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In Exercise 6.93 on page \(430,\) we see that the average number of close confidants in a random sample of 2006 US adults is 2.2 with a standard deviation of \(1.4 .\) If we want to estimate the number of close confidants with a margin of error within ±0.05 and with \(99 \%\) confidence, how large a sample is needed?

Short Answer

Expert verified
After evaluating and rounding up, you will find you need a sample size of approximately 1237 people to obtain an estimate within the ±0.05 margin of error with a \(99\%\) confidence level.

Step by step solution

01

Understand the Given Variables

The exercise provides the standard deviation (\(1.4\)), a desired margin of error (\(0.05\)), and a confidence level of \(99\% .\) Note that for a \(99\%\) confidence level, the Z-score (which can be found on the Z-table or standard normal distribution table) is \(2.576\).
02

Apply the Confidence Interval Formula

The formula we'll use to solve the exercise is the one for the sample size needed for a confidence interval, which is given as: \n Sample size \(n = \left( \frac{Z_{\frac{\alpha}{2}} \cdot \sigma}{E} \right)^2 \). In this formula, \(Z_{\frac{\alpha}{2}}\) is the Z-score, \(\sigma\) is the standard deviation and \(E\) is the margin of error.
03

Plug in the Given Values

We can substitute the provided data into the formula: \(n = \left( \frac{2.576 \cdot 1.4}{0.05} \right)^2\). This needs to be evaluated and the resulting number should be rounded up to the next whole number, because we can't work with fractions of a person.

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Key Concepts

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

Confidence Interval
When researchers conduct studies, they're often interested in understanding a wider population based on a smaller group, or sample, taken from it. To estimate a population parameter like the average number of close confidants among US adults, instead of providing a single number, they give a range of numbers that they believe the actual average is likely to fall within. This range is known as a confidence interval, which gives both the estimated value and a sense of the precision of the estimate. The width of this interval depends on the desired confidence level and variability in the data. A 99% confidence interval is especially wide, implying that there's a high degree of certainty that the population parameter lies within that range. However, to maintain a small margin of error with such a high confidence level typically requires a larger sample size.

Margin of Error
The margin of error reflects the extent of uncertainty involved in the sample estimate. It is the maximum expected difference between the true population parameter and a sample estimate of that parameter. In other words, it tells us how close we can expect our sample statistic to be to the population value. A smaller margin of error indicates greater precision in the sample estimate but usually necessitates a larger sample size to achieve that precision. For example, being within ±0.05 of the true average number of close confidants means researchers can be quite certain that their sample provides an accurate estimate of the actual number, but as a trade-off, they’ll need to collect data from more individuals compared to a larger margin of error.
Standard Deviation
Variability in data is quantified using a measure called the standard deviation. It indicates how much individuals within a dataset deviate from the average or mean value. In the context of sample size determination, the standard deviation plays a crucial role as it affects the width of the confidence interval. A high standard deviation signifies that data points are spread out widely from the mean, potentially increasing the required sample size to achieve a specific margin of error. Conversely, a low standard deviation means data points are clustered closely around the mean, which may allow a smaller sample to still accurately estimate the population parameter with a given level of confidence. The known standard deviation of 1.4 in the close confidants example informs us about the variability in the number of confidants among US adults.
Z-score
The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. For sample size determination, specifically when it's associated with a confidence interval, the Z-score corresponds to the desired confidence level. It tells us how many standard deviations an element is from the mean. To find the required Z-score for a specific confidence level, such as our 99%, statisticians typically use a Z-table. The Z-score of 2.576, for instance, indicates that the desired confidence level will include the mean plus or minus 2.576 standard deviations. This Z-score helps to scale the margin of error according to the level of confidence we wish to achieve, which in turn influences the calculation of the required sample size.

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Most popular questions from this chapter

Using Data 5.1 on page \(375,\) we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. Exercises 6.166 to 6.169 ask you to conduct a hypothesis test using additional data from this study. \(^{40}\) In every case, we are testing $$\begin{array}{ll}H_{0}: & p_{o}=p_{c} \\\H_{a}: & p_{o}>p_{c}\end{array}$$ where \(p_{o}\) and \(p_{c}\) represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Also, in every case, we have \(n_{1}=n_{2}=500 .\) Show all remaining details in the test, using a \(5 \%\) significance level. Effect of Organic Raisins after 15 Days After 15 days, 320 of the 500 fruit flies eating organic raisins are still alive, while 300 of the 500 eating conventional raisins are still alive.

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=15.7, s_{d}=12.2\) \(n_{d}=25 .\)

A survey is planned to estimate the proportion of voters who support a proposed gun control law. The estimate should be within a margin of error of \(\pm 2 \%\) with \(95 \%\) confidence, and we do not have any prior knowledge about the proportion who might support the law. How many people need to be included in the sample?

Systolic Blood Pressure and Survival Status Use technology and the ICUAdmissions dataset to find a \(95 \%\) confidence interval for the difference in systolic blood pressure (Systolic) upon admission to the Intensive Care Unit at the hospital based on survival of the patient (Status with 0 indicating the patient lived and 1 indicating the patient died.) Interpret the answer in context. Is "No difference" between those who lived and died a plausible option for the difference in mean systolic blood pressure? Which group had higher systolic blood pressures on arrival?

Plastic microparticles are contaminating the world's shorelines (see Exercise 6.108\()\), and much of this pollution appears to come from fibers from washing polyester clothes. \({ }^{27}\) The worst offender appears to be fleece, and a recent study found that the mean number of polyester fibers discharged into wastewater from washing fleece was 290 fibers per liter of wastewater, with a standard deviation of 87.6 and a sample size of 120 . (a) Find and interpret a \(99 \%\) confidence interval for the mean number of polyester microfibers per liter of wastewater when washing fleece. (b) What is the margin of error? (c) If we want a margin of error of only ±5 with \(99 \%\) confidence, what sample size is needed?

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