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Chapter 6: Problem 212

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (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 25 from Population 1 with mean 6.2 and standard deviation 3.7 and samples of size 40 from Population 2 with mean 8.1 and standard deviation 7.6

Short Answer

Expert verified
The mean of the distribution of differences in sample means is -1.9 and the standard error is approximately 1.41. The sampling distribution is normal with mean -1.9 and standard deviation 1.41.

Step by step solution

01

Calculate the Mean of Differences in Sample Means

The mean of the distribution of differences in sample means can be found by subtracting the mean of population 2 from the mean of population 1. Thus, \(\mu_{\bar{x}_1 - \bar{x}_2} = \mu_{\bar{x}_1} - \mu_{\bar{x}_2} = 6.2 - 8.1 = -1.9\).
02

Compute the Standard Error

The standard error is the standard deviation of the distribution of differences in sample means. It is calculated as follows: \(\sigma_{\bar{x}_1 - \bar{x}_2} = \sqrt{\left(\frac{\sigma_{1}^{2}}{n_{1}}\right) + \left(\frac{\sigma_{2}^{2}}{n_{2}}\right)} = \sqrt{\left(\frac{3.7^{2}}{25}\right) + \left(\frac{7.6^{2}}{40}\right)} = \sqrt{0.5468 + 1.448} = \sqrt{1.995} = 1.41\) .
03

Apply the Central Limit Theorem

The Central Limit Theorem states that for larger or equal to 30 samples, the distribution of the sample means approximates a normal distribution. Therefore, this condition is satisfied as the sample sizes are 25 and 40. Thus, one can construct a curve showing the shape of the sampling distribution with mean = -1.9 and standard deviation = 1.41. The curve would be normal and centered at -1.9 with points of inflection at -1.9 卤 1.41.

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Key Concepts

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

Mean of Differences
When comparing two populations, it can be insightful to consider the differences in their sample means. We calculate the mean of differences to understand how the average of one sample compares to another. For this particular problem, the mean of the distribution of differences is calculated by subtracting the mean of Population 2 from the mean of Population 1. Mathematically, this is represented as \[\mu_{\bar{x}_1 - \bar{x}_2} = \mu_{\bar{x}_1} - \mu_{\bar{x}_2}\] where \(\mu_{\bar{x}_1}\) is the mean of the first population and \(\mu_{\bar{x}_2}\) is the mean of the second population. In our example, the calculation would look like this: 6.2 - 8.1 = -1.9.
  • This negative value indicates that, on average, samples from Population 1 are less than those from Population 2 by 1.9 units.
  • The result forms the central point for our analysis of differences, guiding further steps.
Standard Error
The standard error is a crucial concept when evaluating the variability of sample means. It tells us how much the sample means are expected to differ from the true population means. To find the standard error of the distribution of differences in sample means, we use the formula: \[\sigma_{\bar{x}_1 - \bar{x}_2} = \sqrt{\left(\frac{\sigma_{1}^{2}}{n_{1}}\right) + \left(\frac{\sigma_{2}^{2}}{n_{2}}\right)}\] Here,
  • \(\sigma_{1}\) and \(\sigma_{2}\) are the standard deviations of Population 1 and 2 respectively.
  • \(n_{1}\) and \(n_{2}\) are the sample sizes from Population 1 and 2.
For the given problem, this calculation evaluates to \(\sqrt{0.5468 + 1.448} = 1.41\). This value suggests that sample means from these populations generally differ by approximately 1.41 units due to sampling variability.
Sampling Distribution
To fully understand the behavior of sample means, it's important to consider the sampling distribution. According to the Central Limit Theorem, when sample sizes are sufficiently large, the distribution of the sample means will approximate a normal distribution. In this case, as one sample size is 25 and the other is 40, the condition for applying the Central Limit Theorem is satisfied. The normal distribution of the sampling means makes it possible for us to visualize it as a bell-curved graph. For our problem, the mean of the sampling distribution is calculated as -1.9, and the standard deviation, known as the standard error, is 1.41.
  • This distribution centers around -1.9, providing a graphical representation.
  • Inflection points at -1.9 卤 1.41 indicate where the distribution changes curvature, providing a comprehensive view of data dispersion.
Thus, the sampling distribution helps in understanding how the difference in sample means fluctuates due to random sampling and underscores the reliability of our statistical conclusions.

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Most popular questions from this chapter

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=8\) and \(n_{2}=10\)

We see in the AllCountries dataset that the percent of the population that is over 65 is 13.4 in Australia and 12.5 in New Zealand. Suppose we take random samples of size 500 from Australia and size 300 from New Zealand, and compute the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{N Z},\) where \(\hat{p}_{A}\) represents the sample proportion of elderly in Australia and \(\hat{p}_{N Z}\) represents the sample proportion of elderly in New Zealand. Find the mean and standard deviation of the differences in sample proportions.

A young statistics professor decided to give a quiz in class every week. He was not sure if the quiz should occur at the beginning of class when the students are fresh or at the end of class when they've gotten warmed up with some statistical thinking. Since he was teaching two sections of the same course that performed equally well on past quizzes, he decided to do an experiment. He randomly chose the first class to take the quiz during the second half of the class period (Late) and the other class took the same quiz at the beginning of their hour (Early). He put all of the grades into a data table and ran an analysis to give the results shown below. Use the information from the computer output to give the details of a test to see whether the mean grade depends on the timing of the quiz. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) $$ \begin{aligned} &\text { Two-Sample T-Test and Cl }\\\ &\begin{array}{lrrrr} \text { Sample } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { Late } & 32 & 22.56 & 5.13 & 0.91 \\ \text { Early } & 30 & 19.73 & 6.61 & 1.2 \end{array} \end{aligned} $$ Difference \(=\mathrm{mu}\) (Late) \(-\mathrm{mu}\) (Early) Estimate for difference: 2.83 \(95 \%\) Cl for difference: (-0.20,5.86) T-Test of difference \(=0\) (vs not \(=\) ): T-Value \(=1.87\) P-Value \(=0.066 \quad \mathrm{DF}=54\)

Involve scores from the high school graduating class of 2010 on the SAT (Scholastic Aptitude Test). The average score on the Writing part of the SAT exam for males is 486 with a standard deviation of \(112,\) while the average score for females is 498 with a standard deviation of 111 (a) If random samples are taken with 100 males and 100 females, find the mean and standard deviation of the distribution of differences in sample means, \(\bar{x}_{m}-\bar{x}_{f},\) where \(\bar{x}_{m}\) represents the sample mean for the males and \(\bar{x}_{f}\) represents the sample mean for the females. (b) Repeat part (a) if the random samples contain 500 males and 500 females. (c) What effect do the different sample sizes have on center and spread of the distribution?

Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=-2.6, s_{d}=4.1\) \(n_{d}=18\)
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