/*! 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 94 Suppose that we obtain independe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that we obtain independent samples of sizes \(n_{1}\) and \(n_{2}\) from two normal populations with equal variances. Use the appropriate pivotal quantity from Section 8.8 to derive a \(100(1-\alpha) \%\) upper confidence bound for \(\mu_{1}-\mu_{2}\)

Short Answer

Expert verified
The upper confidence bound for \(\mu_1 - \mu_2\) is \(\bar{X}_1 - \bar{X}_2 + t_{1-\alpha} S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\).

Step by step solution

01

Define the problem

We want to find a 100(1-\(\alpha\))% upper confidence bound for the difference in means \(\mu_1 - \mu_2\) from two independent samples of sizes \(n_1\) and \(n_2\) from populations with normal distributions and equal variances.
02

State the known pivotal quantity

For two independent samples from normal populations with equal variances, the pivotal quantity is \(T = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\), where \(\bar{X}_1\) and \(\bar{X}_2\) are the sample means, and \(S_p\) is the pooled standard deviation.
03

Determine the critical value

For the upper confidence bound, find the critical value \(t_{1-\alpha}\) from the Student's t-distribution with \(n_1 + n_2 - 2\) degrees of freedom, such that the probability \(P(T \leq t_{1-\alpha}) = 1 - \alpha\).
04

Rearrange the pivotal quantity

Rearrange the pivotal quantity to solve for \(\mu_1 - \mu_2\): \(\bar{X}_1 - \bar{X}_2 + t_{1-\alpha} S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\). This becomes the upper confidence bound on \(\mu_1 - \mu_2\).
05

Express the pooled standard deviation

The pooled standard deviation \(S_p\) is given by \(S_p = \sqrt{\frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}}\), where \(S_1^2\) and \(S_2^2\) are the sample variances.
06

Construct the final expression

Substitute \(S_p\) and the critical value into the upper bound expression. The result is the upper confidence bound for \(\mu_1 - \mu_2\): \(\bar{X}_1 - \bar{X}_2 + t_{1-\alpha} \times \sqrt{\frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}} \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\).

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.

Pivotal Quantity
The concept of a pivotal quantity is central to constructing confidence intervals and conducting hypothesis tests. A pivotal quantity is a function of the data and unknown parameters that has a known distribution, regardless of the parameter values. This property makes it extremely powerful for statistical inference.

In the problem we are looking at, the pivotal quantity used is:
  • \( T = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \)
Here, \( \bar{X}_1 \) and \( \bar{X}_2 \) are the means of two independent samples, \( \mu_1 - \mu_2 \) is the difference in means we are estimating, and \( S_p \) is the pooled standard deviation. The distribution is known as it follows the Student's t-distribution with \( n_1 + n_2 - 2 \) degrees of freedom.

This pivotal quantity allows us to build an interval estimate and make probabilistic statements about the parameter in question.
Student's t-Distribution
The Student's t-distribution is a probability distribution that is very useful when working with small sample sizes or when the population variance is unknown. It is similar to the normal distribution but has heavier tails, which accounts for the additional uncertainty in the estimates.

In our problem, the pivotal quantity follows a Student's t-distribution with specific degrees of freedom:
  • The degrees of freedom are calculated as \( n_1 + n_2 - 2 \), where \( n_1 \) and \( n_2 \) are the sample sizes of the two independent groups.
  • The critical value \( t_{1-\alpha} \) for the upper confidence bound is obtained from this distribution, ensuring the probability \( P(T \leq t_{1-\alpha}) = 1 - \alpha \).
The t-distribution is crucial when determining the width of the confidence interval, as it influences the margin of error based on the chosen level of confidence.
Pooled Standard Deviation
The pooled standard deviation is a weighted average of the standard deviations of two independent samples. It's particularly useful when the assumption of equal population variances is valid across groups.

Mathematically, it is expressed as:
  • \( S_p = \sqrt{\frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}} \)
Here, \( S_1^2 \) and \( S_2^2 \) are the variances of the two samples, and \( n_1 \) and \( n_2 \) are their sizes. The formula combines these variances, adjusted by their respective degrees of freedom, to provide a common estimate. This pooled estimate is then used to normalize the difference of sample means in the pivotal quantity.

Using the pooled standard deviation is important because it gives us a stable estimate of variability across our samples, which helps in accurately determining confidence intervals.
Sample Variances
Sample variances measure the dispersion of data points within a sample. They are essential for understanding the uncertainty or variability present in the data. When constructing a confidence interval for the difference in means, knowing the variability within each sample is a must.

For two samples, the variances are calculated as follows:
  • \( S_1^2 = \frac{1}{n_1 - 1} \sum_{i=1}^{n_1} (X_{1i} - \bar{X}_1)^2 \)
  • \( S_2^2 = \frac{1}{n_2 - 1} \sum_{i=1}^{n_2} (X_{2i} - \bar{X}_2)^2 \)
These calculations involve squaring the difference between each data point and the sample mean, summing these squares, and correcting for bias by dividing by the degrees of freedom.The sample variances play a pivotal role in determining the pooled standard deviation and ultimately affect the width of the confidence interval for \( \mu_1 - \mu_2 \), making accurate variance calculations key for reliable statistical inference.

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

What is the normal body temperature for healthy humans? A random sample of 130 healthy human body temperatures provided by Allen Shoemaker \(^{\star}\) yielded 98.25 degrees and standard deviation 0.73 degrees. a. Give a \(99 \%\) confidence interval for the average body temperature of healthy people. b. Does the confidence interval obtained in part (a) contain the value 98.6 degrees, the accepted average temperature cited by physicians and others? What conclusions can you draw?

A random sample of size 25 was taken from a normal population with \(\sigma^{2}=6\). A confidence interval for the mean was given as \((5.37,7.37) .\) What is the confidence coefficient associated with this interval?

Industrial light bulbs should have a mean life length acceptable to potential users and a relatively small variation in life length. If some bulbs fail too early in their life, users become annoyed and are likely to switch to bulbs produced by a different manufacturer. Large variations above the mean reduce replacement sales; in general, variation in life lengths disrupts the user's replacement schedules. A random sample of 20 bulbs produced by a particular manufacturer produced the following lengths of life (in hours): $$\begin{array}{lllllllll}2100 & 2302 & 1951 & 2067 & 2415 & 1883 & 2101 & 2146 & 2278 & 2019 \\ 1924 & 2183 & 2077 & 2392 & 2286 & 2501 & 1946 & 2161 & 2253 & 1827\end{array}$$ Set up a \(99 \%\) upper confidence bound for the standard deviation of the lengths of life for the bulbs produced by this manufacturer. Is the true population standard deviation less than 150 hours? Why or why not?

Although there are many treatments for bulimia nervosa, some subjects fail to benefit from treatment. In a study to determine which factors predict who will benefit from treatment, Wendy Baell and E. H. Wertheim \(^{\text {? }}\) found that self-esteem was one of the important predictors. The mean and standard deviation of posttreatment self-esteem scores for \(n=21\) subjects were \(\bar{y}=26.6\) and \(s=7.4,\) respectively. Find a \(95 \%\) confidence interval for the true posttreatment self-esteem scores.

A random sample of 985 "likely voters" \(-\) those who are judged to be likely to vote in an upcoming election-were polled during a phone-athon conducted by the Republican Party. Of those contacted, 592 indicated that they intended to vote for the Republican running in the election. a. According to this study, the estimate for \(p\), the proportion of all "likely voters" who will vote for the Republican candidate, is \(p=.601 .\) Find a bound for the error of estimation. b. If the "likely voters" are representative of those who will actually vote, do you think that the Republican candidate will be elected? Why? How confident are you in your decision? c. Can you think of reasons that those polled might not be representative of those who actually vote in the election?

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.