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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 0.01

Short Answer

Expert verified
The required sample size for a margin of error of 0.01 is 10000.

Step by step solution

01

Formula Recognisation

The given formula to determine the sample size for the desired margin of error is \(n=1 /(M E)^{2}\), where ME is the margin of error.
02

Substitution of Values

Substitute the provided margin error ME = 0.01 into the formula. So the sample size \(n = 1 / (0.01)^{2}\).
03

Calculation

On computing \(1 / (0.01)^{2}\), we obtain 10000.

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Key Concepts

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

Margin of Error
When conducting a survey or a study, researchers aim to measure the views or characteristics of a whole population based on a sample from that population. However, the results from the sample are unlikely to match the entire population exactly, which is where the concept of the margin of error comes into play.

The margin of error is a statistic expressing the amount of random sampling error in a survey's results. It represents how much the results could vary if the survey were conducted multiple times. The margin of error is commonly presented as a plus or minus value (e.g., ±3 percent) and indicates the precision of the estimate. A smaller margin of error requires a larger sample size and suggests more confidence in the consistency of the survey results. Understanding the margin of error helps individuals interpret survey results accurately and gauge the level of certainty in the findings.
Confidence Level
The confidence level is a measure of the reliability of the results obtained from a statistical sample. It reflects how often the true population parameter (such as the mean or proportion) would be captured by the interval estimate if the sampling process were repeated numerous times. Typically expressed as a percentage, the confidence level represents the degree of certainty that the sample statistic will fall within a certain range.

For example, a 95% confidence level means that if you were to draw 100 different random samples and compute a confidence interval for each sample, approximately 95 of these intervals would be expected to contain the true population parameter. The confidence level is essential in determining the sample size: higher confidence levels, such as 99% or 95%, often require larger sample sizes to ensure that the margin of error is acceptably low.
Sample Size Formula
To accurately estimate a particular statistic of a population, such as a proportion or a mean, researchers use a sample size formula to determine how large their sample should be to achieve a desired margin of error at a certain confidence level. A fundamental formula used for this purpose, especially when estimating a proportion and assuming a 50% response distribution (the most conservative estimate), is \( n = \frac{1}{(ME)^2} \), where \( n \) is the sample size and \( ME \) is the margin of error.

Using this formula helps researchers to decide on the number of responses or observations needed in their study for reliable and valid results. Given a smaller margin of error and a high confidence level, the required sample size will increase. This is crucial, as underestimating the sample size can lead to inaccurate results, while overestimating can unnecessarily increase the cost and duration of the study.

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Most popular questions from this chapter

Close Confidants and Social Networking Sites Exercise 6.93 introduces a study \(^{48}\) in which 2006 randomly selected US adults (age 18 or older) were asked to give the number of people in the last six months "with whom you discussed matters that are important to you." The average number of close confidants for the full sample was \(2.2 .\) In addition, the study asked participants whether or not they had a profile on a social networking site. For the 947 participants using a social networking site, the average number of close confidants was 2.5 with a standard deviation of 1.4 , and for the other 1059 participants who do not use a social networking site, the average was 1.9 with a standard deviation of \(1.3 .\) Find and interpret a \(90 \%\) confidence interval for the difference in means between the two groups.

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. 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 Potatoes after 20 Days After 20 days, 250 of the 500 fruit flies eating organic potatoes are still alive, while 130 of the 500 eating conventional potatoes are still alive.

Use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of home team wins in soccer, with \(n=120\) and \(\hat{p}=0.583\)

In Exercise \(6.107,\) we see that plastic microparticles are contaminating the world's shorelines and that much of the pollution appears to come from fibers from washing polyester clothes. The same study referenced in Exercise 6.107 also took samples from ocean beaches. Five samples were taken from each of 18 different shorelines worldwide, for a total of 90 samples of size \(250 \mathrm{~mL}\). The mean number of plastic microparticles found per \(250 \mathrm{~mL}\) of sediment was 18.3 with a standard deviation of 8.2 . (a) Find and interpret a \(99 \%\) confidence interval for the mean number of polyester microfibers per \(250 \mathrm{~mL}\) of beach sediment. (b) What is the margin of error? (c) If we want a margin of error of only ±1 with \(99 \%\) confidence, what sample size is needed?

For each scenario, use the formula to find the standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) Samples of size 100 from Population 1 with mean 87 and standard deviation 12 and samples of size 80 from Population 2 with mean 81 and standard deviation 15

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