/*! 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 48 An important aspect of a federal... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An important aspect of a federal economic plan was that consumers would save a substantial portion of the money that they received from an income tax reduction. Suppose that early estimates of the portion of total tax saved, based on a random sampling of 35 economists, had mean 26\% and standard deviation 12\%. a. What is the approximate probability that a sample mean estimate, based on a random sample of \(n=35\) economists, will lie within \(1 \%\) of the mean of the population of the estimates of all economists? b. Is it necessarily true that the mean of the population of estimates of all economists is equal to the percent tax saving that will actually be achieved?

Short Answer

Expert verified
a. The probability is approximately 0.3758. b. No, it's not necessarily true.

Step by step solution

01

Identify the given information

We are given a sample of 35 economists with a sample mean percent tax saving of 26% and a standard deviation of 12%.
02

Determine standard deviation of the sample mean

The standard deviation of the sample mean (standard error) can be calculated using the formula: \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation (12%) and \( n \) is the sample size (35).\[ \sigma_{\bar{x}} = \frac{12}{\sqrt{35}} \approx 2.03\% \]
03

Define the range for sample mean

The problem asks for the probability that the sample mean will lie within 1% of the population mean, therefore, we are interested in the range 25% to 27%.
04

Utilize the Central Limit Theorem

According to the Central Limit Theorem, the sampling distribution of the sample mean is approximately normal for large n. Thus, we can use a normal distribution to approximate probabilities.
05

Calculate z-scores for the boundaries

Calculate the z-scores for 25% and 27% using the population mean 26% and the standard deviation of the sample mean (2.03%): \[ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \] For 25%: \[ z_{25} = \frac{25 - 26}{2.03} \approx -0.49 \] For 27%: \[ z_{27} = \frac{27 - 26}{2.03} \approx 0.49 \]
06

Find probabilities using the standard normal distribution

Using a standard normal distribution table or calculator, find the probability for each z-score: \( P(Z < 0.49) \) and \( P(Z < -0.49) \). These are approximately 0.6879 and 0.3121, respectively.
07

Calculate the probability between the two z-scores

The probability that the sample mean will be within 1% of 26% is the difference: \[ P(25 < \bar{x} < 27) = P(Z < 0.49) - P(Z < -0.49) = 0.6879 - 0.3121 = 0.3758 \]
08

Address the second question

Determine if the mean of the population estimates equals the actual percent tax saving. It is not necessarily true because these estimates are predictions based on a sample, and actual savings could be affected by real-world factors not accounted for in the model.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability
Probability is a fundamental concept in statistics that measures the likelihood that a particular event will occur. In the context of our exercise, we are interested in the probability concerning the sample mean of estimates provided by 35 economists.
When dealing with probability and estimation, we use data from samples to make inferences about a whole population. Here, we aim to find the probability that the economists' sample mean falls within a specific range of the population mean. With probability values, we generally range from 0 to 1, where 0 means the event will not occur, and 1 means it definitely will occur.
In our problem, after calculating the relevant z-scores, we used the standard normal distribution to find probabilities associated with these scores. This allowed us to determine the likelihood of the sample mean being within 1% of the population mean, which helps provide insights into the reliability of the economists' predictions.
Sampling Distribution
Sampling Distribution is a critical concept when discussing the Central Limit Theorem and involves the distribution of sample means over repeated sampling. In layman's terms, it's how the means of different samples from the same population are distributed.
In our exercise, we calculated the standard deviation of the sample mean, also known as the standard error, which helps in understanding the variability of our sample statistic. This was found using the formula: \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]where \( \sigma \) is the population standard deviation, and \( n \) is the sample size. This standard error represents the average distance that any given sample mean is likely to fall from the population mean.
The Central Limit Theorem tells us that for a large enough sample size, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the population's original distribution. In this exercise, this theorem allowed us to use a normal distribution to predict probabilities, which streamlined our solving process.
Z-Score
A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values, measured in terms of standard deviations. Knowing how to calculate and interpret z-scores is crucial for comparing different data points within a dataset.
In our exercise, z-scores were calculated to determine how far the sample mean of tax savings deviates from the population mean. The formula for calculating a z-score is:\[ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \]where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, and \( \sigma_{\bar{x}} \) is the standard error. In the example, we found z-scores for the boundaries of our interest range (25% and 27%), helping us assess the probability of the sample mean falling within 1% of the population mean.
By converting sample means into z-scores, we were able to use the standard normal distribution table to determine these probabilities, effectively turning complex data comparison into a simpler, more interpretable task.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Access the applet Normal Approximation to Binomial Distribution (at https://college.cengage.com/nextbook/statistics/wackerly 966371/student/html/index.html). When the applet is started, it displays the details in Example 7.11 and Figure 7.9 . Initially, the display contains only the binomial histogram and the exact value (calculated using the binomial probability function) for \(p(8)=P(Y=8) .\) Scroll down a little and click the button "Toggle Normal Approximation" to overlay the normal density with mean 10 and standard deviation \(\sqrt{6}=2.449\) the same mean and standard deviation as the binomial random variable \(Y\). You will get a graph superior to the one in Figure \(7.9 .\) a. How many probability mass or density functions are displayed? b. Enter 0 in the box labeled "Begin" and press the enter key. What probabilities do you obtain? c. Refer to part (b). On the line where the approximating normal probability is displayed, you see the expression Normal: $$ P(-0.5<=k<=8.5)=0.2701 $$ Why are the .5s in this expression?

The result in Exercise 7.58 holds even if the sample sizes differ. That is, if \(X_{1}, X_{2}, \ldots, X_{n_{1}}\) and \(Y_{1}, Y_{2}, \ldots, Y_{n_{2}}\) constitute independent random samples from populations with means \(\mu_{1}\) and \(\mu_{2}\) and variances \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\), respectively, then \(\bar{X}-\bar{Y}\) will be approximately normally distributed, for large \(n_{1}\) and \(n_{2},\) with mean \(\mu_{1}-\mu_{2}\) and variance \(\left(\sigma_{1}^{2} / n_{1}\right)+\left(\sigma_{2}^{2} / n_{2}\right)\).The flow of water through soil depends on, among other things, the porosity (volume proportion of voids) of the soil. To compare two types of sandy soil, \(n_{1}=50\) measurements are to be taken on the porosity of soil \(\mathrm{A}\) and \(n_{2}=100\) measurements are to be taken on soil \(\mathrm{B}\). Assume that \(\sigma_{1}^{2}=.01\) and \(\sigma_{2}^{2}=.02 .\) Find the probability that the difference between the sample means will be within. 05 unit of the difference between the population means \(\mu_{1}-\mu_{2}\).

Briggs and King developed the technique of nuclear transplantation in which the nucleus of a cell from one of the later stages of an embryo's development is transplanted into a zygote (a single-cell, fertilized egg) to see if the nucleus can support normal development. If the probability that a single transplant from the early gastrula stage will be successful is \(.65,\) what is the probability that more than 70 transplants out of 100 will be successful?

A machine is shut down for repairs if a random sample of 100 items selected from the daily output of the machine reveals at least \(15 \%\) defectives. (Assume that the daily output is a large number of items.) If on a given day the machine is producing only \(10 \%\) defective items, what is the probability that it will be shut down? [Hint: Use the. 5 continuity correction.]

a. If \(U\) has a \(\chi^{2}\) distribution with \(\nu\) df, find \(E(U)\) and \(V(U)\). b. Using the results of Theorem 7.3 , find \(E\left(S^{2}\right)\) and \(V\left(S^{2}\right)\) when \(Y_{1}, Y_{2}, \ldots, Y_{n}\) is a random sample from a normal distribution with mean \(\mu\) and variance \(\sigma^{2}\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.