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Chapter 7: Problem 8

A sample of size \(n=100\) is taken from a population that has a proportion \(p=1 / 5\) a. Find \(\delta\) such that \(P(|\hat{p}-p| \geq \delta)=0.025\) b. If, in the sample, \(\hat{p}=0.25,\) will the \(95 \%\) confidence interval for \(p\) contain the true value of \(p ?\)

Short Answer

Expert verified
a. \(\delta = 0.0784\); b. Yes, the confidence interval contains the true value.

Step by step solution

01

Understand the Problem Context

We have a sample of size \(n = 100\) from a population with a true proportion \(p = \frac{1}{5} = 0.2\). We need to find \(\delta\) such that the probability of the sample proportion \(|\hat{p} - p| \geq \delta\) is \(0.025\), and then assess whether a \(95\%\) confidence interval would capture the true proportion \(p\) if the sample proportion \(\hat{p} = 0.25\).
02

Recognize the Distribution of the Sample Proportion

The sample proportion \(\hat{p}\) follows a normal distribution \(N(p, \frac{p(1-p)}{n})\) for large samples. Here, \(p = 0.2\) and \(n = 100\), so the standard deviation is \(\sqrt{\frac{0.2 \times (1 - 0.2)}{100}} = 0.04\).
03

Set Up the Probability Equation

We are given \(P(|\hat{p} - p| \geq \delta) = 0.025\). By using the complement, we have \(P(-\delta < \hat{p} - p < \delta) = 0.975\). For a normal distribution, this is equivalent to finding the values that capture the middle 97.5% of the distribution.
04

Calculate the Z-Score for 2.5% Tail

The two-sided tail of 2.5% corresponds to a Z-score of approximately 1.96 (from the standard normal distribution table). Thus, \(\delta = 1.96 \times 0.04 = 0.0784\).
05

Determine the Confidence Interval

For \(\hat{p} = 0.25\), a \(95\%\) confidence interval for \(p\) is given by \((\hat{p} - 1.96 \times 0.04, \hat{p} + 1.96 \times 0.04) = (0.172, 0.328)\).
06

Check if True Proportion is Within the Interval

The true proportion \(p = 0.2\) falls within the confidence interval (0.172, 0.328). This means that the interval does indeed contain the true proportion.

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Key Concepts

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

Sample Proportion
When we collect data from a population, rather than investigate the entire group, we look at a portion known as the sample. In our problem, we took a sample of size 100 from the population.
The sample proportion, represented as \( \hat{p} \), is the fraction of the sample that has a particular characteristic we are interested in. It estimates the true proportion \( p \) in the whole population.
For instance, if in our sample, 25 out of 100 have the characteristic we're studying, then \( \hat{p} = 0.25 \).
  • It's important to know that sample proportion is just an estimate and it can vary from one sample to another.
  • The closer \( \hat{p} \) is to \( p \), the better your sample models the whole population.
Normal Distribution
The concept of normal distribution is crucial when it comes to understanding statistics and data analysis.
It describes how data is spread out across a mean, often in a symmetrical, bell-shaped curve. In large samples, the sample proportion \( \hat{p} \) tends to follow a normal distribution.
This happens because of a principle known as the Central Limit Theorem.
  • Normal distribution has a mean \( \mu \) and a standard deviation \( \sigma \), where most values cluster around the mean.
  • For our sample, the mean is the true proportion \( p = 0.2 \) and the standard deviation gets calculated as \( \sqrt{\frac{p(1-p)}{n}} \).
  • A normal distribution allows us to use various statistical techniques like confidence intervals and hypothesis testing.
Z-Score
Z-score is a statistical measure that expresses the number of standard deviations a data point is from the mean.
In the context of confidence intervals, the Z-score helps determine how far a sample's proportion \( \hat{p} \) can be expected to be from the true proportion \( p \).
  • A Z-score of 0 means the sample proportion is exactly the mean.
  • In our problem, the Z-score of 1.96 was used to find the interval where 97.5% of data lies, helping capture the middle part of the normal bell curve.
  • By knowing the Z-score, we can calculate the Margin of Error (MoE) using \( Z \times \sigma \), which helps us form the confidence interval.
This concept is essential because it translates probabilistic models into actionable insights in real-world scenarios.
Probability
Probability is the branch of mathematics concerned with the likelihood of an event occurring. It's a fundamental concept when dealing with statistics and predictions.
In our context, we had to find \( \delta \) so that the probability \( P( |\hat{p} - p| \geq \delta ) = 0.025 \). This probability tells us about the statistical guarantee of our findings.
  • Probability values are between 0 and 1, where 0 means an event will not occur, and 1 means it definitely will occur.
  • A probability of 0.025 signifies a rare event, occurring just 2.5% of the time.
  • This helps us understand how likely our sample's proportion might miss the real proportion.
Probability empowers us to make educated guesses and decisions based on available data.

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Most popular questions from this chapter

In a wildlife survey, an area of desert land was divided into 1000 squares, or "quadrats," a simple random sample of 50 of which were surveyed. In each surveyed quadrat, the number of birds, \(Y,\) and the area covered by vegetation, \(X\) were determined. It was found that $$\begin{aligned}\sum X_{i} &=3000 \\\\\sum & Y_{i}=150 \\\\\sum X_{i}^{2} &=225,000 \\ \sum Y_{i}^{2} &=650 \\\\\sum X_{i} Y_{i} &=11,000\end{aligned}$$ a. Estimate the ratio of the average number of birds per quadrat to the average vegetation cover per quadrat. b. Estimate the standard error of your estimate and find an approximate \(90 \%\) confidence interval for the population average. c. Estimate the total number of birds and find an approximate \(95 \%\) confidence interval for the population total. d. Suppose that from an aerial survey, the total area covered by vegetation could easily be determined. How could this information be used to provide another estimate of the number of birds? Would you expect this estimate to be better than or worse than that found in part (c)?

A photograph of a large crowd on a beach is taken from a helicopter. The photo is of such high resolution that when sections are magnified, individual people can be identified, but to count the entire crowd in this way would be very time consuming. Devise a plan to estimate the number of people on the beach by using a sampling procedure.

Suppose that the population size \(N\) is not known, but it is known that \(n \leq N\) Show that the following procedure will generate a simple random sample of size \(n .\) Imagine that the population is arranged in a long list that you can read sequentially. a. Let the sample initially consist of the the first \(n\) elements in the list. b. For \(k=1,2, \ldots,\) as long as the end of the list has not been encountered: i. Read the \((n+k)\) -th element in the list. ii. Place it in the sample with probability \(n /(n+k)\) and, if it is placed in the sample, randomly drop one of the existing sample members.

This problem introduces the concept of a one-sided confidence interval. Using the central limit theorem, how should the constant \(k\) be chosen so that the interval \(\left(-\infty, \bar{X}+k s_{\bar{X}}\right)\) is a \(90 \%\) confidence interval for \(\mu-\) i.e., so that \(P(\mu \leq \bar{X}+\) \(\left.k s_{\bar{X}}\right)=.9 ?\) This is called a one-sided confidence interval. How should \(k\) be chosen so that \(\left(\bar{X}-k s_{\bar{X}}, \infty\right)\) is \(95 \%\) one-sided confidence interval?

True or false? a. The center of a \(95 \%\) confidence interval for the population mean is a random variable. b. A \(95 \%\) confidence interval for \(\mu\) contains the sample mean with probability .95 c. A \(95 \%\) confidence interval contains \(95 \%\) of the population. d. Out of one hundred \(95 \%\) confidence intervals for \(\mu, 95\) will contain \(\mu\).
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