/*! 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 188 In Exercises 6.188 to 6.191 , us... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 6.188 to 6.191 , use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=75.2, s_{1}=10.7, n_{1}=30\) and \(\bar{x}_{2}=69.0, s_{2}=8.3, n_{2}=20 .\)

Short Answer

Expert verified
The best estimate for \(\mu_{1}-\mu_{2}\) is 6.2 with a standard error of 2.68. The margin of error is 5.61. Thus, the \(95\%\) confidence interval for \(\mu_{1}-\mu_{2}\) is \((0.59, 11.81)\).

Step by step solution

01

Compute the best estimate.

The best estimate for \(\mu_{1}-\mu_{2}\) is the difference between the sample means. We denote this as \(D\), so our best estimate is calculated as \(D = \bar{x}_{1} - \bar{x}_{2} = 75.2 - 69.0 = 6.2\).
02

Calculate the standard error of difference.

The standard error of difference is calculated using the formula: \(SE_{D} = sqrt(\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}})\) where \(s_{1}\) and \(s_{2}\) are the sample standard deviations and \(n_{1}\) and \(n_{2}\) are the sample sizes. Plugging in the given values, we get \(SE_{D} = sqrt(\frac{10.7^{2}}{30} + \frac{8.3^{2}}{20}) = 2.68\).
03

Identify the t-value corresponding to the given level of confidence.

To construct a \(95\%\) confidence interval, we would refer the t-table for degrees of freedom \(df = min(n_{1}, n_{2}) - 1 = min(30,20) - 1 = 19\). The t-value for \(95\%\) confidence and \(19\) degrees of freedom is approximately \(2.093\).
04

Calculate the margin of error.

The margin of error (MoE) is calculated as: \(MoE = t \times SE_{D}\) where \(t\) is the t-value and \(SE_{D}\) is the standard error of difference. By plugging in the values from steps 2 and 3, we get \(MoE = 2.093 \times 2.68 = 5.61\).
05

Construct the confidence interval.

The confidence interval is then \(D \pm MoE\) where \(D\) is our best estimate and \(MoE\) is the margin of error. Thus our \(95\%\) confidence interval is \(6.2 \pm 5.61\), or \((0.59, 11.81)\).

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.

Confidence Interval
A confidence interval is a range of values, derived from sample statistics, that is likely to contain the value of an unknown population parameter. When we talk about a 95% confidence interval, it means there is a 95% chance that the interval contains the true difference in the population means, \( \mu_{1} - \mu_{2} \).

This concept is particularly useful in determining the uncertainty associated with a sample estimate, known as the best estimate, of a population parameter. Viewed as an interval estimate, it gives an idea of how far these estimates could be from the actual parameter in the whole population.

Key points to remember:
  • It requires sample data to compute.
  • The confidence level (in this exercise, 95%) indicates the probability that the interval calculated from the sample contains the true population parameter.
  • This interval is determined using the sample statistic, standard error, and a critical t-value.
Confidence intervals provide a range for the population mean difference which, through statistical inference, we believe contains the true mean difference with a certain level of certainty.
Difference in Means
The difference in means quantifies the gap between the central tendencies of two groups or samples. It is essentially the difference between the average values, or means, of two samples:\( \bar{x}_{1} - \bar{x}_{2} \).

In the context of this exercise, it serves as a "best estimate" for the difference in the overall populations from which these samples were drawn.

Here’s what you should know:
  • The best estimate helps to compare two different groups based on the sample data.
  • This calculation is particularly useful when analyzing whether two populations have distinctly different characteristics.
  • The computed difference, \( D \), was found to be \( 6.2 \) in our example.
Detecting a significant difference between groups can shed light on underlying effects or treatment impacts, especially when the interval does not include zero, suggesting that a real difference likely exists.
Standard Error
The standard error provides a measure of the variability or spread of sample means and helps us understand how sample means differ from the actual population mean. In the context of comparing means of two groups, the standard error of difference is more relevant.

It is calculated using the formula: \( SE_{D} = \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}} \) This formula takes into account the variability within each group and their respective sample sizes.

Some crucial aspects include:
  • The standard error decreases as sample sizes increase, providing a more precise estimate of the population parameter.
  • In this exercise, the standard error was calculated to be \( 2.68 \).
  • A larger standard error suggests more variability in the estimate, leading to a wider confidence interval.
By understanding the standard error, you gain insights into the reliability of the sample’s mean estimate and the confidence interval, thereby assessing the precision of the 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

Effect of Splitting the Bill Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The 24 people paying individually had a mean cost of 37.29 Israeli shekels with a standard deviation of 12.54 , while the 24 people splitting the bill had a higher mean cost of 50.92 Israeli shekels with a standard deviation of 14.33. The raw data can be found in SplitBill and both distributions are reasonably bell-shaped. Use this information to find and interpret a \(95 \%\) confidence interval for the difference in mean meal cost between these two situations.

In Exercises 6.150 and \(6.151,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section. Sample A has a count of 90 successes with \(n=120\) and Sample \(\mathrm{B}\) has a count of 180 successes with \(n=300\).

Gender Bias In a study \(^{52}\) examining gender bias, a nationwide sample of 127 science professors evaluated the application materials of an undergraduate student who had ostensibly applied for a laboratory manager position. All participants received the same materials, which were randomly assigned either the name of a male \(\left(n_{m}=63\right)\) or the name of a female \(\left(n_{f}=64\right) .\) Participants believed that they were giving feedback to the applicant, including what salary could be expected. The average salary recommended for the male applicant was \(\$ 30,238\) with a standard deviation of \(\$ 5152\) while the average salary recommended for the (identical) female applicant was \(\$ 26,508\) with a standard deviation of \(\$ 7348\). Does this provide evidence of a gender bias, in which applicants with male names are given higher recommended salaries than applicants with female names? Show all details of the test.

(a) Find the relevant sample proportions in each group and the pooled proportion. (b) Complete the hypothesis test using the normal distribution and show all details. Test whether people with a specific genetic marker are more likely to have suffered from clinical depression than people without the genetic marker, using the information that \(38 \%\) of the 42 people in a sample with the genetic marker have had clinical depression while \(12 \%\) of the 758 people in the sample without the genetic marker have had clinical depression.

We examine the effect of different inputs on determining the sample size needed to obtain a specific margin of error when finding a confidence interval for a proportion. Find the sample size needed to give a margin of error to estimate a proportion within \(\pm 3 \%\) with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. (Assume no prior knowledge about the population proportion \(p\).) Comment on the relationship between the sample size and the confidence level desired.

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.