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Suppose a sample of size \(n\) is to be drawn from a normal distribution where \(\sigma\) is known to be 14.3. How large does \(n\) have to be to guarantee that the length of the \(95 \%\) confidence interval for \(\mu\) will be less than \(3.06 ?\)

Short Answer

Expert verified
The sample size 'n' required to guarantee that the length of the 95% confidence interval for the mean will be less than 3.06 with a known standard deviation of 14.3 is 81.

Step by step solution

01

Identify the given values

Identify all the given information from the problem. In this case, the standard deviation (σ) is given as 14.3, the Z-score (Z) at a 95% confidence level is approximately 1.96 (found from a standard normal distribution table or calculator), and the desired confidence interval length is less than 3.06.
02

Set up the formula

Set up the formula for the length of the confidence interval and substitute the known values into the formula. In this case, we have the formula:\(Length = Z \times \frac{2 \sigma}{\sqrt{n}}\)Insert the desired length, Z-score and the standard deviation into the equation:\(3.06 > 1.96 \times \frac{2\times 14.3}{\sqrt{n}}\)
03

Solve for n

Rearrange the equation and solve for n. This results in a quadratic equation that can be solved by algebraic methods. Start by dividing both sides by 1.96 and inverting the inequality (it changes direction because we divide by a number under the root which is in the denominator):\(\frac{3.06}{1.96} < \frac{2\times 14.3}{\sqrt{n}}\)Then square both sides and solve for n:\(n > \left(\frac{2\times 14.3}{3.06/1.96}\right)^2\)
04

Calculate n

Calculate the value for n. Since 'n' has to be a whole number (you can't sample a fraction of a subject), round up to the next highest whole number. This gives the smallest n that satisfies the condition.\[\text{We get: } n > 80.85\text{, so } n = 81 \]

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Key Concepts

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

Normal Distribution
Understanding the normal distribution is crucial when analyzing data that is assumed to follow a bell-shaped curve, which is common in natural and social phenomena. A normal distribution, also known as a Gaussian distribution, is characterized by its symmetric 'bell curve' shape, where the mean, median, and mode of the distribution are equal.

The normal distribution is important in statistics because it allows for the determination of probabilities and the calculation of confidence intervals for population parameters. In the given exercise, the sample is drawn from a population that follows a normal distribution, which enables the use of Z-scores to find the confidence interval for the population mean \( \mu \). When sample data is normally distributed, the tools of inferential statistics, such as confidence intervals, become more applicable and reliable.
Sample Size Determination
Determining the right sample size is key to obtaining accurate estimates while conducting statistical analyses. Sample size determination involves choosing the number of observations or replicates to include in a statistical sample. The goal is to make a sample as representative of the population as possible, within the constraints of time and cost.

Several factors influence sample size determination, including the desired confidence level, the acceptable margin of error, and the population's standard deviation. In the exercise, the objective is to find the smallest sample size (\( n \) that ensures that the confidence interval for the population mean does not exceed a specified width. A larger sample size typically results in a narrower confidence interval, which implies a more precise estimate of the population mean.
Standard Deviation
Standard deviation (\( \sigma \) is a critical measure in statistics that quantifies the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean (also called the expected value) of the set, while a high standard deviation indicates that the values are spread out over a wider range.

In the context of the exercise, the known standard deviation (14.3) represents the variability in the population. Accurate knowledge of the standard deviation is necessary for creating confidence intervals, as it directly impacts the width of the interval. The standard deviation is a part of the formula used to calculate the required sample size to achieve a certain precision for the interval.
Z-score
A 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. The Z-score is a crucial tool in statistics as it allows for standardization, enabling the comparison of scores from different data sets. Furthermore, it's instrumental in the calculation of the probability of a score occurring within a normal distribution and is a key part of finding confidence intervals.

In this exercise, the Z-score represents the standardized distance from the mean required to capture the central 95% of the normal distribution. The Z-score for a 95% confidence level is approximately 1.96, meaning that the sample mean is expected to fall within 1.96 standard deviations of the population mean 95% of the time. This value is utilized in the formula to determine the minimum sample size needed to ensure that the confidence interval width remains within acceptable limits.

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Most popular questions from this chapter

Suppose that \(W_{1}\) is a random variable with mean \(\mu\) and variance \(\sigma_{1}^{2}\) and \(W_{2}\) is a random variable with mean \(\mu\) and variance \(\sigma_{2}^{2}\). From Example 5.4.3, we know that \(c W_{1}+(1-c) W_{2}\) is an unbiased estimator of \(\mu\) for any constant \(c>0\). If \(W_{1}\) and \(W_{2}\) are independent, for what value of \(c\) is the estimator \(c W_{1}+(1-c) W_{2}\) most efficient?

In 1927 , the year he hit sixty home runs, Babe Ruth batted 356 , having collected 192 hits in 540 official at-bats (150). Based on his performance that season, construct a \(95 \%\) confidence interval for Ruth's probability of getting a hit in a future at-bat.

Suppose a random sample of size \(n\) is drawn from a normal pdf where the mean \(\mu\) is known but the variance \(\sigma^{2}\) is unknown. Use the method of maximum likelihood to find a formula for \(\theta^{2}\). Compare your answer to the maximum likelihood estimator found in Example \(5.2 .5 .\)

Consider, again, the scenario described in Example \(5.8 .2\) - a binomial random variable \(X\) has parameters \(n\) and \(\theta\), where the latter has a beta prior with integer parameters \(r\) and \(s\). Integrate the joint pdf \(p_{X}(k \mid \theta) f_{\theta}(\theta)\) with respect to \(\theta\) to show that the marginal pdf of \(X\) is given by $$ p_{X}(k)=\frac{\left(\begin{array}{c} k+r-1 \\ k \end{array}\right)\left(\begin{array}{c} n-k+s-1 \\ n-k \end{array}\right)}{\left(\begin{array}{c} n+r+s-1 \\ n \end{array}\right)}, \quad k=0,1, \ldots, n $$

. Five hundred adults are asked whether they favor a bipartisan campaign finance reform bill. If the true proportion of the electorate is \(52 \%\) in favor of the legislation, what are the chances that fewer than half of those in the sample support the proposal? Use a \(Z\) transformation to approximate the answer.

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