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The \(z\) confidence interval and test are based on the sampling distribution of the sample mean \(x^{-} \bar{x}\). The National Survey of Student Engagement (NSSE) asks college seniors to rate how much their experience at their institution has contributed to their ability to think critically and analytically. Ratings are on a scale of 1 to 7 , with 7 being the highest (best) rating. Suppose scores for all seniors in 2013 are Normal with mean \(\mu=3.3\) and standard deviation \(\sigma=0.8\). (a) You take an SRS of 100 seniors. According to the \(99.7\) part of the 68-95-99.7 rule, about what range of ratings do you expect to see in your sample? (b) You look at many SRSs of size 100. About what range of sample mean ratings \(\mathrm{x}^{-} \bar{x}\) do you expect to see?

Short Answer

Expert verified
Individual ratings are expected to range from 0.9 to 5.7, and sample means from 3.06 to 3.54.

Step by step solution

01

Identify 99.7% range for individuals

The 99.7% part of the 68-95-99.7 rule applies to individual values in a Normal distribution. Given that the distribution of ratings is Normal with mean \(\mu = 3.3\) and standard deviation \(\sigma = 0.8\), we can apply this rule by calculating \(\mu - 3\sigma\) and \(\mu + 3\sigma\). This will give us the range of individual ratings that we expect to contain 99.7% of the data. Thus:\[ 3.3 - 3(0.8) = 0.9 \]\[ 3.3 + 3(0.8) = 5.7 \]So, the 99.7% of individual ratings are expected to lie between 0.9 and 5.7.
02

Calculate Standard Error for Sample Mean

The standard deviation of the sample mean, or standard error (SE), is calculated using the population standard deviation \(\sigma = 0.8\) divided by the square root of the sample size \(n=100\). Therefore:\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{0.8}{\sqrt{100}} = \frac{0.8}{10} = 0.08 \]
03

Identify 99.7% range for sample means

For the sample mean \(\bar{x}\), we apply the 99.7% rule using the calculated standard error. Since the sample means are normally distributed around the population mean \(\mu = 3.3\), we find:\[ \mu - 3(SE) \quad \text{and} \quad \mu + 3(SE) \]\[ 3.3 - 3(0.08) = 3.3 - 0.24 = 3.06 \]\[ 3.3 + 3(0.08) = 3.3 + 0.24 = 3.54 \]So, we expect the sample means \(\bar{x}\) to lie between 3.06 and 3.54 in 99.7% of such samples.

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Key Concepts

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

Understanding Sampling Distribution
A sampling distribution refers to the probability distribution of a given statistic based on a random sample. In simpler terms, when we take several samples from a population, each sample has a mean. If we plot these means, we will have what is known as a sampling distribution of the sample mean. This distribution gives us insight into how the sample mean varies from sample to sample.

It's important because it allows us to make inferences about the entire population based on a sample. For example, we know that if we draw a large number of samples from a population, the sampling distribution of the sample means will be normally distributed if the population itself is normally distributed. This concept is crucial for constructing confidence intervals and hypothesis testing.
Decoding Standard Error
The standard error (SE) is a measure of the variability or dispersion of a sample statistic. Specifically, for the sample mean, it shows us how much we can expect the sample mean to differ from the actual population mean.

The formula to calculate the standard error of the sample mean is:
  • SE = \( \frac{\sigma}{\sqrt{n}} \)
where \( \sigma \) is the standard deviation of the population, and \( n \) is the sample size.

Lower values of the standard error indicate that the sample mean is a more precise estimate of the population mean. If you take larger samples, the SE decreases, meaning your estimate becomes more reliable.
Explaining the 68-95-99.7 Rule
The 68-95-99.7 rule, also known as the Empirical Rule, is a handy guide to understanding the spread of data in a normal distribution.

Here is what this rule tells us:
  • Approximately 68% of the data falls within one standard deviation of the mean.
  • Approximately 95% falls within two standard deviations of the mean.
  • Approximately 99.7% falls within three standard deviations of the mean.
This rule is particularly useful when trying to estimate the range of expected sample means or individual values.

For example, in our case where ratings of student engagement are studied, approximately 99.7% of the ratings are expected to fall within the range defined by three standard deviations on either side of the mean.
Unraveling the Sample Mean
The sample mean, often represented by \( \bar{x} \), is calculated by summing up all the observed values in a sample and then dividing by the number of observations. It serves as an estimate of the population mean, \( \mu \).

Calculating this mean is straightforward:
  • \( \bar{x} = \frac{\sum x}{n} \)
where \( \sum x \) is the sum of all sample values and \( n \) is the number of values.

In the context of the problem, the sample mean from different samples will vary, but thanks to the sampling distribution concept, these means will gravitate around the population mean, giving us a powerful tool for statistical inference.

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Most popular questions from this chapter

The Environmental Protection Agency (EPA) fuel economy ratings say that the 2016 Toyota Prius hybrid car gets 50 miles per gallon (mpg) on the highway. Deborah wonders whether the actual long-term average highway mileage \(\mu\) of her new Prius is less than \(50 \mathrm{mpg}\). She keeps careful records of gas mileage for 3000 miles of highway driving. Her result is \(x^{-} \bar{x}=49.2 \mathrm{mpg}\). What are her null and alternative hypotheses? (a) \(H_{0}: \mu=50, H_{a}: \mu<50\) (b) \(H_{0}: \mu=50, H_{a}: \mu>50\) (c) \(H_{0}: \mathrm{x}^{-} \bar{x}=50, H_{a}: \mathrm{x}^{-} \bar{x}<50\) (d) \(H_{0}: \mathrm{x}^{-} \bar{x}=50, H_{a}: \mathrm{x}^{-} \bar{x}>50\)

In a 2013 study, researchers compared various measurements on overweight firstborn and second-born middle-aged men. \({ }^{7}\) They found that first- borns had a significantly higher weight \((P=0.013\) ) than second-borns, but no significant difference in total cholesterol \((P=0.74)\). Explain carefully why \(P=0.013\) means there is evidence that first-born middle-aged men may have higher weights than second-borns and why \(P=0.74\) provides no evidence that first-born middle-aged men may have different total cholesterol levels than second-borns.

Choose a college freshman at random and ask what type of high school he or she attended. Here is the distribution of results: 13 $$ \begin{array}{l|llllll} \hline \text { Type } & \begin{array}{l} \text { Regular } \\ \text { Public } \end{array} & \begin{array}{l} \text { Public } \\ \text { Charter } \end{array} & \begin{array}{l} \text { Public } \\ \text { Magnet } \end{array} & \begin{array}{l} \text { Private } \\ \text { Religious } \end{array} & \begin{array}{l} \text { Private } \\ \text { Independent } \end{array} & \begin{array}{l} \text { Home } \\ \text { School } \end{array} \\ \hline \text { Probability } & 0.758 & 0.029 & 0.035 & 0.109 & 0.063 & 0.006 \\\ \hline \end{array} $$ What is the conditional probability that a college freshman was home schooled, given that he or she did not attend a regular public high school?

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