/*! 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 82 Use the t-distribution to find a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the t-distribution to find a confidence interval for a mean \(\mu\) given the relevant sample results. Give the best point estimate for \(\mu,\) the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A \(95 \%\) confidence interval for \(\mu\) using the sample results \(\bar{x}=84.6, s=7.8,\) and \(n=42\)

Short Answer

Expert verified
The best point estimate for the mean (\(\mu\)) is 84.6, the margin of error is approximately 2.43, and the 95% confidence interval for the mean (\(\mu\)) is (82.17, 87.03).

Step by step solution

01

Compute t-score

A t-score is used here because the population standard deviation is unknown. For a 95% confidence level and \(n-1=41\) degrees of freedom, a t-score can be found using a t-distribution table or statistical software. The t-score is approximately 2.021.
02

Calculate Standard Error

The standard error is given by formula \(SE=\frac{s}{\sqrt{n}}\), where \(s=7.8\) is the sample standard deviation and \(n=42\) is the sample size. Substituting these values into the formula gives \(SE=\frac{7.8}{\sqrt{42}}\), which is approximately 1.202.
03

Calculate Margin of Error

The margin of error is calculated using the formula \(ME= t*SE\), where \(t\) corresponds to the t-score and \(SE\) is the standard error. Substitute the previously calculated values into this formula to get \(ME= 2.021* 1.202\), which is approximately 2.43.
04

Find Confidence Interval

Finally, to get the confidence interval for the mean, subtract and add the margin of error from/to the sample mean. The sample mean \(\bar{x}\) is 84.6. The confidence interval is therefore given by \(\bar{x} ± ME\), or \(84.6 ± 2.43\). This results in a confidence interval of (82.17, 87.03).

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.

Sample Mean
The sample mean is a crucial concept when it comes to estimating population parameters because it provides the best point estimate for a population mean, denoted as \( \mu \). In the context of data analysis, the sample mean, \( \bar{x} \), is simply the average of your sampled data points. It can be calculated by summing all the sample values and then dividing by the number of values in the sample.

For example, in the case of finding a 95% confidence interval as given, the sample mean is \( 84.6 \). This means that if you take the average of all the data points in our sample, the result is exactly \( 84.6 \).
This is your central figure, from which the confidence interval, or range where the true population mean is expected to lie, is developed. By itself, the sample mean gives you a glimpse of the population's central tendency. However, to understand how precise this estimate is, we need to delve further into calculating margins of error and standard error.
Margin of Error
The margin of error is a key component in determining the confidence interval around a sample mean. It reflects the extent of possible variation or uncertainty in the estimation of the population mean. Essentially, the margin of error offers a buffer around your sample mean, suggesting that the true population mean is likely contained within this range.

For calculating the margin of error, you need to consider both the t-score and the standard error. In the provided example, a t-score of approximately 2.021 is used. The t-score corresponds to the confidence level you're interested in (such as 95%) and the degrees of freedom associated with your sample size.
The margin of error (\( ME \)) is then calculated using the formula:
  • \( ME = t \times SE \)
where \( SE \) represents the standard error. In this exercise, putting all components together, we compute \( ME = 2.021 \times 1.202 \), which yields approximately \( 2.43 \). This means we add and subtract this margin from the sample mean, framing the confidence interval within which the true population mean likely falls.
Standard Error
The standard error is a measure that gives insight into how much the sample mean is expected to fluctuate from the true population mean. It is essentially the standard deviation of the sampling distribution of the sample mean. A smaller standard error suggests that the sample mean is a more precise estimator of the population mean.

To compute the standard error, you use the formula:
  • \( SE = \frac{s}{\sqrt{n}} \)
where \( s \) is the sample standard deviation and \( n \) is the sample size. In our exercise, the sample standard deviation is given as \( 7.8 \) and the sample size \( n \) is \( 42 \). Using these values in our formula gives:

  • \( SE = \frac{7.8}{\sqrt{42}} \approx 1.202 \)

This calculation shows that the sample mean can be off from the true mean by about \( 1.202 \). Understanding the standard error is imperative as it directly impacts the width of the confidence interval: less standard error narrows your confidence interval, leading to greater precision.

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

Using Data 5.1 on page \(375,\) we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. ask you to conduct a hypothesis test using additional data from this study. \(^{40}\) In every case, we are testing $$\begin{array}{ll}H_{0}: & p_{o}=p_{c} \\\H_{a}: & p_{o}>p_{c}\end{array}$$ where \(p_{o}\) and \(p_{c}\) represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Also, in every case, we have \(n_{1}=n_{2}=500 .\) Show all remaining details in the test, using a \(5 \%\) significance level. Effect of Organic Bananas after 15 Days After 15 days, 345 of the 500 fruit flies eating organic bananas are still alive, while 320 of the 500 eating conventional bananas are still alive.

Do Ovulating Women Affect Men's Speech? Studies suggest that when young men interact with a woman who is in the fertile period of her menstrual cycle, they pick up subconsciously on subtle changes in her skin tone, voice, and scent. A study introduced in Exercise \(\mathrm{B} .23\) suggests that men may even change their speech patterns around ovulating women. The men were randomly divided into two groups with one group paired with a woman in the fertile phase of her cycle and the other group with a woman in a different stage of her cycle. The same women were used in the two different stages. For the men paired with a less fertile woman, 38 of the 61 men copied their partner's sentence construction in a task to describe an object. For the men paired with a woman at peak fertility, 30 of the 62 men copied their partner's sentence construction. The experimenters hypothesized that men might be less likely to copy their partner during peak fertility in a (subconscious) attempt to attract more attention to themselves. Use the normal distribution to test at a \(5 \%\) level whether the proportion of men copying sentence structure is less when the woman is at peak fertility.

Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution less than -1.4 if the samples have sizes \(n_{1}=30\) and \(n_{2}=40\)

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 \(\bar{x}_{2}=69.0, s_{2}=8.3, n_{2}=20 .\)

Use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of home team wins in soccer, with \(n=120\) and \(\hat{p}=0.583\)

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.