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Let \(\mu\) denote the true \(\mathrm{pH}\) of a chemical compound. A sequence of \(n\) independent sample pH determinations will be made. Suppose each sample \(\mathrm{pH}\) is a random variable with expected value \(\mu\) and standard deviation .1. How many determinations are required if we wish the probability that the sample average is within \(.02\) of the true \(\mathrm{pH}\) to be at least \(.95\) ? What theorem justifies your probability calculation?

Short Answer

Expert verified
97 determinations are required, justified by the Central Limit Theorem.

Step by step solution

01

Identify the problem statement

We are trying to find the sample size \(n\) needed to ensure that the probability of the sample average being within \(0.02\) of the true pH is at least \(0.95\). This involves assessing the sampling distribution of the sample mean.
02

Determine applicable statistical concept

The problem involves calculating the required sample size to achieve a specific probability, relating to the sample mean's proximity to the true mean. According to the Central Limit Theorem, the distribution of the sample mean will be approximately normal with mean \(\mu\) and standard deviation \(\sigma/\sqrt{n}\), where \(\sigma\) is the standard deviation of the sample.
03

Apply the Central Limit Theorem

Given that the standard deviation of the sample is \(0.1\), the standard deviation of the sample mean will be \(\frac{0.1}{\sqrt{n}}\). The probability question translates to a normal distribution problem involving the distance of the sample mean from the true mean \(\mu\).
04

Use the Z-score for the probability

We want \(P(|\bar{X} - \mu| < 0.02) \geq 0.95\). For a normal distribution, \(P(|Z| < 1.96) \approx 0.95\), where \(Z\) is the standard normal variable. Therefore, solve for \(n\) in the inequality \(1.96 \times \frac{0.1}{\sqrt{n}} \leq 0.02\).
05

Solve for sample size n

Rearrange the inequality \[ 1.96 \times \frac{0.1}{\sqrt{n}} \leq 0.02 \] to find \(n\). Multiply both sides by \(\sqrt{n}\) and divide by \(0.02\), we get \[ \sqrt{n} \geq \frac{1.96 \times 0.1}{0.02} \]. Calculate \(n\): \[ \sqrt{n} \geq 9.8 \] therefore, \(n \geq 96.04\). Since \(n\) must be a whole number, round up to \(n = 97\).
06

Identify relevant theorem

The Central Limit Theorem is the theorem that justifies this probability calculation by approximating the distribution of the sample mean as a normal distribution when \(n\) is large.

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Key Concepts

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

Sample Size Calculation
When determining the required number of samples, we are calculating the sample size needed to ensure a desired level of accuracy in our estimation. In this case, we need our sample mean to be within a specific range of the true mean with high probability. More specifically, we're ensuring that our sample mean is within 0.02 units of the true population mean (\(\bar{X} \approx \mu\)).The steps include:
  • Identifying the precision needed (here, within 0.02 of the true mean).
  • Using the given probability (at least 0.95) and standard deviation of the sample (\(0.1\)).
  • Integrating these requirements with the properties of the normal distribution via the Z-score.
Finally, the calculated sample size should be rounded up to the nearest whole number to assure that the precision requirement is met. In our exercise, we found that 97 samples are required.
Sampling Distribution
The concept of a sampling distribution can be perplexing, but it's essential for understanding the behavior of sample means. Essentially, a sampling distribution is the probability distribution of a statistic (like a sample mean) based on a random sample. Consider taking multiple samples from the same population; each will have its own sample mean. The distribution of these sample means forms the sampling distribution. According to the Central Limit Theorem, as the sample size increases, this distribution becomes normal, even if the underlying data is not.
  • It focuses on the mean of a sample rather than the data itself.
  • The standard deviation of this distribution (\(\sigma/\sqrt{n}\)) shrinks as sample size increases.
Understanding sampling distributions is crucial for determining how accurate a sample mean is in representing the population mean.
Normal Distribution
Normal distribution is a fundamental statistical concept, often referred to as the "bell curve." It's essential in statistics because many phenomena in nature, like heights and test scores, naturally follow a normal distribution. Key properties are:
  • Symmetrical, with a peak at the mean (\(\mu\)).
  • The standard deviation (\(\sigma\)) determines the width of the curve.
  • Most data points lie within three standard deviations of the mean.
In the context of sampling, when we refer to the sampling distribution of the sample mean, we are often interested in how this distribution approximates normality (thanks to the Central Limit Theorem). Our exercise relies on this property to make informed calculations about sample size and precision.
Z-score
The Z-score is a dimensionless statistical measure that describes a data point's relation to the mean of a group of points. It's quantified in terms of standard deviations from the mean. In other words, a Z-score tells us how far, and in what direction, a particular value deviates from the mean.Key points about Z-scores:
  • A Z-score can be positive or negative, indicating whether the data point is above or below the mean.
  • In a normal distribution, the probability of a Z-score falling within certain ranges can be calculated easily (e.g., 95% of data lies within the range defined by Z=±1.96).
  • Z-scores help compare data points from different normal distributions.
In this exercise, we use the Z-score (\(Z = 1.96\)) to determine how large a sample needs to be to achieve our probability requirement. This score corresponds to the percentage of data (95%) within certain boundaries around the mean of the distribution.

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Most popular questions from this chapter

Suppose we take a random sample of size \(n\) from a continuous distribution having median 0 so that the probability of any one observation being positive is .5. We now disregard the signs of the observations, rank them from smallest to largest in absolute value, and then let \(W=\) the sum of the ranks of the observations having positive signs. For example, if the observations are \(-.3,+.7\), \(+2.1\), and \(-2.5\), then the ranks of positive observations are 2 and 3, so \(W=5\). In Chapter \(14, W\) will be called Wilcoxon's signed-rank statistic. W can be represented as follows: $$ \begin{aligned} W &=1 \cdot Y_{1}+2 \cdot Y_{2}+3 \cdot Y_{3}+\cdots+n \cdot Y_{n} \\ &=\sum_{i=1}^{n} i \cdot Y_{i} \end{aligned} $$ where the \(Y_{i}^{\prime}\) s are independent Bernoulli rv's, each with \(p=.5\left(Y_{i}=1\right.\) corresponds to the observation with rank \(i\) being positive). Compute the following: a. \(E\left(Y_{i}\right)\) and then \(E(W)\) using the equation for \(W\) [Hint: The first \(n\) positive integers sum to \(n(n+1) / 2 .]\) b. \(V\left(Y_{i}\right)\) and then \(V(W)\) [Hint: The sum of the squares of the first \(n\) positive integers is \(n(n+1)(2 n+1) / 6 .]\)

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