Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Use the information given in Exercises 9-15 to find the necessary confidence interval for the population mean \(\mu .\) Interpret the interval that you have constructed. A \(90 \%\) confidence interval, \(n=50, \bar{x}=21.9\), \(s^{2}=3.44\)

Short Answer

Expert verified
Answer: The 90% confidence interval for the population mean is approximately (21.48, 22.32).

Step by step solution

01

Calculate the sample standard deviation

The sample variance is given as \(s^2=3.44\). To find the sample standard deviation, we take the square root of the variance. So, the sample standard deviation is: \(s=\sqrt{s^2}=\sqrt{3.44} \approx 1.85\)
02

Determine the degrees of freedom and critical t-value

The degrees of freedom are calculated as \(df = n-1 = 50 - 1 = 49\). To find the critical t-value for \(90\%\) confidence, we will consult a t-distribution table or use a calculator with the inverse t-distribution function. The critical t-value is \(t_{0.05, 49} \approx 1.68\), corresponding to the \(0.05\) level of significance in a two-tailed test with 49 degrees of freedom.
03

Calculate the margin of error

The margin of error is calculated using the critical t-value and the sample standard deviation: Margin of error \(= t_{0.05, 49} \times \frac{s}{\sqrt{n}} \approx 1.68 \times \frac{1.85}{\sqrt{50}} \approx 0.42\)
04

Construct the confidence interval

To find the confidence interval, we will subtract and add the margin of error to the sample mean: Lower limit \(= \bar{x} - \text{Margin of error} = 21.9 - 0.42 = 21.48\) Upper limit \(= \bar{x} + \text{Margin of error} = 21.9 + 0.42 = 22.32\) So, the \(90\%\) confidence interval for \(\mu\) is \((21.48, 22.32)\).
05

Interpret the interval

The interval \((21.48, 22.32)\) means that we are \(90\%\) confident that the population mean \(\mu\) lies within this range. In other words, if we were to conduct this study many times using different samples, the true population mean would lie within this interval \(90\%\) of the time.

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 Vaia!

Key Concepts

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

Population Mean
When we speak about the population mean, we're referring to the average value of a data set that includes every member of the population. It's symbolized by the Greek letter \(\mu\), and is a parameter that is typically unknown and that we're trying to estimate using sample data. In our example, we try to infer the population mean from the sample mean \(\bar{x} = 21.9\), which serves as our best estimate based on the given sample of 50 observations.

Understanding the population mean is crucial because it's a key characteristic of any set of data, representing the central point around which the values of the data set are distributed.
Sample Standard Deviation
The sample standard deviation is a measure of the amount of variation or dispersion of a set of values. In our example, the calculation starts from the sample variance \(s^2 = 3.44\). Taking the square root of this variance gives us the sample standard deviation \(s \approx 1.85\).

It's important to distinguish between the sample standard deviation and the population standard deviation, as the former is a statistic used to estimate the latter. Especially when working with small sample sizes, using the sample standard deviation correctly is vital to accurately estimating the uncertainty of the population mean.
Degrees of Freedom
The concept of degrees of freedom (df) relates to the number of independent values in a calculation that are free to vary. When calculating a sample statistic like the sample standard deviation, one value is used to estimate the population mean, which constrains our calculation. Therefore, the degrees of freedom are calculated as \(df = n - 1 \), where \(n\) is the sample size. In this case, with a sample size of 50, we have 49 degrees of freedom \(df = 50 - 1 = 49\).

Degrees of freedom are used in statistical tests, including the t-distribution, to account for the sample size and help control the type I error rate, providing a more accurate distribution model for statistical inference.
T-distribution
The t-distribution is a probability distribution that is symmetric and bell-shaped like the normal distribution but has heavier tails. It's particularly useful when dealing with small samples (usually n < 30) or when the population standard deviation is unknown, as is often the case in practice.

The t-distribution varies according to the degrees of freedom; the larger the sample size (and thus the more degrees of freedom), the closer the t-distribution resembles the normal distribution. To construct our confidence interval, we used a critical t-value found by referencing the t-distribution with our calculated degrees of freedom (49 in this example). The critical t-value \(t_{0.05, 49} \approx 1.68\) represents the point in the tails of the distribution: there's a 5% chance that the t-value will be beyond this point on either side in a two-tailed test.

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

A random sample of \(n\) measurements has been selected from a population with unknown mean \(\mu\) and known standard deviation \(\sigma=10 .\) Calculate the width of a \(95 \%\) confidence interval for \mu for the sample sizes given. What effect do the changing sample sizes have on the width of the interval? \(n=200\)

Construct a \(95 \%\) confidence interval for the difference in the population means. Then find a point estimate for the difference in the population means and calculate the margin of error: Compare your results. Can you conclude that there is a difference in the two population means? $$\begin{aligned}&\text { } n_{1}=35, n_{2}=45, \bar{x}_{1}=36.8, \bar{x}_{2}=33.6, s_{1}=4.9\\\ &s_{2}=3.4\end{aligned}$$

If 36 measurements of the specific gravity of aluminum had a mean of 2.705 and a standard deviation of .028 , find the point estimate for the actual specific gravity of aluminum and calculate the margin of error.

Calculate the margin of error in estimating a binomial proportion \(p\) for the sample sizes given in Exercises \(11-14\). Use \(p=.5\) to calculate the standard error of the estimator, and comment on how an increased sample size affects the margin of error. \(n=30\)

A pediatrician randomly selected 50 six-month-old boys from her practice's database and recorded an average weight of 8.0 kilograms with a standard deviation of 0.30 kilogram. She also recorded an average length of 67.3 centimeters with a standard deviation of 0.64 centimeter. a. Find a \(95 \%\) confidence interval for the average weight of all six-month- old boys. b. Find a \(99 \%\) confidence interval for the average length of all six-month- old boys. c. What do you have to assume about the pediatrician's database in order to make inferences about all sixmonth-old boys?

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.

Sign-up for free