. For a margin of error of 卤5, the sample size is approximately 138. Lastly, for the smallest margin of error, 卤1, the sample size goes up to about 3456. The smaller the margin of error, the larger the sample size needed thus indicating that for more precision a larger sample size is required.

Step by step solution

01

Setup the equation

Firstly, get the z-score, \(Z_{\frac{\alpha}{2}}\), for the desired 95% confidence level. For a two-tail test, the z-score is approximately 1.96. Secondly, solve the formula for the sample size \(n = \frac{(Z_{\frac{\alpha}{2}} \cdot \sigma)^{2}}{E^{2}}\) for each of the three different margin of errors.
02

Calculate sample size for margin of error 卤10

Substitute the values into the equation, where \(Z_{\frac{\alpha}{2}}\) is 1.96, \(\sigma\) is 30, and \(E\) is 10. Thus, calculate the sample size, where \(n = \frac{(1.96 \cdot 30)^{2}}{10^{2}}\).
03

Calculate sample size for margin of error 卤5

Repeat the previous process but this time use \(E=5\), thus \(n = \frac{(1.96 \cdot 30)^{2}}{5^{2}}\).
04

Calculate sample size for margin of error 卤1

Again, substitute the values into the equation, this time with \(E=1\), and solve for \(n\), resulting in \(n = \frac{(1.96 \cdot 30)^{2}}{1^{2}}\).
05

Comment on relationship

After obtaining all results, it can be noticed that the smaller the margin of error, the larger the calculated sample size. This explains that precision comes with the need for a larger sample size as each incremental decrease in margin of error requires an increase in the sample size.

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Key Concepts

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

Margin of Error
The margin of error is a crucial concept in statistics, helping us understand how much uncertainty there is in sample estimates of a population parameter. Essentially, it tells us how accurate our sample results are likely to be. When calculating the required sample size, the margin of error directly impacts how confident we can be about our estimates. A smaller margin of error indicates a desire for more precise results.
However, achieving a smaller margin of error typically requires a larger sample size.
  • This relationship means that as we request more precision and, hence, decrease our margin of error, the number of samples needed increases substantially.
  • For example, a margin of error of 卤10 may require significantly fewer samples than a margin of 卤1.
Understanding this relationship helps researchers balance resource constraints, as collecting a larger sample can be more time-consuming and costly.
Confidence Interval
A confidence interval provides a range of values which is likely to contain a population parameter with a certain degree of confidence. For instance, a 95% confidence interval means we can be 95% certain that the actual parameter lies within this range.
The width of the confidence interval is influenced by a few factors, including the margin of error and sample size.
  • The greater the confidence level (for example, moving from 95% to 99%), the wider the confidence interval will be, indicating more assurance but less precision.
  • Conversely, a narrower interval with the same confidence level signifies more precision but typically requires a larger sample size.
This balance impacts research planning as it affects how researchers estimate the scope of study results in practical settings.
Standard Deviation
Standard deviation serves as a measure of variability or dispersion within a set of data. Importantly, it quantifies the amount by which individual data points differ from the average of the data set.
In sample size determination, the estimated standard deviation (\(\tilde{\sigma}\)) is used to gauge how spread out the data is likely to be. This impacts how we compute sample size because:
  • A larger standard deviation suggests a greater spread in the data, requiring us to collect more samples to accurately capture this variation.
  • Conversely, a smaller standard deviation implies a more concentrated dataset, possibly reducing the needed sample size for the same level of confidence and margin of error.
Using standard deviation effectively in sample size calculations helps improve the accuracy of statistical inferences and decisions based on data analysis.

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Most popular questions from this chapter

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.05 .

Survival Status and Heart Rate in the ICU The dataset ICUAdmissions contains information on a sample of 200 patients being admitted to the Intensive Care Unit (ICU) at a hospital. One of the variables is HeartRate and another is Status which indicates whether the patient lived (Status \(=0)\) or died (Status \(=1\) ). Use the computer output to give the details of a test to determine whether mean heart rate is different between patients who lived and died. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) Two-sample \(\mathrm{T}\) for HeartRate \(\begin{array}{lrrrr}\text { Status } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 0 & 160 & 98.5 & 27.0 & 2.1 \\ 1 & 40 & 100.6 & 26.5 & 4.2 \\ \text { Difference } & = & m u(0) & -\operatorname{mu}(1) & \end{array}\) Estimate for difference: -2.13 \(95 \% \mathrm{Cl}\) for difference: (-11.53,7.28) T-Test of difference \(=0\) (vs not \(=\) ): T-Value \(=-0.45\) P-Value \(=0.653\) DF \(=60\)

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\). A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=3.7, s_{d}=\) 2.1, \(n_{d}=30\)

Examine the results of a study \(^{45}\) investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics that has been linked to multiple health problems including hormone disruption. The study included 8,877 people who recorded all the food they ate over a 24 -hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in \(\mathrm{ng} / \mathrm{mL}\) ) in the urine: DEHP and DiNP. Find and interpret a \(95 \%\) confidence interval for the difference, \(\mu_{F}-\mu_{N},\) in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DiNP in the 3095 participants who had eaten fast food was \(\bar{x}_{F}=10.1\) with \(s_{F}=38.9\) while the mean for the 5782 participants who had not eaten fast food was \(\bar{x}_{N}=7.0\) with \(s_{N}=22.8\)

In Exercises 6.32 and 6.33, find a \(95 \%\) confidence interval for the proportion two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the normal distribution and the formula for standard error. Compare the results. Proportion of home team wins in soccer, using \(\hat{p}=0.583\) with \(n=120\)
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