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In Exercises 6.111 to \(6.114,\) what sample size is needed to give the desired margin of error in estimating a population mean with the indicated level of confidence? A margin of error within ±5 with \(95 \%\) confidence, assuming a previous sample had \(s=18\)

Short Answer

Expert verified
In order to attain a margin of error within ±5 with 95% confidence, assuming a previous sample had s = 18, a sample size of 19 is required.

Step by step solution

01

Understand the Concepts

In this problem, we are working with the concept of margin of error which is a measure of accuracy in statistical study. We are provided with a margin of error, a confidence level, and a standard deviation and we are asked to find out the required sample size.
02

Use the Formula for Margin of Error

We are going to use the formula for the margin of error: \(E = z \cdot \frac{s}{\sqrt{n}}\) here, where E is the margin of error, z is the z-score, s is the sample standard deviation, and n is the sample size. From the given problem, we know that our margin of error (E) is ±5, our confidence level is 95% (this corresponds to a z-score of 1.96 which is found on a z-score table), and our standard deviation (s) from the previous sample is 18.
03

Solving for Sample Size

Substitute the given values into the formula and then solve for n: 5 = 1.96(18)/\(\sqrt{n}\). Square both sides to get rid of the square root: 25 = 1.38416n. Then, divide each side by 1.38416 to solve for n: n ≈ 18.06. Since we can't have a fraction of a sample, we always round up to ensure our sample is large enough. So we need a sample size of 19.

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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 sample survey, one critical aspect is estimating the population mean. The 'margin of error' plays a vital role in this estimation. It represents a range above and below the sample mean in which the true population mean is expected to fall. Imagine it as a safety buffer around the result of a survey that tells you how close your sample results are likely to be to the actual population value.

In the context of our exercise, a margin of error of ±5 means the true population mean is expected to be no more than 5 units away from our sample mean, either above or below it. To calculate it mathematically, the standard formula used is: \(E = z \cdot \frac{s}{\sqrt{n}}\), where \(E\) stands for margin of error, \(z\) represents the z-score correlated to our confidence level, \(s\) is the standard deviation, and \(n\) is the sample size. By fixing the margin of error, we constrain the range within which we expect the true mean to fall, which is essential for ensuring the precision of a study's findings.
Confidence Level
The confidence level is another essential concept in statistics, which signals how sure we can be about our estimates. It's a measure of reliability, often expressed as a percentage, like in our exercise where we have a 95% confidence level. This value indicates that we are 95% certain our interval contains the true population mean.

The confidence level corresponds to a z-score, which is a statistical measurement that describes a value's relationship to the mean of a group of values. For a 95% confidence level, the z-score is approximately 1.96. This means that if we were to take many samples, about 95% of the time, the true population mean would fall within the margin of error we've calculated around the sample mean. Selecting a confidence level is a balance between certainty and the width of the confidence interval: a higher confidence level requires a larger margin of error or a larger sample size to maintain the same margin of error.
Standard Deviation
Standard deviation is a measure that tells us how much the values in a set of data vary, on average, from the mean of the data set. It is a core concept in statistics that gives us an idea of the spread or dispersion of the sample. The larger the standard deviation, the more spread out the data points are.

In the exercise, the standard deviation (\(s\)) is given as 18, which is a previous sample's calculation. This number helps us determine the variability of the population from which the sample was drawn. When we calculate the sample size needed for a specific margin of error, the standard deviation is pivotal as it directly impacts the sample size needed. A higher standard deviation would typically require a larger sample size to achieve the same margin of error for the same confidence level.

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Most popular questions from this chapter

According to the 2006 Australia Census, \(^{43} 25.5 \%\) of Australian women over the age of 25 had a college degree, while the percentage for Australian men was \(21.4 \% .\) Suppose we select random samples of 200 women and 200 men from this population and look at the differences in proportions with college degrees, \(\hat{p}_{f}-\hat{p}_{m}\), in those samples. (a) Describe the distribution (center, spread,shape) for the difference in sample proportions. Include a rough sketch of the distribution with values labeled on the horizontal axis. (b) What is the chance that the proportion with college degrees in the men's sample is actually more than the proportion in the women's sample? (Hint: Think about what must be true about \(\hat{p}_{f}-\hat{p}_{m}\) when this happens.)

Is a Normal Distribution Appropriate? In Exercises 6.13 and \(6.14,\) indicate whether the Central Limit Theorem applies so that the sample proportions follow a normal distribution. In each case below, does the Central Limit Theorem apply? (a) \(n=80\) and \(p=0.1\) (b) \(n=25\) and \(p=0.8\) (c) \(n=50\) and \(p=0.4\) (d) \(n=200\) and \(p=0.7\)

Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=-2.6, s_{d}=4.1\) \(n_{d}=18\)

Ron flips a coin \(n_{1}\) times and Freda flips a coin \(n_{2}\) times. We can assume all coin flips are fair: The coin has an equal chance of landing heads or tails. In each of the following cases, state whether inference for a difference in proportions is appropriate using the methods of this section. If so, give the mean and standard error for the distribution of the difference in proportions \(\left(\hat{p}_{1}-\hat{p}_{2}\right)\) and state whether the normal approximation is appropriate. (a) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Freda's flips that land heads; \(n_{1}=100\) and \(n_{2}=50\). (b) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Ron's flips that land tails; \(n_{1}=100\). (c) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Freda's flips that land tails; \(n_{1}=200\) and \(n_{2}=200\). (d) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land tails and \(\hat{p}_{2}\) be the proportion of Freda's flips that land tails; \(n_{1}=5\) and \(n_{2}=10\).

In the mid-1990s a Nabisco marketing campaign claimed that there were at least 1000 chips in every bag of Chips Ahoy! cookies. A group of Air Force cadets collected a sample of 42 bags of Chips Ahoy! cookies, bought from locations all across the country, to verify this claim. \({ }^{41}\) The cookies were dissolved in water and the number of chips (any piece of chocolate) in each bag were hand counted by the cadets. The average number of chips per bag was \(1261.6,\) with standard deviation 117.6 chips. (a) Why were the cookies bought from locations all over the country? (b) Test whether the average number of chips per bag is greater than 1000 . Show all details. (c) Does part (b) confirm Nabisco's claim that every bag has at least 1000 chips? Why or why not?

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