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91Ó°ÊÓ

We prefer the \(t\) procedures to the z procedures for inference about a population mean because (a) \(z\) requires that you know the observations are from a Normal population, while \(t\) does not. (b) \(z\) requires that you know the population standard deviation \(\sigma\), while \(t\) does not. (c) \(z\) requires that you can regard your data as an SRS from the population, while \(t\) does not.

Short Answer

Expert verified
(b) T procedures don't require knowing the population standard deviation.

Step by step solution

01

Understanding the Requirement of Z Procedures

Z procedures are used for inference about a population mean when we know the population standard deviation (\(\sigma\)). This is often a limitation since in practice, the population standard deviation is rarely known.
02

Noting the Requirements of T Procedures

T procedures are useful when the population standard deviation is unknown, which is a common situation. Instead of \(\sigma\), the sample standard deviation \(s\) is used in calculations. Both t and z procedures assume data is from an SRS (Simple Random Sample) and normally distributed.
03

Comparing Options (a), (b), and (c)

Option (a) is incorrect as both procedures assume data is from a normal distribution if sample size is small; large sample sizes relax this assumption. Option (c) is also incorrect as both procedures require data to be from an SRS. Option (b) is correct; the key difference is that z requires known \(\sigma\), while t does not.
04

Concluding the Correct Option

The primary advantage of using t procedures over z procedures is related to the knowledge of population standard deviation. Since t procedures rely on the sample standard deviation rather than the population standard deviation, they are more flexible and widely applicable.

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Key Concepts

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

Population Mean
The population mean is a fundamental concept in statistics and represents the average of all individual data points in a given population. Think of it as the center of your data. For instance, if you had a population of test scores, the population mean would be the average score. This mean reflects the overall tendency or average outcome you would expect from that population.
In practice, calculating the exact population mean can be difficult, especially when dealing with large populations. Instead, we often estimate the population mean using sample data. This is where concepts like the sample mean, calculated by averaging the data points in our sample, come into play.
Remember, the main goal of statistical analysis, especially when using tests like the t-test, is to make inferences about the population mean based on sample data.
Normal Distribution
A normal distribution, often called a bell curve due to its shape, is a probability distribution that is symmetrical about its mean. Most data points cluster around the central peak, and the probabilities taper off equally on both sides.
Some key characteristics of a normal distribution include:
  • The mean, median, and mode are all the same and located at the center of the distribution.
  • It is completely defined by its mean and standard deviation.
  • About 68% of the data falls within one standard deviation (\(σ\)) of the mean, about 95% within two, and about 99.7% within three.

In statistics, many tests, including the t-test and z-test, assume that data is normally distributed. However, when sample sizes are large, the Central Limit Theorem allows for the assumption of normality to be relaxed, meaning the distribution of the sample means approximates normality, even if the underlying population distribution is not strictly normal.
Population Standard Deviation
The population standard deviation, denoted by \(σ\), is a measure of the amount of variation or dispersion in a population. Simply put, it tells you how spread out the data points are around the mean. If you have a small standard deviation, it indicates that the data points tend to be very close to the mean. Conversely, a large standard deviation means that the data points are spread out over a wider range of values.
In many real-world situations, the population standard deviation is unknown and can be challenging to determine. This is where inferential statistics come in, allowing us to estimate this standard deviation by using the sample standard deviation. The t-test is particularly useful for such situations as it relies on the sample standard deviation when the population standard deviation is not available. This makes it very practical for analyzing real-world data.
Sample Standard Deviation
The sample standard deviation is denoted by \(s\) and is used to quantify the amount of variation within a sample. It plays a crucial role in statistics when the population standard deviation is unknown, which is commonly the case. The sample standard deviation is calculated by taking the square root of the variance, which is the average of the squared differences from the sample mean.
Using the sample standard deviation is essential when performing a t-test. The t-test employs \(s\) instead of the unknown population standard deviation to ascertain if there is a significant difference between the sample mean and the hypothesized population mean. Thus, \(s\) provides us with an estimation based on the data available from our sample, offering a glimmer into the population's variability in lieu of \(σ\). Understanding this concept is pivotal as it forms the backbone of making reliable statistical inferences when the population standard deviation is not accessible.

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Most popular questions from this chapter

The Trial Urban District Assessment (TUDA) is a government-sponsored study of student achievement in large urban school districts. TUDA gives a mathematics test scored from 0 to 500 . A score of 262 is a "basic" mathematics level and a score of 299 is "proficient." Scores for a random sample of 1100 eighth- graders in Dallas had \(x \bar{x}=271\) with standard error 1.3. 16 (a) We don't have the 1100 individual scores, but use of the \(t\) procedures is surely safe. Why? (b) Give a \(99 \%\) confidence interval for the mean score of all Dallas eighthgraders. (Be careful: the report gives the standard error of \(x^{-} \bar{x}\), not the standard deviation s.) (c) Urban children often perform below the basic level. Is there good evidence that the mean for all Dallas eighth-graders is more than the basic level?

Does a football filled with helium travel farther than one filled with ordinary air? To test this, the Columbus Dispatch conducted a study. Two identical footballs, one filled with helium and one filled with ordinary air, were used. A casual observer was unable to detect a difference in the two footballs. A novice kicker was used to punt the footballs. A trial consisted of kicking both footballs in a random order. The kicker did not know which football (the helium-filled or the air-filled football) he was kicking. The distance of each punt was recorded. Then another trial was conducted. A total of 39 trials were run. Here are the data for the 39 trials, in yards that the footballs traveled. The difference (helium minus air) is the response variable. \({ }^{25}\) $$ \begin{array}{l|rrrrrrrrrr} \hline \text { Helium } & 25 & 16 & 25 & 14 & 23 & 29 & 25 & 26 & 22 & 26 \\ \text { Air } & 25 & 23 & 18 & 16 & 35 & 15 & 26 & 24 & 24 & 28 \\ \hline \text { Difference } & 0 & -7 & 7 & -2 & -12 & 14 & -1 & 2 & -2 & -2 \\\ \hline \text { Helium } & 12 & 28 & 28 & 31 & 22 & 29 & 23 & 26 & 35 & 24 \\ \text { Air } & 25 & 19 & 27 & 25 & 34 & 26 & 20 & 22 & 33 & 29 \\ \hline \text { Difference } & -13 & 9 & 1 & 6 & -12 & 3 & 3 & 4 & 2 & -5 \\ \hline \text { Helium } & 31 & 34 & 39 & 32 & 14 & 28 & 30 & 27 & 33 & 11 \\ \text { Air } & 31 & 27 & 22 & 29 & 28 & 29 & 22 & 31 & 25 & 20 \\ \hline \text { Difference } & 0 & 7 & 17 & 3 & -14 & -1 & 8 & -4 & 8 & -9 \\ \hline \text { Helium } & 26 & 32 & 30 & 29 & 30 & 29 & 29 & 30 & 26 & \\ \hline \text { Air } & 27 & 26 & 28 & 32 & 28 & 25 & 31 & 28 & 28 & \\ \hline \text { Difference } & -1 & 6 & 2 & -3 & 2 & 4 & -2 & 2 & -2 & \\ \hline \end{array} $$ (a) Examine the data. Is it reasonable to use the \(t\) procedures? (b) If your conclusion in part (a) is Yes, do the data give convincing evidence that the helium-filled football travels farther than the air-filled football?

Because the \(t\) procedures are robust, the most important condition for their safe use is that (a) the sample size is at least \(15 .\) (b) the population distribution is exactly Normal. (c) the data can be regarded as an SRS from the population.

How much oil will ultimately be produced by wells in a given field is key information in deciding whether to drill more wells. Here are the estimated total amounts of oil recovered from 64 wells in the Devonian Richmond Dolomite area of the Michigan basin, in thousands of barrels: \({ }^{29}\) $$ \begin{array}{llllllll} 21.7 & 53.2 & 46.4 & 42.7 & 50.4 & 97.7 & 103.1 & 51.9 43.4 & 69.5 & 156.5 & 34.6 & 37.9 & 12.9 & 2.5 & 31.4 \\ 79.5 & 26.9 & 18.5 & 14.7 & 32.9 & 196 & 24.9 & 118.2 \\ 82.2 & 35.1 & 47.6 & 54.2 & 63.1 & 69.8 & 57.4 & 65.6 \\ 56.4 & 49.4 & 44.9 & 34.6 & 92.2 & 37.0 & 58.8 & 21.3 \\ 36.6 & 64.9 & 14.8 & 17.6 & 29.1 & 61.4 & 38.6 & 32.5 \\ 12.0 & 28.3 & 204.9 & 44.5 & 10.3 & 37.7 & 33.7 & 81.1 \\ 12.1 & 20.1 & 30.5 & 7.1 & 10.1 & 18.0 & 3.0 & 2.0 \end{array} $$ Take these wells to be an SRS of wells in this area. (a) Give a \(95 \%\) confidence interval for the mean amount of oil recovered from all wells in this area. (b) Make a graph of the data. The distribution is very skewed, with several high outliers. A computer-intensive method that gives accurate confidence intervals without assuming any specific shape for the distribution gives a \(95 \%\) confidence interval of \(40.28\) to \(60.32\). How does the \(t\) interval compare with this? Should the \(t\) procedures be used with these data?

A study of commuting times reports the travel times to work of a random sample of 1000 employed adults in Seattle. \({ }^{2}\) The mean is \(\mathrm{x}^{-} \bar{x}=\) 37.9 minutes and the standard deviation is \(s=27.2\) minutes. What is the standard error of the mean?

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