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91Ó°ÊÓ

Multiple choice: Why z? The reason we use a z-score from a normal distribution in constructing a large-sample confidence interval for a proportion is that a. For large random samples the sampling distribution of the sample proportion is approximately normal. b. The population distribution is normal. c. For large random samples the data distribution is approximately normal. d. For any \(n\) we use the \(t\) distribution to get a confidence interval, and for large \(n\) the \(t\) distribution looks like the standard normal distribution.

Short Answer

Expert verified
The correct option is a: For large random samples, the sampling distribution of the sample proportion is approximately normal.

Step by step solution

01

Understanding the Question

The question asks why we use a z-score when constructing a large-sample confidence interval for a proportion. We need to understand the properties of sample proportions and their distributions.
02

Review of Sampling Distribution

For a large sample size, the sampling distribution of the sample proportion \( \hat{p} \) becomes approximately normal, due to the Central Limit Theorem. This applies even if the population distribution itself is not normal, as long as the sample size is sufficiently large.
03

Evaluate Each Option

Evaluate each option against the understanding from Step 2: - Option a describes the sampling distribution of the sample proportion becoming approximately normal for large samples. - Option b implies the population distribution is normal, which is not required. - Option c is incorrect as it suggests the data distribution is approximately normal, not the sampling distribution. - Option d discusses the t-distribution, which is used for small samples and not exactly relevant for large samples with proportions.
04

Select the Correct Answer

Option a, which states that for large random samples the sampling distribution of a sample proportion is approximately normal, aligns with the need for using z-scores.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem is a powerful statistical concept, crucial for understanding why we can use certain methods in statistics, especially when it comes to distributions. Essentially, it states that when you have a large enough sample size, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the original distribution of the population. This is transformative because:
  • It allows statisticians to make inferences about a population from a sample.
  • Even if the population distribution is skewed or unknown, a large enough sample will yield a normal distribution of sample means.
In practical terms, this is why the Central Limit Theorem is so fundamental. It underpins methods like constructing confidence intervals and hypothesis testing, making it easier to draw meaningful conclusions from sample data.
The beauty of the Central Limit Theorem lies in its simplicity and robustness. It works nearly universally, assuming your sample size is adequately large.
Sampling Distribution
Sampling distribution is another core concept that is directly tied to the Central Limit Theorem. It refers to the probability distribution of a given statistic based on a random sample. For example, if you repeatedly take samples and calculate the mean of each, the distribution of those means is called the sampling distribution of the mean.
Several characteristics are key to understanding sampling distributions:
  • Mean: The average of the sampling distribution will generally equal the mean of the population.
  • Standard Deviation: Known as the standard error, it’s typically smaller than the population standard deviation and decreases with larger sample sizes.
  • Shape: According to the Central Limit Theorem, for large enough samples, this distribution will be normal, even if the population distribution is not.
Sampling distribution is crucial in understanding how sample statistics can be used to make inferences about population parameters, a foundational aspect of statistical analysis.
Confidence Interval
A confidence interval provides a range of values that is likely to contain a population parameter, like a mean or proportion. When we talk about confidence intervals, we're really discussing how sure we are that the interval captures the true parameter value.
For example, a 95% confidence interval means that if we were to take 100 different samples and compute a confidence interval for each, we would expect about 95 of them to contain the true parameter. Key aspects of confidence intervals include:
  • Level of Confidence: The probability that the interval contains the true parameter. Higher confidence levels result in wider intervals.
  • Interval Width: Reflects the uncertainty about the parameter estimate. Influenced by sample size – larger samples produce narrower intervals for the same confidence level.
  • Formula: For proportions, a common formula is \[ \hat{p} \pm z \left( \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right) \]where \( \hat{p} \) is the sample proportion, \( z \) is the z-score, and \( n \) is the sample size.
Confidence intervals are a primary tool in statistics for assessing estimation precision and reliability.
z-score for Proportions
The z-score for proportions is a measure used when estimating percentages or proportions in a population. It's particularly useful for creating confidence intervals and hypothesis tests for proportion metrics.
  • Why Z-Scores?: When you have a large sample size, the sampling distribution of the sample proportion tends to be normally distributed—which is where the z-score comes into play.
  • Calculation: A z-score indicates how many standard deviations an element is from the mean, calculated by the formula \[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \]where \( \hat{p} \) is the observed sample proportion, \( p \) is the hypothetical population proportion you're testing against, and \( n \) is the sample size.
  • Applying Z-Scores: They make it easier to determine the probability of observing an extreme value and are used to derive confidence intervals for proportions.
Understanding z-scores for proportions helps statisticians accurately assess and predict population behaviors based on sample data, which is fundamental for data-driven decision-making.

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Most popular questions from this chapter

Heights of seedlings Exercise 8.6 reported heights (in \(\mathrm{mm}\) ) of \(55.5,60.3,60.6,62.1,65.5,\) and 69.2 for six seedlings fourteen days after germination. a. Using software or a calculator, verify that the \(95 \%\) confidence interval for the population mean is (57.3,67.1) . b. Name two things you could do to get a narrower interval than the one in part a. c. Construct a \(99 \%\) confidence interval. Why is it wider than the \(95 \%\) interval? d. On what assumptions is the interval in part a based? Explain how important each assumption is.

Accept a credit card? A bank wants to estimate the proportion of people who would agree to take a credit card they offer if they send a particular mailing advertising it. For a trial mailing to a random sample of 100 potential customers, 0 people accept the offer. Can they conclude that fewer than \(10 \%\) of their population of potential customers would take the credit card? Answer by finding an appropriate \(95 \%\) confidence interval.

Number of children For the question, "How many children have you ever had?" use the GSS Web site sda berkeley.edu/GSS with the variable CHILDS to find the sample mean and standard deviation for the 2008 survey. a. Show how to obtain a standard error of 0.04 for a random sample of 2020 adults. b. Construct a \(95 \%\) confidence interval for the population mean. Can you conclude that the population mean is less than \(2.0 ?\) Explain. c. Discuss the assumptions for the analysis in part \(\mathrm{b}\) and whether that inference seems to be justified.

t-scores a. Show how the \(t\) -score for a \(95 \%\) confidence interval changes as the sample size increases from 10 to 20 to 30 to infinity. b. What does the answer in part a suggest about how the \(t\) distribution compares to the standard normal distribution?

Why called "degrees of freedom"? You know the sample mean \(\bar{x}\) of \(n\) observations. Once you know \((n-1)\) of the observations, show that you can find the remaining one. In other words, for a given value of \(\bar{x},\) the values of \((n-1)\) observations determine the remaining one. In summarizing scores on a quantitative variable, having \((n-1)\) degrees of freedom means that only that many observations are independent. (If you have trouble with this, try to show it for \(n=2,\) for instance showing that if you know that \(\bar{x}=80\) and you know that one observation is \(90,\) then you can figure out the other observation. The \(d f\) value also refers to the divisor in \(\left.s^{2}=\Sigma(x-\bar{x})^{2} /(n-1) .\right)\)

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