/*! 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 The following information was ob... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 18

The following information was obtained from two independent samples selected from two populations with unknown but equal standard deviations. $$ \begin{array}{lll} n_{1}=55 & \bar{x}_{1}=90.40 & s_{1}=11.60 \\ n_{2}=50 & \bar{x}_{2}=86.30 & s_{2}=10.25 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\)

Short Answer

Expert verified
The point estimate of \(\mu_{1}-\mu_{2}\) is 4.1. The 99 % confidence interval for \(\mu_{1}-\mu_{2}\) would be found using the formula given in Step 5.

Step by step solution

01

Calculate the point estimate

The point estimate of \(\mu_{1}-\mu_{2}\) is the difference of sample means, which is \(\bar{x}_{1}-\bar{x}_{2}=90.40-86.30=4.1\).
02

Compute the Pooled Sample Variance

The pooled sample variance (s_{p}^{2}) is a weighted average of the sample variances. It is computed as follows. s_{p}^{2} = \([(n_1 - 1) * s_1^{2} + (n_2 - 1) * s_2^{2}\]\/\[(n_1 + n_2 - 2)\] = \([(55 - 1) * 11.60^{2} + (50 - 1) * 10.25^{2}\]\/\[(55 + 50 - 2)\]\).
03

Compute Standard Error

The standard error (s_{\mu_1 - \mu_2}) of the difference in sample means is given by the square root of the sum of the variances divided by their respective sample sizes. s_{\mu_1 - \mu_2} = \[\sqrt{\[(s_1^2/n_1) + (s_2^2/n_2)\]}\] = \[\sqrt{\[(s_p^2/55) + (s_p^2/50\]}.
04

Define Confidence Interval

A 99 % confidence interval for the difference in population means \(\mu_1 - \mu_2\) can be expressed as \((\mu_1 - \mu_2) \pm ts_{\mu_1 - \mu_2}\), where t is the t-score associated with the desired level of confidence from the t-distribution with \(n_1 + n_2 - 2\) degrees of freedom.
05

Compute Confidence Interval

Substitute the calculated values from previous steps and compute the confidence interval as follows: 4.1 \(\pm\) (t * s_{\mu_1 - \mu_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 91影视!

Key Concepts

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

Point Estimate
The point estimate is a simple yet crucial concept in statistics. It represents a single value that estimates a population parameter based on sample data. In the context of comparing two samples, the point estimate of the difference between the population means \(\mu_1 - \mu_2\) is obtained by subtracting the sample mean of the second group from the sample mean of the first group.
This gives us a straightforward approach to gauge the difference between populations.
For example: \(\bar{x}_1 - \bar{x}_2 = 90.40 - 86.30 = 4.1\), where the result, 4.1, suggests that the first sample is, on average, 4.1 units higher than the second.
Pooled Sample Variance
When dealing with two sample groups, it may be necessary to pool their variances to better estimate the common population variance when the standard deviations are assumed to be equal. The pooled sample variance, \(s_p^2\), is a weighted average that considers the sample sizes and individual variances.
The formula is:
  • \( s_p^2 = \frac{(n_1 - 1) \, s_1^2 + (n_2 - 1) \, s_2^2}{n_1 + n_2 - 2} \)

This approach leads to a more balanced estimation, particularly when the two sample sizes are unequal.
The pooled variance plays a pivotal role in further calculations, such as standard error and confidence intervals.
Standard Error
The standard error of the difference between two means is an important component in estimating how much the sample mean differences might vary from the actual population mean difference. It accounts for both the natural variability within each sample and the sample sizes.
The formula is:
  • \( s_{\mu_1 - \mu_2} = \sqrt{\left(\frac{s_p^2}{n_1}\right) + \left(\frac{s_p^2}{n_2}\right)} \)

The standard error helps to quantify the precision of the point estimate.
A smaller standard error indicates a more precise estimate, reinforcing confidence in the results of sample analyses.
t-Distribution
The t-distribution is employed in statistics when estimating population parameters based on sample data, especially with smaller sample sizes where the normal distribution may not apply. This distribution is symmetrical, like the normal distribution, but has heavier tails.
It is often used to find the t-score, which helps in determining confidence intervals and hypothesis testing.
  • The degrees of freedom for the t-distribution in our case is \( n_1 + n_2 - 2 \).

When constructing a confidence interval for the difference in means, the t-score defines the range around the point estimate where the true population mean difference likely lies.
This scoring is particularly important for constructing accurate intervals with a certain level of confidence, such as 99% in this example.

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

Two local post offices are interested in knowing the average number of Christmas cards that are mailed out from the towns that they serve. A random sample of 80 households from Town A showed that they mailed an average of \(28.55\) Christmas cards with a standard deviation of \(10.30 .\) The corresponding values of the mean and standard deviation produced by a random sample of 58 households from Town B were \(33.67\) and \(8.97\) Christmas cards. Assume that the distributions of the numbers of Christmas cards mailed by all households from both these towns have the same population standard deviation. a. Construct a \(95 \%\) confidence interval for the difference in the average numbers of Christmas cards mailed by all households in these two towns. b. Using a \(10 \%\) significance level, can you conclude that the average number of Christmas cards mailed out by all households in Town \(\mathrm{A}\) is different from the corresponding average for Town B?

Manufacturers of two competing automobile models, Gofer and Diplomat, each claim to have the lowest mean fuel consumption. Let \(\mu_{1}\) be the mean fuel consumption in miles per gallon (mpg) for the Gofer and \(\mu_{2}\) the mean fuel consumption in mpg for the Diplomat. The two manufacturers have agreed to a test in which several cars of each model will be driven on a 100 -mile test run. Then the fuel consumption, in mpg, will be calculated for each test run. The average of the \(m p g\) for all 100 -mile test runs for each model gives the corresponding mean. Assume that for each model the gas mileages for the test runs are normally distributed with \(\sigma=2 \mathrm{mpg} .\) Note that each car is driven for one and only one 100 -mile test run. a. How many cars (i.e., sample size) for each model are required to estimate \(\mu_{1}-\mu_{2}\) with a \(90 \%\) confidence level and with a margin of error of estimate of \(1.5 \mathrm{mpg}\) ? Use the same number of cars (i.c., sample size) for each model. b. If \(\mu_{1}\) is actually \(33 \mathrm{mpg}\) and \(\mu_{2}\) is actually \(30 \mathrm{mpg}\), what is the probability that five cars for each model would yield \(\bar{x}_{1} \geq \bar{x}_{2}\) ?

Assuming that the two populations have unequal and unknown population standard deviations, construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) for the following. $$ \begin{array}{lll} n_{1}=48 & \bar{x}_{1}=.863 & s_{1}=.176 \\ n_{2}=46 & \bar{x}_{2}=.796 & s_{2}=.068 \end{array} $$

The manager of a factory has devised a detailed plan for evacuating the building as quickly as possible in the event of a fire or other emergency. An industrial psychologist believes that workers actually leave the factory faster at closing time without following any system. The company holds fire drills periodically in which a bell sounds and workers leave the building according to the system. The evacuation time for each drill is recorded. For comparison, the psychologist also records the evacuation time when the bell sounds for closing time each day. A random sample of 36 fire drills showed a mean evacuation time of \(5.1\) minutes with a standard deviation of \(1.1\) minutes. A random sample of 37 days at closing time showed a mean evacuation time of \(4.2\) minutes with a standard deviation of \(1.0\) minute. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Test at a \(5 \%\) significance level whether the mean evacuation time is smaller at closing time than during fire drills.

Describe the sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) for two independent samples when \(\sigma_{1}\) and \(\sigma\), are known and either both sample sizes are large or both populations are normally distributed. What are the mean and standard deviation of this sampling distribution?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation
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.