/*! 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 15 The superintendent of a large sc... [FREE SOLUTION] | 91Ó°ÊÓ

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The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter \(\lambda\). Use the accompanying data on absences for 50 days to derive a large-sample CI for \(\lambda\). [Hint: The mean and variance of a Poisson variable both cqual \(\lambda, s 0\) $$ Z=\frac{X-\lambda}{\sqrt{\lambda / n}} $$ has approximately a standard normal distribution. Now proceed as in the derivation of the interval for \(p\) by making a probability statement (with probability \(1-\alpha\) ) and solving the resulting inequalities for \(\lambda\) (see the argument just after \((8.10))\) ]. \begin{tabular}{l|lll} Number of absences & 0 & 1 & 2 \\ \hline Frequency & 1 & 4 & 8 \end{tabular}

Short Answer

Expert verified
The 95% CI for \(\lambda\) is approximately (1.07, 2.01).

Step by step solution

01

Calculate the Sample Mean

First, note the provided frequencies and calculate the sample mean (\(\bar{x}\)). The formula for the sample mean in this case is \(\bar{x} = \frac{\Sigma (x_i \cdot f_i)}{n}\), where \(x_i\) represents the number of absences and \(f_i\) the frequency. Here, we have \(x_i\) values: 0, 1, 2 and corresponding \(f_i\) values: 1, 4, 8. Calculate: \(\bar{x} = \frac{(0\times 1 + 1\times 4 + 2\times 8)}{13} = \frac{20}{13} \approx 1.54\).
02

Determine the Value of n

Determine the sample size \(n\). Sum the frequencies to find \(n\): \(n = 1 + 4 + 8 = 13\).
03

Compute the Sample Variance

For the Poisson distribution, the theoretical variance \(\lambda\) should equal the mean. However, let's compute the sample variance for completion: \(s^2 = \frac{\Sigma ((x_i - \bar{x})^2 \cdot f_i)}{n-1}\). This is just a consistency check with the expected variance \(\lambda\).
04

Setting Up the Confidence Interval

The given hint provides a formula for a standard normal variable: \(Z=\frac{X-\lambda}{\sqrt{\lambda / n}}\) is approximately standard normal. Instead, considering the large-sample CI for \(\lambda\), we use the normal approximation since both mean and variance are \(\lambda\).
05

Confidence Interval Formula

For a CI of \(1-\alpha\), utilize: \(\lambda = \bar{x} \pm z_{\alpha/2}\sqrt{\bar{x}/n}\). Using a standard normal table, find \(z_{\alpha/2}\) for a common confidence level, like 95%, \(\alpha/2 = 0.025\), \(z_{0.025} \approx 1.96\).
06

Calculate the Confidence Interval

Plug \(\bar{x} \approx 1.54\), \(n = 13\), \(z_{0.025} \approx 1.96\) into the formula: \[\lambda = 1.54 \pm 1.96 \sqrt{\frac{1.54}{13}}\]. Compute the interval.

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Key Concepts

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

Confidence Interval
A Confidence Interval (CI) gives us a range within which we expect the true parameter, in this case, the Poisson parameter \(\lambda\), to lie. When we talk about a CI, it gives you an idea of the reliability of the estimate you have made. Here, the CI is constructed around the sample mean, since for a Poisson distribution, the mean is equal to its parameter \(\lambda\). In this exercise, we aim to construct a CI for \(\lambda\) based on the sample mean from the absences data. The formula used is derived by approximating the distribution of the sample mean with a normal distribution. This becomes valid especially when dealing with large sample sizes. For the CI, we use the formula:
  • \(\lambda = \bar{x} \pm z_{\alpha/2}\sqrt{\frac{\bar{x}}{n}}\)
Here, \(z_{\alpha/2}\) is the critical value from the standard normal distribution for our chosen level of confidence. Most commonly, a 95% confidence level is chosen, where \(z_{0.025}\approx 1.96\).
The CI tells us that we can be 95% confident that the true \(\lambda\) is within this derived interval.
Sample Mean
The Sample Mean is a central concept in statistics and is used as an estimate of the population mean. Here, it serves as an estimate for the Poisson parameter \(\lambda\) since, for a Poisson distribution, the mean has the same value as \(\lambda\). To find the sample mean, you multiply each observed value (e.g., number of absences) by its frequency, sum those products, and then divide by the total number of observations, \(n\). In this problem, the sample mean \(\bar{x}\) is calculated as follows:
  • \(\bar{x} = \frac{(0\times 1 + 1\times 4 + 2\times 8)}{13} = \frac{20}{13} \approx 1.54\)
This mean stands for the average number of absences per day based on the data collected from the 50 recorded days. Calculating the sample mean is a key step in constructing the confidence interval for \(\lambda\).
This value not only estimates the central tendency of our data but is also pivotal for further statistical inference, such as constructing confidence intervals or hypothesis testing.
Standard Normal Distribution
The Standard Normal Distribution is a special case of a normal distribution with a mean of 0 and a standard deviation of 1. It is often denoted as \(Z\). This distribution is fundamental as it allows for standardizing data to compare different datasets or variables. When we standardize data, we convert an original data value \(X\) to a standardized value \(Z\) using the formula:
  • \(Z = \frac{X - \mu}{\sigma}\)
where \(\mu\) is the mean and \(\sigma\) is the standard deviation of the original data. In this exercise, we approximate the sampling distribution of our sample mean using a normal distribution because of the Central Limit Theorem, which facilitates using \(Z\) for large sample sizes.The Standard Normal Distribution provides the critical values \(z_{\alpha/2}\) necessary for constructing the confidence interval. For a 95% confidence level, this critical value is approximately 1.96. This process helps transform any normal random variable to a standard form, making it much easier to calculate probabilities and critical values for hypothesis testing and confidence intervals.
Sample Variance
Sample Variance measures the spread or variability of sample data points around the mean. In the context of a Poisson distribution, the variance is theoretically equal to \(\lambda\), the parameter of the distribution. However, calculating the sample variance provides a robustness check, confirming if our assumption for the variance's similarity to the mean holds true.To compute the sample variance \(s^2\), use the formula:
  • \(s^2 = \frac{\Sigma ((x_i - \bar{x})^2 \cdot f_i)}{n-1}\)
where \(x_i\) represents observed values, \(\bar{x}\) is the sample mean, and \(f_i\) is the frequency. This calculation ensures that the variability of data is as expected.
Since Poisson's mean and variance are equal, you expect the sample variance to be approximately the same as the sample mean. Nonetheless, knowing the spread of data through the sample variance is crucial in determining how much data points differ from the mean, providing a deeper insight into the data distribution's nature.

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Most popular questions from this chapter

Here are the names of 12 orchestra conductors and their performance times in minutes for Beethoven's Ninth Symphony: \(\begin{array}{llll}\text { Bermstein } & 71.03 & \text { Furtwängler } & 74.38 \\ \text { Leinsdorf } & 65.78 & \text { Ormandy } & 64.72 \\ \text { Solti } & 74.70 & \text { Szell } & 66.22 \\ \text { Bohm } & 72.68 & \text { Karajan } & 66.90 \\ \text { Masur } & 69.45 & \text { Rattle } & 69.93 \\\ \text { Steinberg } & 68.62 & \text { Tennstedt } & 68.40\end{array}\) a. Check to see that normality is a reasonable assumption for the performance time distribution. b. Compute a \(95 \%\) CI for the population standard deviation, and interpret the interval. c. Supposedly, classical music is \(100 \%\) determined by the composer's notation, including all timings. Based on your results, is this true or false?

Consider a normal population distribution with the value of \(\sigma\) known. a. What is the confidence level for the interval \(\pi \pm 2.81 \sigma / \sqrt{n}\) ? b. What is the confidence level for the interval \(\pi \pm 1.44 \sigma / \sqrt{n}\) ? c. What value of \(z_{x / 2}\) in the CI formula (8.5) results in a confidence level of \(99.7 \%\) ? d. Answer the question posed in part (c) for a confidence level of \(75 \%\).

A sample of 66 obese adults was put on a lowcarbohydrate diet for a year. The average weight loss was \(11 \mathrm{lb}\) and the standard deviation was \(19 \mathrm{lb}\). Calculate a \(99 \%\) lower confidence bound for the true average weight loss. What does the bound say about confidence that the mean weight loss is positive?

A study was done on 41 first-year medical students to see if their anxiety levels changed during the first semester. One measure used was the level of serum cortisol, which is associated with stress. For each of the 41 students the level was compared during finals at the end of the semester against the level in the first week of classes. The average difference was \(2.08\) with a standard deviation of \(7.88\). Find a \(95 \%\) lower confidence bound for the population mean difference \(\mu\). Does the bound suggest that the mean population stress change is necessarily positive?

Determine the following: a. The 95 th percentile of the chi-squared distribution with \(v=10\) b. The 5 th percentile of the chi-squared distribution with \(v=10\) c. \(P\left(10.98 \leq \chi^{2} \leq 36.78\right)\), where \(\chi^{2}\) is a chisquared rv with \(v=22\) d. \(P\left(x^{2}<14.611\right.\) or \(\left.\chi^{2}>37.652\right)\), where \(x^{2}\) is a chi-squared rv with \(v=25\)

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