/*! 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 65 If random samples of the given s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 40 from a population with mean 250 and standard deviation 80

Short Answer

Expert verified
The standard error of the distribution of sample means is 12.66.

Step by step solution

01

Define the Parameters

For this exercise, the population mean (µ) is 250 but isn't needed for the calculations. The standard deviation (σ) is 80, and the sample size (n) is 40.
02

Calculate the Square Root of the Sample Size

In this step, calculate the square root of the sample size (\(n\)). Using a calculator, the square root of 40 is approximately 6.32.
03

Perform the Division

Finally, divide the standard deviation (\(σ\)) by the square root of the sample size. Therefore, \(\(σ\)/\(\sqrt{n}\) = 80 / 6.32 = 12.66 (rounded to two decimal places)

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Key Concepts

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

Sampling Distribution
When we talk about a sampling distribution, we are referring to a probability distribution of a statistic obtained through a large number of samples drawn from a specific population. To put it plainly, if you were to take many samples of the same size from a given population, calculate a statistic for each sample (like the mean), and then create a graph of all those statistics, you would have what's known as a sampling distribution.

The characteristics of this distribution, such as its shape, mean, and spread, can tell us a lot about what we can expect to find in the population as a whole. Particularly, the Central Limit Theorem, an important concept in statistics, tells us that the sampling distribution of the sample mean will approach a normal distribution as the sample size gets larger, regardless of the population's distribution, provided that the samples are independent and identically distributed.
Sample Mean
The sample mean, often represented by \( \bar{x} \), is simply the average of all the values in a sample. It serves as an estimate of the population mean (\(\mu\)).

The accuracy of the sample mean as a representation of the population mean largely depends on the size of the sample and the variability of the population. Larger samples tend to give a more accurate estimate of the population mean, given that they provide more data points. It's also worth noting that the sample mean can have its own distribution, as seen when multiple samples are taken from the population – this is the sampling distribution of the sample mean we discussed earlier.
Standard Deviation
Standard deviation, symbolized as \( \sigma \) for a population and \( s \) for a sample, is a measure of dispersion or variability within a set of data.

It tells us how spread out the numbers are in a data set. A low standard deviation means that the data points are generally close to the mean, while a high standard deviation indicates that the data points are spread out over a larger range of values.

Understanding standard deviation helps us grasp how much variability we can expect from a sample and how comparable or diverse the individual observations are. This, in turn, has a direct impact on the reliability of statistical inferences drawn from the data.
Population Mean
The population mean, often symbolized by \( \mu \), is the average of all the values in a population. To clarify, when statisticians mention 'population,' they're talking about the complete set of data or measurements that one could potentially observe or analyze.

The population mean is a fixed value for any given population, but it might not always be known. When it's not known, we use sample data to estimate it. The population mean can differ from a sample mean of a subset of that population simply due to natural variability or sampling error, but the aim is to get the sample mean as close as possible to the actual population mean.

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Most popular questions from this chapter

In Exercise \(6.107,\) we see that plastic microparticles are contaminating the world's shorelines and that much of the pollution appears to come from fibers from washing polyester clothes. The same study referenced in Exercise 6.107 also took samples from ocean beaches. Five samples were taken from each of 18 different shorelines worldwide, for a total of 90 samples of size \(250 \mathrm{~mL}\). The mean number of plastic microparticles found per \(250 \mathrm{~mL}\) of sediment was 18.3 with a standard deviation of 8.2 . (a) Find and interpret a \(99 \%\) confidence interval for the mean number of polyester microfibers per \(250 \mathrm{~mL}\) of beach sediment. (b) What is the margin of error? (c) If we want a margin of error of only ±1 with \(99 \%\) confidence, what sample size is needed?

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.

Effect of Splitting the Bill Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The 24 people paying individually had a mean cost of 37.29 Israeli shekels with a standard deviation of 12.54 , while the 24 people splitting the bill had a higher mean cost of 50.92 Israeli shekels with a standard deviation of 14.33. The raw data can be found in SplitBill and both distributions are reasonably bell-shaped. Use this information to find and interpret a \(95 \%\) confidence interval for the difference in mean meal cost between these two situations.

Is Gender Bias Influenced by Faculty Gender? Exercise 6.215 describes a study in which science faculty members are asked to recommend a salary for a lab manager applicant. All the faculty members received the same application, with half randomly given a male name and half randomly given a female name. In Exercise \(6.215,\) we see that the applications with female names received a significantly lower recommended salary. Does gender of the evaluator make a difference? In particular, considering only the 64 applications with female names, is the mean recommended salary different depending on the gender of the evaluating faculty member? The 32 male faculty gave a mean starting salary of \(\$ 27,111\) with a standard deviation of \(\$ 6948\) while the 32 female faculty gave a mean starting salary of \(\$ 25,000\) with a standard deviation of \(\$ 7966 .\) Show all details of the test.

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 Potatoes after 20 Days After 20 days, 250 of the 500 fruit flies eating organic potatoes are still alive, while 130 of the 500 eating conventional potatoes are still alive.

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