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Answer true or false. Explain your answer. Every random sample of the same size from a given population will produce exactly the same confidence interval for \(\mu\).

Short Answer

Expert verified
False. Different random samples can produce different confidence intervals.

Step by step solution

01

Understanding Confidence Interval

A confidence interval is a range of values that is used to estimate the true parameter of a population, often the mean (). It is calculated in such a way that it has a specified probability (confidence level) of containing the true parameter.
02

Random Sampling

When you take a random sample from a population, the sample statistics may vary from one sample to another. This means that different random samples could yield different sample means, which, in turn, affects the confidence interval for .
03

Components of a Confidence Interval

The confidence interval formula is generally given by: \[CI = \bar{x} \pm z(\frac{\sigma}{\sqrt{n}})\]where \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the confidence level, \(\sigma\) is the standard deviation, and \(n\) is the sample size. If \(\bar{x}\) differs in different samples, the CI will differ too.
04

Conclusion

Since different random samples can produce different sample means and hence different confidence intervals, the statement is false. Not every random sample of the same size from a population will produce the same confidence interval for .

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Key Concepts

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

Random Sampling
Understanding random sampling is crucial in statistics because it plays a significant role in ensuring the accuracy and reliability of results derived from data. Random sampling refers to a method where every individual or item in a population has an equal chance of being selected as part of the sample. This process ensures that samples are unbiased and adequately represent the population.

One of the main goals of random sampling is to obtain a sample that is representative of the entire population. By ensuring each population member has an equal opportunity to be included, random sampling helps to mitigate any unintended bias, which can skew results.

However, it is important to note that even with random sampling, the sample statistics such as the mean or standard deviation can vary from one random sample to another. This statistical variability is a natural part of the sampling process and underscores the importance of considering multiple random samples for analysis.
Sample Mean
The sample mean is a key element in statistical analysis and is essentially the average of values in a sample. It is represented by \(\bar{x}\) and is calculated by summing up all data points in a sample and dividing by the number of observations.

Mathematically, it is expressed as:

\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]

where \(x_i\) represents each data point in the sample and \(n\) is the size of the sample.

The sample mean is not only fundamental in summarizing data but also in estimating the population mean \(\mu\). It forms the basis of the confidence interval calculations, helping us determine how the sample mean relates to the broader population mean.
  • The closer the sample mean to the population mean, the more representative that sample is of the whole population.
  • Variations in sample mean values across different random samples can influence the confidence intervals calculated from those samples.
These variations highlight the importance of random sampling and the statistical variability inherent in sampling.
Statistical Variability
Statistical variability, sometimes called statistical dispersion, refers to the extent to which data points in a statistical distribution or sample differ from each other. It denotes the variability within a dataset and is a critical concept when analyzing how random sampling can yield different results.

Several measures help quantify statistical variability, including variance and standard deviation. Variability is crucial for several reasons:
  • It enables understanding of the spread and distribution of data.
  • Higher variability indicates a wider spread of data, which can result in less precise estimates.
  • Understanding variability is important in determining the confidence interval width. More variability typically leads to wider confidence intervals.
In the context of random sampling, each sample may exhibit different levels of variability. Therefore, it's typical for confidence intervals derived from different random samples of the same size to vary, as variability impacts the computation of these intervals.

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Most popular questions from this chapter

Campus Life: Coeds What percentage of your campus student body is female? Let \(p\) be the proportion of women students on your campus. (a) If no preliminary study is made to estimate \(p\), how large a sample is needed to be \(99 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of \(0.05\) from \(p ?\) (b) The Statistical Abstract of the United States, 112 th Edition, indicates that approximately \(54 \%\) of college students are female. Answer part (a) using this estimate for \(p\).

Large U.S. Companies: Foreign Revenue For large U.S. companies, what percentage of their total income comes from foreign sales? A random sample of technology companies (IBM, Hewlett-Packard, Intel, and others) gave the following information. $$ \begin{aligned} &\text { Technology companies, } \% \text { foreign revenue: } x_{1} ; n_{1}=16\\\ &\begin{array}{llllllll} 62.8 & 55.7 & 47.0 & 59.6 & 55.3 & 41.0 & 65.1 & 51.1 \\ 53.4 & 50.8 & 48.5 & 44.6 & 49.4 & 61.2 & 39.3 & 41.8 \end{array} \end{aligned} $$ Another independent random sample of basic consumer product companies (Goodyear, Sarah Lee, H.J. Heinz, Toys " \(q\) "Us) gave the following information. $$ \begin{aligned} &\text { Basic consumer product companies, } \% \text { foreign revenue: } x_{2} ; n_{2}=17\\\ &\begin{array}{llllllll} 28.0 & 30.5 & 34.2 & 50.3 & 11.1 & 28.8 & 40.0 & 44.9 \\ 40.7 & 60.1 & 23.1 & 21.3 & 42.8 & 18.0 & 36.9 & 28.0 \end{array}\\\ &\begin{aligned} &40.7 \\ &32.5 \end{aligned} \end{aligned} $$ (Reference: Forbes Top Companies.) Assume that the distributions of percentage foreign revenue are mound-shaped and symmetric for these two company types. (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 51.66, s_{1} \approx 7.93, \bar{x}_{2} \approx 33.60\), and \(s_{2} \approx 12.26\). (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2}\). Find an \(85 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Interpretation Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(85 \%\) level of confidence, do technology companies have a greater percentage foreign revenue than basic consumer product companies? (d) Check Requirements Which distribution (standard normal or Student's \(t)\) did you use? Why?

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of \(\bar{x}=\$ 6.88\) per 100 pounds of watermelon. Assume that \(\sigma\) is known to be \(\$ 1.92\) per 100 pounds (Reference: Agricultural Statistics, U.S. Department of Agriculture). (a) Find a \(90 \%\) confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. What is the margin of error? (b) Sample Size Find the sample size necessary for a \(90 \%\) confidence level with maximal margin of error \(E=0.3\) for the mean price per 100 pounds of watermelon. (c) A farm brings 15 tons of watermelon to market. Find a \(90 \%\) confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds.

Basic Computation: Confidence Interval for \(p_{1}-p_{2}\) Consider two independent binomial experiments. In the first one, 40 trials had 15 successes. In the second one, 60 trials had 6 successes. (a) Check Requirements Is it appropriate to use a normal distribution to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation Based on the confidence interval you computed, can you be \(95 \%\) confident that \(p_{1}\) is more than \(p_{2} ?\) Explain.

Psychology: Self-Esteem Female undergraduates in randomized groups of 15 took part in a self-esteem study ("There's More to Self-Esteem than Whether It Is High or Low: The Importance of Stability of Self-Esteem," by M. H. Kernis et al., Journal of Personality and Social Psychology, Vol. 65, No. 6). The study measured an index of self-esteem from the points of view competence, social acceptance, and physical attractiveness. Let \(x_{1}, x_{2}\), and \(x_{3}\) be random variables representing the measure of self-esteem through \(x_{1}\) (competence), \(x_{2}\) (social acceptance), and \(x_{3}\) (attractiveness). Higher index values mean a more positive influence on self-esteem. $$ \begin{array}{ccccc} \hline \text { Variable } & \text { Sample Size } & \text { Mean } \bar{x} & \text { Standard Deviation } s & \text { Population Mean } \\ \hline x_{1} & 15 & 19.84 & 3.07 & \mu_{1} \\ x_{2} & 15 & 19.32 & 3.62 & \mu_{2} \\ x_{3} & 15 & 17.88 & 3.74 & \mu_{3} \\ \hline \end{array} $$ (a) Find an \(85 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (b) Find an \(85 \%\) confidence interval for \(\mu_{1}-\mu_{3} .\) (c) Find an \(85 \%\) confidence interval for \(\mu_{2}-\mu_{3}\). (d) Interpretation Comment on the meaning of each of the confidence intervals found in parts (a), (b), and (c). At the \(85 \%\) confidence level, what can you say about the average differences in influence on self-esteem between competence and social acceptance? between competence and attractiveness? between social acceptance and attractiveness?

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