/*! 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 3 In Exercises 6.1 to \(6.6,\) if ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion, find the standard error of the distribution of sample proportions. Samples of size 60 from a population with proportion 0.90

Short Answer

Expert verified
The standard error of the distribution of sample proportions for samples of size 60 from a population with a proportion of 0.90 is approximately 0.038.

Step by step solution

01

Identify the given values

From the problem, we're given that the sample size (n) is 60 and the population proportion (p) is 0.90.
02

Substitute the values into the standard error formula

Plugging these values into the standard error formula, we have \(SE = \sqrt{\frac{0.90(1-0.90)}{60}}\).
03

Perform the calculations

First, calculate the expression in the numerator. Since \(1 – 0.90 = 0.10\), the expression within the square root becomes \(0.90 * 0.10 / 60 = 0.0015\). The square root of 0.0015 is approximately 0.03782.

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 Proportions
Understanding sample proportions is crucial when delving into statistical studies. The sample proportion, commonly represented as \( \hat{p} \), is an estimate of the population proportion \( p \) derived from a sample taken from the population. Consider it as a snapshot of what you're trying to measure across the entire population. For instance, if you want to know the percentage of left-handed students in a school, you could take a sample and find the proportion of left-handed students in that sample. This proportion is your sample proportion.

It's important to note that the sample proportion can vary from sample to sample. This variability is due to the random nature of sample selection. Ideally, the sample should be representative of the population to ensure that the sample proportion is a good estimate of the population proportion. Ensure this representation by using random sampling methods and a sufficient sample size to mitigate the risk of bias and decrease the margin of error.
Population Proportion
The population proportion represents the proportion of individuals in the entire population that have a particular attribute. It is denoted by \( p \) and is a fixed value, although in practice it's almost always unknown and needs to be estimated from the sample data.

In the context of the given exercise, the population proportion of 0.90 indicates that if we were to look at the entire population, we would expect 90% of the observations to have the characteristic of interest. This is a key parameter in many statistical procedures and is fundamental to hypothesis testing and confidence interval estimation. For accuracy in these procedures, it's critical to have a good estimate of the population proportion, which brings us back to the importance of sample proportions as an estimate of \( p \).
Standard Error Formula
The standard error (SE) is a statistical term that measures the accuracy with which a sample distribution represents a population using standard deviation. In the context of sample proportions, the standard error formula is expressed as \( SE = \sqrt{\frac{p(1 - p)}{n}} \), where \( p \) is the population proportion and \( n \) is the sample size.

Using the Standard Error Formula

In our exercise, you've been given a sample size of 60 and a population proportion of 0.90. By substituting these values into the formula, you calculate the SE to assess how much the sample proportion is expected to vary from the true population proportion. The resulting standard error lets us gauge the potential error or uncertainty in our sample estimate, enabling us to understand the reliability of our sample proportion and to construct confidence intervals around it.

A key factor to remember is that the standard error decreases as the sample size increases, meaning that larger samples tend to yield more precise estimates of the population proportion. Conversely, a smaller sample size would increase the standard error, suggesting a less reliable estimate. This inverse relationship is fundamental in designing studies and interpreting statistical data.

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 Raisins after 20 Days After 20 days, 275 of the 500 fruit flies eating organic raisins are still alive, while 170 of the 500 eating conventional raisins are still alive.

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 \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=10.1, s_{1}=2.3, n_{1}=50\) and \(\bar{x}_{2}=12.4, s_{2}=5.7, n_{2}=50 .\)

In Exercises 6.32 and 6.33, find a \(95 \%\) confidence interval for the proportion two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the normal distribution and the formula for standard error. Compare the results. Proportion of home team wins in soccer, using \(\hat{p}=0.583\) with \(n=120\)

We examine the effect of different inputs on determining the sample size needed. Find the sample size needed to give a margin of error within ±3 with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. Assume that we use \(\tilde{\sigma}=30\) as our estimate of the standard deviation in each case. Comment on the relationship between the sample size and the confidence level desired.

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=-2.6, s_{d}=4.1\) \(n_{d}=18\)

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.