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91Ó°ÊÓ

Compute the standard error for sample means from populations all with mean \(\mu=100\) and with standard deviations \(\sigma=5, \sigma=25,\) and \(\sigma=75\) using a sample size of \(n=100 .\) Discuss the effect of the population standard deviation on the standard error of the sample means.

Short Answer

Expert verified
The standard errors for \(\sigma = 5, 25, 75\) are \(0.5, 2.5, 7.5\) respectively. It's observed that as the population standard deviation increases, so does the standard error.

Step by step solution

01

Compute for \(\sigma = 5\)

The standard error when standard deviation is 5 is found by substituting \(\sigma = 5\) and \(n = 100\) into the standard error formula. This gives \(SE = \frac{5}{\sqrt{100}} = 0.5\)
02

Compute for \(\sigma = 25\)

The standard error when standard deviation is 25 is found by substituting \(\sigma = 25\) and \(n = 100\) into the standard error formula. This gives \(SE = \frac{25}{\sqrt{100}} = 2.5\)
03

Compute for \(\sigma = 75\)

The standard error when standard deviation is 75 is found by substituting \(\sigma = 75\) and \(n = 100\) into the standard error formula. This gives \(SE = \frac{75}{\sqrt{100}} = 7.5\)
04

Discuss The Effect of the Standard Deviation On The Standard Error

The standard error increases as the standard deviation increases. This is because the standard error is directly proportional to the standard deviation; this means that as the standard deviation increases, the standard error also increases, hence reflecting a greater spread of the populationsample data around the mean.

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Key Concepts

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

Understanding Sample Size
Sample size refers to the number of observations or data points collected in a study or experiment. It plays a crucial role in determining the reliability and precision of the statistical analysis. In simpler terms, think of sample size as the number of people you ask for their opinion in a survey. The larger your sample size, the more accurate your results tend to be.
When it comes to calculating the standard error, the sample size is essential. The standard error (SE) formula is: \[ SE = \frac{\sigma}{\sqrt{n}} \] Here, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. Notice how the square root of \(n\) is in the denominator? This means as the sample size increases, the standard error decreases.
  • Larger sample sizes give a smaller standard error.
  • This means the sample mean will be closer to the true population mean.
In summary, don't skimp on collecting enough data! A larger sample size makes your findings more reliable.
Importance of Population Standard Deviation
Population standard deviation is a measure of the variability or spread of scores in a dataset. It tells us how much individual data points differ from the mean of the population.
Why is this important for calculating standard error? Because the population standard deviation directly affects it. Using the formula for standard error: \[ SE = \frac{\sigma}{\sqrt{n}} \] You can see \(\sigma\) (the standard deviation) is in the numerator. So, a higher standard deviation leads to a higher standard error.
  • Larger \(\sigma\) means data is more spread out.
  • This results in a higher standard error, indicating more variability in sample means.
Therefore, understanding the standard deviation helps in predicting the expected variability in repeated sampling. Smaller standard deviations lead to more precise estimates.
Mean Calculation and Its Role
Mean calculation is the process of adding up all the data points in a dataset and dividing by the number of data points. It gives us the average value. The mean is a central point around which data can be analyzed.
When you have multiple samples from a population, calculating the mean allows you to summarize data with a single value. In the context of standard error, the mean is a vital component because it helps us understand how much sample means fluctuate.
  • The population mean (\(\mu\)) is the target value sample means aim to estimate.
  • With enough samples, their means tend to form a normal distribution around \(\mu\).
By examining the mean, researchers can determine how representative a sample is of the population. Hence, mean calculation is the cornerstone of statistical analysis, guiding you in making predictions about a larger data set.

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Most popular questions from this chapter

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 50 from a population with proportion 0.25

Is a Normal Distribution Appropriate? In Exercises 6.13 and \(6.14,\) indicate whether the Central Limit Theorem applies so that the sample proportions follow a normal distribution. In each case below, does the Central Limit Theorem apply? (a) \(n=80\) and \(p=0.1\) (b) \(n=25\) and \(p=0.8\) (c) \(n=50\) and \(p=0.4\) (d) \(n=200\) and \(p=0.7\)

Professor A and Professor \(\mathrm{B}\) are teaching sections of the same introductory statistics course and decide to give common exams. They both have 25 students and design the exams to produce a grade distribution that follows a bell curve with mean \(\mu=75\) and standard deviation \(\sigma=10\) (a) Suppose students are randomly assigned to the two classes and the instructors are equally effective. Describe the center, spread, and shape of the distribution of the difference in class means, \(\bar{x}_{A}-\bar{x}_{B},\) for the common exams. (b) Based on the distribution in part (a), how often should one of the class means differ from the other class by three or more points? (Hint: Look at both the tails of the distribution.) (c) How do the answers to parts (a) and (b) change if the exams are much harder than expected so the distribution for each class is \(N(60,10)\) rather that \(N(75,10) ?\)

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) 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 lie detector trials in which the technology misses a lie, with \(n=48\) and \(\hat{p}=0.354\)

If random samples of the given sizes are drawn from populations with the given proportions: (a) Find the mean and standard error of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 100 from population \(A\) with proportion 0.20 and samples of size 50 from population \(B\) with proportion 0.30

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