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Chapter 7: Problem 92

From each of two normal populations with identical means and with standard deviations of 6.40 and \(7.20,\) independent random samples of 64 observations are drawn. Find the probability that the difference between the means of the samples exceeds. 6 in absolute value.

Short Answer

Expert verified
The probability is practically 0.

Step by step solution

01

Define the Problem

We have two independent normal populations with identical means. The goal is to find the probability that the absolute difference between the two sample means exceeds 6. This involves calculating the probability that the difference is greater than 6 or less than -6.
02

Set Up the Formula for Standard Error

Since the samples are independent, we use the formula for the standard error of the difference between two sample means. The standard error (SE) is calculated as \( \sqrt{\frac{\sigma_1^2}{n} + \frac{\sigma_2^2}{n}} \), where \( \sigma_1 = 6.40 \) and \( \sigma_2 = 7.20 \), and \( n = 64 \).
03

Calculate the Standard Error

Calculate the standard error using the formula from Step 2: \( SE = \sqrt{\frac{6.40^2}{64} + \frac{7.20^2}{64}} \). This simplifies to \( SE = \sqrt{0.64 + 0.81} = \sqrt{1.45} \approx 1.204 \).
04

Set Up Probability for the Sample Mean Difference

The sample mean difference follows a normal distribution with mean 0 (since population means are identical) and standard deviation equal to the SE. We want to find \( P(|\bar{X}_1 - \bar{X}_2| > 6) \) which is equivalent to \( P(\bar{X}_1 - \bar{X}_2 > 6) + P(\bar{X}_1 - \bar{X}_2 < -6) \).
05

Standardize the Difference

Use the standard normal distribution to find these probabilities. Standardize the threshold of 6: \( z = \frac{6}{1.204} \approx 4.98 \). Find the z-score for 6 (or -6 for the negative side).
06

Calculate Probabilities

Use standard normal tables or a calculator to find \( P(Z > 4.98) \) and \( P(Z < -4.98) \). Since the normal distribution is symmetric, both probabilities are equal. \( P(|\bar{X}_1 - \bar{X}_2| > 6) = 2 \times P(Z > 4.98) \). For \( z = 4.98 \), the probability is extremely small, practically 0.

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics. It's often called the bell curve due to its characteristic shape. What makes the normal distribution particularly important is its properties:
  • It is symmetric around the mean.
  • It follows the empirical rule where approximately 68% of data falls within one standard deviation from the mean, 95% within two, and 99.7% within three.
  • The mean, median, and mode are equal.
Normal distributions appear naturally in many situations in the real world and are a crucial part of statistical theory because many statistical tests rely on them. In our exercise, we assume that the samples from the two populations are normally distributed, which simplifies the calculations of mean differences and probabilities.
Standard Error
The standard error measures the variability of a sample statistic. It provides insight into how much the sample mean would differ from the true population mean.
  • The formula for the standard error (SE) of the difference between two sample means is:\[ SE = \sqrt{\frac{\sigma_1^2}{n} + \frac{\sigma_2^2}{n}} \]
  • Here, \( \sigma_1 \) and \( \sigma_2 \) are the standard deviations of each population, and \( n \) is the sample size.
The SE is crucial in this problem because it helps standardize the difference between the sample means, allowing us to calculate probabilities and understand the likelihood of the observed sample mean differences.
Probability Calculations
Probability calculations in the context of normal distributions involve determining how likely it is for a random variable to take on a specific value or range of values. For this exercise:
  • The focus is on finding the probability that the sample mean difference exceeds a certain threshold (6 in this case).
  • To do this, we make use of the properties of the normal distribution, which allows us to convert the sample mean difference to a standard normal form.
The probability that the absolute difference is greater than 6 involves not just the direct calculation but also understanding the symmetry of the normal distribution to simplify the process.
Z-score
A Z-score is a standardized score that indicates how many standard deviations an element is from the mean. To compute a Z-score, you use the formula:\[ z = \frac{X - \mu}{SE} \]where:
  • \( X \) is the value we are interested in (e.g., 6).
  • \( \mu \) is the mean of the distribution, which is 0 in this case since we are considering a difference.
  • \( SE \) is the standard error of the difference.
In our case, a Z-score of 4.98 for the difference suggests that this outcome is far out in the tails of the normal distribution. Probabilities associated with extreme Z-scores like this indicate highly unlikely events, as reflected by the approximate zero probability in this context.

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Most popular questions from this chapter

A retail dealer sells three brands of automobiles. For brand A, her profit per sale, \(X\) is normally distributed with parameters \(\left(\mu_{1}, \sigma_{1}^{2}\right) ;\) for brand \(\mathrm{B}\) her profit per sale \(Y\) is normally distributed with parameters \(\left(\mu_{2}, \sigma_{2}^{2}\right) ;\) for brand \(C\), her profit per sale \(W\) is normally distributed with parameters \(\left(\mu_{3},\right.\) \(\sigma_{3}^{2}\) ). For the year, two-fifths of the dealer's sales are of brand \(\mathrm{A}\), one-fifth of brand \(\mathrm{B}\), and the remaining two- fifths of brand C. If you are given data on profits for \(n_{1}, n_{2},\) and \(n_{3}\) sales of brands \(A\) B, and \(C\), respectively, the quantity \(U=.4 \bar{X}+.2 \bar{Y}+.4 \bar{W}\) will approximate to the true average profit per sale for the year. Find the mean, variance, and probability density function for \(U .\) Assume that \(X, Y,\) and \(W\) are independent.

What does the sampling distribution of the sample mean look like if samples are taken from an approximately normal distribution? Use the applet Sampling Distribution of the Mean (at https://college.cengage.com/nextbook/statistics/wackerly 966371/student/html/index.html) to complete the following. The population to be sampled is approximately normally distributed with \(\mu=16.50\) and \(\sigma=6.03\) (these values are given above the population histogram and denoted \(M\) and S, respectively). a. Use the button "Next Obs" to select a single value from the approximately normal population. Click the button four more times to complete a sample of size \(5 .\) What value did you obtain for the mean of this sample? Locate this value on the bottom histogram (the histogram for the values of \(Y\) ). b. Click the button "Reset" to clear the middle graph. Click the button "Next Obs" five more times to obtain another sample of size 5 from the population. What value did you obtain for the mean of this new sample? Is the value that you obtained equal to the value you obtained in part (a)? Why or why not? c. Use the button "1 Sample" eight more times to obtain a total of ten values of the sample mean. Look at the histogram of these ten means. i. What do you observe? ii. How does the mean of these \(10 \bar{y}\) -values compare to the population mean \(\mu\) ? d. Use the button "1 Sample" until you have obtained and plotted 25 realized values for the sample mean \(\bar{Y}\), each based on a sample of size 5 . i. What do you observe about the shape of the histogram of the 25 values of \(\bar{y}_{i}\) $$ i=1,2, \ldots, 25 ? $$ ii. How does the value of the standard deviation of the \(25 \bar{y}\) values compare with the theoretical value for \(\sigma_{Y}\) obtained in Example 5.27 where we showed that, if \(Y\) is computed based on a sample of size \(n,\) then \(V(\bar{Y})=\sigma^{2} / n ?\) e. Click the button "1000 Samples" a few times, observing changes to the histogram as you generate more and more realized values of the sample mean. What do you observe about the shape of the resulting histogram for the simulated sampling distribution of \(Y\) ? f. Click the button "Toggle Normal" to overlay (in green) the normal distribution with the same mean and standard deviation as the set of values of \(Y\) that you previously generated. Does this normal distribution appear to be a good approximation to the sampling distribution of \(Y ?\)

The times that a cashier spends processing individual customer's order are independent random variables with mean 2.5 minutes and standard deviation 2 minutes. What is the approximate probability that it will take more than 4 hours to process the orders of 100 people?

Suppose that independent samples (of sizes \(n_{i}\) ) are taken from each of \(k\) populations and that population \(i\) is normally distributed with mean \(\mu_{i}\) and variance \(\sigma^{2}, i=1,2, \ldots, k\). That is, all populations are normally distributed with the same variance but with (possibly) different means. Let \(\bar{X}_{i}\) and \(S_{i}^{2}, i=1,2, \ldots, k\) be the respective sample means and variances. Let \(\theta=c_{1} \mu_{1}+c_{2} \mu_{2}+\cdots+c_{k} \mu_{k},\) where \(c_{1}, c_{2}, \ldots, c_{k}\) are given constants. a. Give the distribution of \(\hat{\theta}=c_{1} \bar{X}_{1}+c_{2} \bar{X}_{2}+\cdots+c_{k} \bar{X}_{k}\). Provide reasons for any claims that you make. b. Give the distribution of $$\frac{\mathrm{SSE}}{\sigma^{2}}, \quad \text { whereSSE }=\sum_{i=1}^{k}\left(n_{i}-1\right) S_{i}^{2}$$. Provide reasons for any claims that you make. c. Give the distribution of $$\frac{\hat{\theta}-\theta}{\sqrt{\left(\frac{c_{1}^{2}}{n_{1}}+\frac{c_{2}^{2}}{n_{2}}+\cdots+\frac{c_{k}^{2}}{n_{k}}\right) \mathrm{MSE}}}, \quad \text { whereMSE }=\frac{\mathrm{SSE}}{n_{1}+n_{2}+\cdots+n_{k}-k}$$ Provide reasons for any claims that you make.

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?
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