/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 What is the formula for the stan... [FREE SOLUTION] | 91Ó°ÊÓ

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What is the formula for the standard error of the normal approximation to the \(\hat{p}\) distribution? What is the mean of the \(\hat{p}\) distribution?

Short Answer

Expert verified
The standard error of \( \hat{p} \) is \( \sqrt{\frac{p(1-p)}{n}} \) and its mean is \( p \).

Step by step solution

01

Understanding the Normal Approximation to the sampling distribution of Sample Proportion \( \hat{p} \)

The sample proportion \( \hat{p} \) is used to estimate the true population proportion \( p \). When the sample size is large enough, the distribution of \( \hat{p} \) can be approximated by a normal distribution due to the Central Limit Theorem.
02

Formula for the Standard Error of \( \hat{p} \)

The standard error of the sample proportion \( \hat{p} \) measures how much \( \hat{p} \) is expected to vary from sample to sample. It is computed by the formula:\[ SE(\hat{p}) = \sqrt{\frac{p(1-p)}{n}} \]where \( n \) is the sample size, and \( p \) is the population proportion.
03

Identifying the Mean of the \( \hat{p} \) Distribution

The mean of the \( \hat{p} \) distribution is the population proportion \( p \), because \( \hat{p} \) is an unbiased estimator of \( p \). Hence, the expectation or mean of \( \hat{p} \) is \( p \):\[ \text{Mean of } \hat{p} = p \]

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Key Concepts

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

Standard Error
When we talk about the standard error in the context of sampling distributions, it's essentially a measure of how spread out the sample statistics are around the true population parameter. In simple terms, it's like a "yardstick" that tells us how much an estimate from a sample might differ from the true value in the whole population. This is crucial in statistics, especially when dealing with the variability of sample proportions.
For the sample proportion \( \hat{p} \), the standard error is calculated using the formula: \[ SE(\hat{p}) = \sqrt{\frac{p(1-p)}{n}} \] Here, \( p \) is the population proportion, and \( n \) is the sample size. The standard error decreases as the sample size increases, indicating that larger samples provide more precise estimates.
Normal Approximation
The concept of normal approximation arises from the Central Limit Theorem, which is a fundamental theorem in statistics. This theorem implies that when the sample size is sufficiently large, the distribution of the sample proportion \( \hat{p} \) becomes approximately normal. This is true even if the population from which the sample is drawn does not have a normal distribution.
Normal approximation is particularly useful because it allows us to apply methods designed for normal distributions, simplifying analysis and calculations. This approximation is appropriate when:
  • \( np \geq 5 \)
  • \( n(1-p) \geq 5 \)
These conditions ensure that the sample size is large enough for the normal approximation to be valid.
Sample Proportion
The sample proportion \( \hat{p} \) is an estimator of the true proportion \( p \) in the population. It is calculated by dividing the number of successes by the total number of observations in the sample. This statistic provides a practical way to estimate population parameters based on a smaller subset of data.
More formally, if a sample contains \( x \) successes and \( n \) total observations, the sample proportion is given by: \[ \hat{p} = \frac{x}{n} \] The sample proportion \( \hat{p} \) is an unbiased estimator of the population proportion, meaning on average, it provides a true reflection of \( p \). Likewise, the mean of the \( \hat{p} \) distribution equals the population proportion \( p \). This unbiased quality enhances the reliability of \( \hat{p} \) when making inferences about the population.

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Most popular questions from this chapter

Another application for exponential distributions (see Problem 18 ) is supply/demand problems. The operator of a pumping station in a small Wyoming town has observed that demand for water on a typical summer afternoon is exponentially distributed with a mean of 75 cfs (cubic feet per second). Let \(x\) be a random variable that represents the town's demand for water (in cfs). What is the probability that on a typical summer afternoon, this town will have a water demand \(x\) (a) more than 60 cfs (i.e., \(60 < x < \infty\) )? Hint: \(e^{-\infty}=0\) (b) less than 140 cfs (i.e., \(0 < x < 140\) )? (c) between 60 and 100 cfs? (d) Brain teaser How much water \(c\) (in cfs) should the station pump to be \(80 \%\) sure that the town demand \(x\) (in cfs) will not exceed the supply \(c ?\) Hint: First explain why the equation \(P(0< x < c)=0.80\) represents the problem as stated. Then solve for \(c .\)

Maintenance The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was observed over a long period of time to be approximately normally distributed, with mean \(\mu=\$ 615\) and standard deviation \(\sigma=\$ 42\). (a) If \(\$ 646 dollars is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount? (b) Inverse Normal Distribution How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only \)0.10 ?$

Porphyrin is a pigment in blood protoplasm and other body fluids that is significant in body energy and storage. Let \(x\) be a random variable that represents the number of milligrams of porphyrin per deciliter of blood. In healthy adults, \(x\) is approximately normally distributed with mean \(\mu=38\) and standard deviation \(\sigma=12\) (see reference in Problem 25 ). What is the probability that : (a) \(x\) is less than \(60 ?\) (b) \(x\) is greater than \(16 ?\) (c) \(x\) is between 16 and \(60 ?\) (d) \(x\) is more than \(60 ?\) (This may indicate an infection, anemia, or another type lness.)

The manager of Motel 11 has 316 rooms in Palo Alto, California. From observation over a long period of time, she knows that on an average night, 268 rooms will be rented. The long-term standard deviation is 12 rooms. This distribution is approximately mound-shaped and symmetric. (a) For 10 consecutive nights, the following numbers of rooms were rented each night: $$\begin{array}{l|cccccc} \hline \text { Night } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Number of rooms } & 234 & 258 & 265 & 271 & 283 & 267 \\ \hline & & & & & & \\ \hline \text { Night } & 7 & 8 & 9 & 10 & & \\ \hline \text { Number of rooms } & 290 & 286 & 263 & 240 & & \\ \hline \end{array}$$ Make a control chart for the number of rooms rented each night, and plot the preceding data on the control chart. Interpretation Looking at the control chart, would you say the number of rooms rented during this 10-night period has been unusually low? unusually high? about what you expected? Explain your answer. Identify all out-of-control signals by type (I, II, or III). Make a control chart for the number of rooms rented each night, and plot the preceding data on the control chart. Interpretation Looking at the control chart, would you say the number of rooms rented during this 10-night period has been unusually low? unusually high? about what you expected? Explain your answer. Identify all out-of-control signals by type (I, II, or III). (b) For another 10 consecutive nights, the following numbers of rooms were rented each night: $$\begin{array}{l|cccccc} \hline \text { Night } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Number of rooms } & 238 & 245 & 261 & 269 & 273 & 250 \\ \hline & & & & & & \\ \hline \text { Night } & 7 & 8 & 9 & 10 & & \\ \hline \text { Number of rooms } & 241 & 230 & 215 & 217 & & \\ \hline \end{array}$$ Make a control chart for the number of rooms rented each night, and plot the preceding data on the control chart. Would you say the room occupancy has been high? low? about what you expected? Explain your answer. Identify all out-of- control signals by type (I, II, or IIII).

Find the indicated probability, and shade the corresponding area under the standard normal curve. $$P(-2.37 \leq z \leq 0)$$

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