/*! 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 21 Using the sample information giv... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Using the sample information given in Exercises \(20-21\), give the best point estimate for the population mean \(\mu\) and calculate the margin of error: A random sample of \(n=75\) observations from a quantitative population produced \(\bar{x}=29.7\) and \(s^{2}=10.8\).

Short Answer

Expert verified
Answer: The point estimate for the population mean is 29.7, and the margin of error is approximately 0.76.

Step by step solution

01

Identify the given information

In this problem, we are given the following information: - Sample size: \(n = 75\) - Sample mean: \(\bar{x} = 29.7\) - Sample variance: \(s^{2} = 10.8\)
02

Calculate the point estimate for the population mean

The best point estimate for the population mean \(\mu\) is simply the sample mean, \(\bar{x}\). Therefore, the point estimate for the population mean is: \(\mu = 29.7\).
03

Calculate the sample standard deviation

In order to calculate the margin of error, we need to first find the sample standard deviation. We can do this by taking the square root of the sample variance: \(s = \sqrt{s^2} = \sqrt{10.8} \approx 3.29\).
04

Determine the appropriate t-value

The margin of error can be calculated using the t-distribution. To determine the appropriate t-value, we first need to determine the degrees of freedom (df) and use the desired confidence level. We will use a common confidence level of 95% for this problem. Degrees of freedom: \(df = n-1 = 75-1 = 74\). Now, using a t-table or calculator, we can find the t-value corresponding to a 95% confidence level and 74 degrees of freedom: \(t_{\alpha/2} \approx 1.99\).
05

Calculate the margin of error

Finally, we can calculate the margin of error using the following formula: \(E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}\). Plugging in the values we found: \(E = 1.99 \times \frac{3.29}{\sqrt{75}} \approx 1.99 \times 0.38 \approx 0.76\).
06

Interpret the results

The point estimate for the population mean is \(\mu = 29.7\), and the margin of error is \(E \approx 0.76\). Thus, we can say with 95% confidence that the true population mean lies within the interval \((29.7 - 0.76, 29.7 + 0.76)\), or \((28.94, 30.46)\).

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.

Point Estimate
When working with statistics, a point estimate is essentially a single value that serves as an approximation of a larger unknown quantity – in most cases, this refers to a population parameter. Think of it as using a small sample to make an educated guess about a much larger group.

For instance, if we're interested in the average height of trees in a forest, it would be impractical to measure every single tree. Instead, we measure a few and use the average of these measurements as our point estimate for the population mean height. In the provided exercise, the sample mean \( \bar{x} = 29.7 \) is the best point estimate we have for the population mean \( \mu \).
Population Mean
The population mean \( \mu \) is a key concept in statistics that refers to the average of all measurements in a full population. If we had the time and resources to measure every member of our population, we'd get the exact mean. However, since that is often not the case, the point estimate attempts to approximate \( \mu \) rather than determining it exactly. The population mean is the central reference point for many statistical analyses and helps to understand the central tendency of the data.
Sample Standard Deviation
When it comes to summarizing data, the sample standard deviation (represented by \( s \) in equations) is crucial because it quantifies the extent to which individual data points in a sample differ from the sample mean. In simpler terms, it measures how spread out the numbers in your sample are.

It's the square root of the sample variance, which is the average of the squared differences between each data point and the mean. If the numbers are close to the mean, the standard deviation will be small; if they're spread out over a wide range, the standard deviation will be much larger. In our exercise, we first find the sample variance \( s^2 = 10.8 \) and then calculate the sample standard deviation to be about \( 3.29 \) using the square root.
t-distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, like the normal distribution, but it has heavier tails. This distribution becomes important especially when dealing with small sample sizes or when the population standard deviation is unknown.

It is used for estimating population parameters and its shape is determined by the degrees of freedom. As the sample size increases, the t-distribution looks more like a normal distribution. In the context of our exercise, the t-distribution is critical for determining the t-value needed to calculate the margin of error. To find our margin of error, we need the t-value that corresponds to our confidence level and degrees of freedom—which were 95% and 74 respectively.
Degrees of Freedom
Degrees of freedom often abbreviated as df in statistics, are a measure of the amount of variation in a set of values that is 'free' to vary. Imagine you have a set of numbers and their mean; if you know the mean and all but one of the numbers, you can determine that last number. Essentially, it's one less than the number of values you have because the last value must comply to achieve the known mean.

In statistical analyses, the degrees of freedom are typically calculated as the number of observations minus the number of necessary parameters estimated. In the case of the sample standard deviation, it’s the sample size minus one. For our problem, the degrees of freedom were \( df = n-1 = 75-1 = 74 \) because we used the sample (of size 75) to estimate our population parameter.

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

To determine whether there is a significant difference in the average weights of boys and girls beginning kindergarten, random samples of 50 5-year-old boys and 50 5-year-old girls produced the following information. \(^{17}\) $$\begin{array}{llcc}\hline & & \text { Standard } & \text { Sample } \\\& \text { Mean } & \text { Deviation } & \text { Size } \\\\\hline \text { Boys } & 19.4 \mathrm{~kg} & 2.4 & 50 \\\\\text { Girls } & 17.0 \mathrm{~kg} & 1.9 & 50 \\\\\hline\end{array}$$ Find a \(99 \%\) lower confidence bound for the difference in the average weights of 5 -year-old boys and girls. Can you conclude that on average 5 -year-old boys weigh more than 5-year-old girls?

Using the sample information given in Exercises \(22-23,\) give the best point estimate for the binomial proportion \(p\) and calculate the margin of error. A random sample of \(n=500\) observations from a binomial population produced \(x=450\) successes.

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? $$\begin{aligned}&\text { } n_{1}=35, n_{2}=45, \bar{x}_{1}=36.8, \bar{x}_{2}=33.6, s_{1}=4.9\\\ &s_{2}=3.4\end{aligned}$$

What \(i s\) normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{11}\) in the Journal of Statistical Education, had a mean of \(98.25^{\circ}\) Fahrenheit and a standard deviation of \(0.73^{\circ}\) Fahrenheit. a. Construct a \(99 \%\) confidence interval for the average body temperature of healthy people. b. Does the confidence interval constructed in part a contain the value \(98.6^{\circ}\) Fahrenheit, the usual average temperature cited by physicians and others? If not, what conclusions can you draw?

Do you think that the United States should pursue a program to send humans to Mars? An opinion poll conducted by the Associated Press indicated that \(49 \%\) of the 1034 adults surveyed think that we should pursue such a program. a. Estimate the true proportion of Americans who think that the United States should pursue a program to send humans to Mars. Calculate the margin of error. b. The question posed in part a was only one of many questions concerning our space program that were asked in the opinion poll. If the Associated Press wanted to report one sampling error that would be valid for the entire poll, what value should they report?

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.