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In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 60 from a population with proportion 0.90

Short Answer

Expert verified
The mean of the distribution of sample proportions is 0.90, and the standard error is 0.0386. The Central Limit Theorem states that with a large sample size, the sample proportions will have approximately a normal distribution with the sample size of 60 being large enough.

Step by step solution

01

Calculate Mean

The mean (\(\mu\)) of the distribution of sample proportions is given by \(\mu = p\), where \(p\) is the population proportion. Here, \(p = 0.90\). Thus, \(\mu = 0.90\).
02

Compute Standard Error

The Standard Error (\(SE\)) of the distribution of sample proportions is given by \(SE = \sqrt{\frac{p(1-p)}{n}}\), where \(p\) is the population proportion and \(n\) is the sample size. Here, \(p = 0.90\) and \(n = 60\). So, \(SE = \sqrt{\frac{0.90(1-0.90)}{60}} = \sqrt{\frac{0.090}{60}} = 0.0386\).
03

Drawing Distribution Curve

According to the Central Limit Theorem, if the sample size is large enough, the distribution of the sample proportions is approximately normal. Here, \(n = 60\) is large enough. The mean (\(\mu = 0.90\)) lies in the center of the curve, and the standard error (\(SE = 0.0386\)) determines how spread out the data distribution is.

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Key Concepts

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

Mean of Sample Proportions
The mean of sample proportions is a central value used in statistics to understand the average outcome of different samples drawn from a given population. It is represented using the symbol \( \mu \).
Specifically, the mean \( \mu \) of sample proportions is equal to the population proportion \( p \). In simpler terms, if the population is known to possess a certain characteristic, the average proportion of samples possessing this characteristic will be equivalent to \( p \).
For example, if we have a population proportion \( p = 0.90 \), this means that 90% of the population shares a particular trait. Thus, the mean of the sample proportions \( \mu \) will also be 0.90.
  • The concept simplifies the analysis and allows us to make predictions about various samples drawn from the population.
  • It serves as a foundation in understanding the behavior of samples, especially when the sample size is reasonably large.
Standard Error
Standard error is a vital concept in statistics, which reflects the variability or dispersion of sample proportions around the true population proportion. It provides a measure of how much sample proportions are expected to vary from one sample to another.
The standard error \( SE \) of the sample proportion is calculated using the formula: \[ SE = \sqrt{\frac{p(1-p)}{n}} \]where:\( p \) is the population proportion and \( n \) is the sample size.
For a population proportion of 0.90 and sample size of 60, the standard error would be:\( SE = \sqrt{\frac{0.90 \times (1-0.90)}{60}} = 0.0386 \).
  • A smaller standard error indicates that the sample proportion is more consistent with the population proportion.
  • The larger the sample size, the smaller the standard error, implying more precise estimates of the population proportion.
Normal Distribution
Normal distribution is a key concept in statistics, often emerging when the Central Limit Theorem (CLT) is applied. The theorem states that, given a sufficiently large sample size, the sampling distribution of the sample mean will approximate a normal distribution, even if the underlying distribution is not normal.
In this context, for sample proportions, we apply the CLT to understand their distribution when samples are drawn.
With the Central Limit Theorem, if you have a large enough sample size (like \( n = 60 \) in our scenario), the distribution of the sample proportions will be approximately normal. This approximation holds true especially when \( n \times p \) and \( n \times (1-p) \) are both greater than 5.
This normal distribution is centered around the mean \( \mu = 0.90 \) and has a spread defined by the standard error \( SE = 0.0386 \). If you visualize this, you'd see a bell-shaped curve with the peak at 0.90 and a spread that spans about 0.0386 units above and below this mean.
  • The normal curve helps in making statistical inferences and predictions about the data.
  • It's a crucial tool for conducting hypothesis tests and constructing confidence intervals.

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Most popular questions from this chapter

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 \(p=0.7\)

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A sample with \(n=10, \bar{x}=508.5,\) and \(s=21.5\)

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