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Consider two populations for which \(\mu_{1}=30\), \(\sigma_{1}=2, \mu_{2}=25\), and \(\sigma_{2}=3\). Suppose that two independent random samples of sizes \(n_{1}=40\) and \(n_{2}=50\) are selected. Describe the approximate sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) (center, spread, and shape).

Short Answer

Expert verified
The sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) is centred around 5 (mean=5), has a spread indicated by standard deviation (approx. 0.529) and the shape will be approximately normal due to the Central Limit Theorem.

Step by step solution

01

Finding Mean of Sampling Distribution of Difference in Sample Means

Mean of the sampling distribution of the difference in sample means, \(\mu_{\bar{x}_{1}-\bar{x}_{2}}\), is given by the difference in population means. In this case, \(\mu_{\bar{x}_{1}-\bar{x}_{2}} = \mu_{1} - \mu_{2} = 30 - 25 = 5\).
02

Finding Standard Deviation of Sampling Distribution of Difference in Sample Means

The standard deviation of the sampling distribution of the difference in sample means, \(\sigma_{\bar{x}_{1}-\bar{x}_{2}}\), is found by squaring the standard deviations, dividing by the respective sample sizes, adding the results and taking the square root. So \(\sigma_{\bar{x}_{1}-\bar{x}_{2}} = \sqrt{{\sigma_{1}^2/n_{1} + \sigma_{2}^2/n_{2}}} = \sqrt{{2^2/40 + 3^2/50}} = \sqrt{{0.1 + 0.18}} = \sqrt{0.28} \approx 0.529\).
03

Describing the Shape of the Distribution

Given the large sample sizes, we can say that, as per Central Limit Theorem, the sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) will be approximately normal, regardless of the shape of the original population distributions.

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Key Concepts

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

Population Means
When discussing statistical analysis, understanding the concept of a population mean is fundamental. The population mean, denoted as \( \mu \), is the average of all the measurements in a complete population. For example, if we were to measure the heights of every person in a city, the population mean would be the average height of everyone taken into account.

This concept is essential when comparing two different populations, as in the exercise concerning two populations with means \( \mu_{1}=30 \) and \( \mu_{2}=25 \) respectively. When we take random samples from each of these populations and calculate the sample means, the expectation would be that these sample means should estimate the population means if the sample size is sufficiently large and random.

Understanding population means is critical when working with sampling distributions because the mean of these sampling distributions is expected to align with the population means, especially as we calculate the difference between the sample means.
Standard Deviation
Standard deviation is a statistical measurement of variability or dispersion within a set of data. It is symbolized by \( \sigma \) in the case of population standard deviation and \( s \) for sample standard deviation. A smaller standard deviation indicates that the data points tend to be close to the mean, while a larger standard deviation means that the data points are spread out over a wider range of values.

In the context of the presented exercise, two populations have standard deviations \( \sigma_{1}=2 \) and \( \sigma_{2}=3\). This information is used to ascertain the spread or dispersion of the sampling distribution's difference in sample means. A noteworthy aspect is the calculation of the standard deviation for the difference between the two sample means \( \sigma_{\bar{x}_{1}-\bar{x}_{2}} \), which involves the variances (squared standard deviations) of each population, divided by their respective sample sizes, and then taking the square root of the sum.

Grasping the concept of standard deviation is crucial because it is an instrumental component in interpreting the spread or uncertainty within sampling distributions, which essentially allows us to understand how much variation we might expect when comparing two sample means.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in the field of statistics. It states that, given a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the original distribution of the population. This is true for sample sizes typically greater than 30, although the exact threshold depends on the distribution in question.

In practice, this means that if we repeatedly take large samples and calculate their means, these sample means will tend to form a normal distribution around the true population mean. In our exercise, the samples sizes are \( n_{1}=40 \) and \( n_{2}=50 \) which are both greater than 30, thus the CLT can be applied guaranteeing that the sampling distribution of the difference in sample means \( \bar{x}_{1}-\bar{x}_{2} \) will also be approximately normal.

The application of CLT in our exercise simplifies our understanding of the sampling distribution, providing us with a predictable and well-understood structure, which is the normal distribution. This is a powerful tool as it allows statisticians and researchers to make inferences about population parameters, such as the difference between population means, using the properties of the normal distribution.

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Most popular questions from this chapter

A hotel chain is interested in evaluating reservation processes. Guests can reserve a room by using either a telephone system or an online system that is accessed through the hotel's web site. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. Based on these data, is it reasonable to conclude that the proportion who are satisfied is higher for those who reserve a room online? Test the appropriate hypotheses using \(\alpha=.05 .\)

Each person in a random sample of 228 male teenagers and a random sample of 306 female teenagers was asked how many hours he or she spent online in a typical week (Ipsos, January 25, 2006). The sample mean and standard deviation were \(15.1\) hours and \(11.4\) hours for males and \(14.1\) and \(11.8\) for females. a. The standard deviation for each of the samples is large, indicating a lot of variability in the responses to the question. Explain why it is not reasonable to think that the distribution of responses would be approximately normal for either the population of male teenagers or the population of female teenagers. Hint: The number of hours spent online in a typical week cannot be negative. b. Given your response to Part (a), would it be appropriate to use the two- sample \(t\) test to test the null hypothesis that there is no difference in the mean number of hours spent online in a typical week for male teenagers and female teenagers? Explain why or why not. c. If appropriate, carry out a test to determine if there is convincing evidence that the mean number of hours spent online in a typical week is greater for male teenagers than for female teenagers. Use a .05 significance level.

The article "Fish Oil Staves Off Schizophrenia" (USA Today. February 2, 2010) describes a study in which 81 patients age 13 to 25 who were considered atrisk for mental illness were randomly assigned to one of two groups. Those in one group took four fish oil capsules daily. The other group took a placebo. After 1 year, \(5 \%\) of those in the fish oil group and \(28 \%\) of those in the placebo group had become psychotic. Is it appropriate to use the two-sample \(z\) test of this section to test hypotheses about the difference in the proportions of patients receiving the fish oil and the placebo treatments who became psychotic? Explain why or why not.

The press release titled "Keeping Score When It counts: Graduation Rates and Academic Progress Rates" (The Institute for Diversity and Ethics in Sport. March 16,2009 ) gave the 2009 graduation rates for African-American basketball players and for white basketball players at every NCAA Division I university with a basketball program. Explain why it is not necessary to use a paired \(t\) test to determine if the mean graduation rate for African-American basketball players differs from the mean graduation rate for white basketball players for Division I schools.

The authors of the paper "Adolescents and MP3 Players: Too Many Risks, Too Few Precautions" (Pediatrics 12009 ] e953-e958) concluded that more boys than girls listen to music at high volumes. This conclusion was based on data from independent random samples of 764 Dutch boys and 748 Dutch girls age 12 to \(19 .\) Of the boys, 397 reported that they almost always listen to music at a high volume setting. Of the girls, 331 reported listening to music at a high volume setting. Do the sample data support the authors' conclusion that the proportion of Dutch boys who listen to music at high volume is greater than this proportion for Dutch girls? Test the relevant hypotheses using a \(.01\) significance level.

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