/*! 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 66 If random samples of the given s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 66

If random samples of the given size are drawn from a population with the given mean and standard deviation, find the standard error of the distribution of sample means. Samples of size 75 from a population with mean 60 and standard deviation 32

Short Answer

Expert verified
The standard error of the distribution of sample means is approximately 3.69.

Step by step solution

01

Identify the Given Values

First, identify the given values from the problem. The population standard deviation (蟽) is 32, The sample size (n) is 75.
02

Apply the Formula for Standard Error

Then, use the formula for the standard error of the mean, which is given by: Standard error = 蟽 / sqrt(n)Substitute 蟽 with 32 and n with 75 in the formula. So it becomes:Standard error = 32 / sqrt(75)
03

Calculate the Standard Error

Now, calculate the standard error.By doing the calculation, the standard error equals to approximately 3.69.

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 91影视!

Key Concepts

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

Understanding the Population Mean
The population mean, often represented by the Greek letter \( \mu \), is a key concept in statistics that refers to the average value of all observations in an entire population. It is calculated by summing all the values in the population and dividing by the total number of values. In the exercise provided, the population mean is given as 60.

This means if we were to add up all the individual data points in the population and then divide by the number of these data points, the result would be 60. It is important to note that the population mean includes every single possible observation and is not based on a sample from the population.
Deciphering Standard Deviation
Standard deviation, symbolized by \( \sigma \) for the population and \( s \) for a sample, measures how much the values in a data set vary, or 'deviate', from the mean. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

In the context of our exercise, the population standard deviation is 32. This tells us that on average, the individual data points in the population are 32 units away from the population mean of 60. It's a measure of variability or diversity in the set of values within the population.
Sample Size Significance
Sample size, denoted as \( n \), plays a crucial role in statistical analysis. It refers to the number of observations or data points selected from a population to estimate the characteristics of the whole population. The larger the sample size, the more accurate the estimation of the population parameters tends to be.

In the given exercise, the sample size is 75. This size impacts the calculation of the standard error, which is a measure of how much we expect sample means to deviate from the population mean. Larger samples generally lead to a smaller standard error, indicating that the sample mean is likely a better estimate of the population mean.

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

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 survey respondents who say exercise is important, with \(n=1000\) and \(\hat{p}=0.753\)

A survey of 1000 adults in the US conducted in March 2011 asked "Do you favor or oppose 'sin taxes' on soda and junk food?" The proportion in favor of taxing these foods was \(32 \% .10\) (a) Find a \(95 \%\) confidence interval for the proportion of US adults favoring taxes on soda and junk food. (b) What is the margin of error? (c) If we want a margin of error of only \(1 \%\) (with \(95 \%\) confidence \()\), what sample size is needed?

Do Hands Adapt to Water? Researchers in the UK designed a study to determine if skin wrinkled from submersion in water performed better at handling wet objects. \(^{62}\) They gathered 20 participants and had each move a set of wet objects and a set of dry objects before and after submerging their hands in water for 30 minutes (order of trials was randomized). The response is the time (seconds) it took to move the specific set of objects with wrinkled hands minus the time with unwrinkled hands. The mean difference for moving dry objects was 0.85 seconds with a standard deviation of 11.5 seconds. The mean difference for moving wet objects was -15.1 seconds with a standard deviation of 13.4 seconds. (a) Perform the appropriate test to determine if the wrinkled hands were significantly faster than unwrinkled hands at moving dry objects. (b) Perform the appropriate test to determine if the wrinkled hands were significantly faster than unwrinkled hands at moving wet objects.

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 卤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 \(H_{a}: \mu_{1}>\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllllll} \hline \text { Situation } & 1 & 125 & 156 & 132 & 175 & 153 & 148 & 180 & 135 & 168 & 157 \\ \text { Situation } & 2 & 120 & 145 & 142 & 150 & 160 & 148 & 160 & 142 & 162 & 150 \\ \hline \end{array} $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation
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.