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The reason we use t procedures instead of z procedures when carrying out a test about a population mean is that (a) z can be used only for large samples. (b) z requires that you know the population standard deviation S. (c) z requires you to regard your data as an SRS from the population. (d) z applies only if the population distribution is perfectly Normal. (e) z can be used only for confidence intervals.

Short Answer

Expert verified
We use t-procedures because z requires the population standard deviation, option (b).

Step by step solution

01

Understand the Context

When we carry out tests about a population mean, we need to choose between using t-procedures and z-procedures. The choice depends on certain assumptions about the population and the sample data.
02

Identify the Requirement for t-procedures

T-procedures are particularly used when the population standard deviation (\( \sigma \)) is unknown. Instead of \( \sigma \), we use the sample standard deviation \( s \), which allows us to use the t-distribution.
03

Assess the Options

Consider each option given in the exercise:- (a) Incorrect: z-procedures can be used for both large and small samples if the population standard deviation is known and certain conditions are met.- (b) Correct: z-procedures require knowledge of the population standard deviation \( \sigma \).- (c) Incorrect: While a SRS is a requirement for both t and z tests, it doesn't differentiate between them.- (d) Incorrect: z-procedures can be used for approximately Normal populations, especially with large samples due to the Central Limit Theorem.- (e) Incorrect: z-procedures can be used for hypothesis testing as well as confidence intervals.
04

Choose the Correct Answer

Based on the assessment, the correct reason for using t-procedures over z-procedures is option (b), which states that z-procedures require knowledge of the population standard deviation \( \sigma \) , typically unknown in practice.

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Key Concepts

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

Population Mean Testing
Testing for a population mean is a common statistical procedure used to make inferences about the average of an entire population based on a sample. Often, we want to determine if the mean of a population is equal to a specific value or if it differs in some way. To carry out these tests, we may choose between t-procedures and z-procedures depending on the data available to us.One key consideration is whether the population standard deviation, denoted as \( \sigma \), is known. When \( \sigma \) is known, z-procedures are generally preferred because they can provide more precise results. However, in most real-world situations, \( \sigma \) is unknown, making it necessary to use the sample standard deviation to estimate it. Using the sample standard deviation, we apply t-procedures, which utilize the t-distribution. This distribution helps account for the additional uncertainty that results from estimating the population standard deviation rather than knowing it exactly.It's also essential to assess whether the sample is large enough or if the population distribution approximates normality, as these factors influence whether we can confidently use normal approximations in our testing.
Sample Standard Deviation
The sample standard deviation, denoted as \( s \), plays a critical role in statistical analysis, particularly when testing a population mean where the population standard deviation \( \sigma \) is unknown.It is a measure of how spread out the values in a sample are and is calculated from the data in that sample. Importantly, \( s \) helps to estimate \( \sigma \), allowing us to perform statistical tests like the t-test, which compensates for not having direct access to the precise population parameter.When calculating the sample standard deviation, each data point's deviation from the sample mean is squared, summed, and then averaged (mean average). This average is done by dividing by one less than the sample size (n-1), known as the degrees of freedom. The square root of this average gives us the sample standard deviation \( s \).
  • Allows for estimations when \( \sigma \) is unknown.
  • Utilized in t-procedures providing a robust analysis framework.
  • Reflects sample variability, integral to determining the reliability of inferential statistics.
The sample standard deviation is foundational in bridging sample observations with predictions about the entire population, ensuring more accurate analysis and conclusions.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental statistical principle that plays a crucial role in sampling and inferential statistics. It posits that when you take a sufficiently large sample size from any population, regardless of the original distribution of the population, the distribution of the sample means will approach a normal distribution. This holds true even if the population distribution is not normal, provided the sample size is large enough, typically more than 30 observations. This theorem justifies the use of normal approximation techniques for hypothesis testing and constructing confidence intervals, particularly when dealing with large samples. It helps to simplify analysis by allowing statisticians to use the normal distribution as a model, which is mathematically well-understood and convenient. Key outcomes of the CLT include:
  • Sample mean distribution tends towards normality with large samples.
  • Applicable to almost any population distribution shape.
  • Facilitates use of z-procedures and t-procedures in practical applications.
The Central Limit Theorem thus provides the underpinnings for much of statistical inference, enabling more effective and efficient conclusions about population parameters based on samples.

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Most popular questions from this chapter

One-sided test Suppose you carry out a significance test of \(H_{0} : \mu=5\) versus \(H_{a} : \mu>5\) based on a sample of size \(n=20\) and obtain \(t=1.81\) (a) Find the P-value for this test using (i) Table B and (ii) your calculator. What conclusion would you draw at the 5% significance level? At the 1% significance level? (b) Redo part (a) using an alternative hypothesis of \(H_{a} : \mu \neq 5\)

A researcher plans to conduct a significance test at the \(\alpha=0.01\) significance level. She designs her study to have a power of 0.90 at a particular alternative value of the parameter of interest. The probability that the researcher will commit a Type II error for the particular alternative value of the parameter at which she computed the power is (a) 0.01. (b) 0.10. (c) 0.89. (d) 0.90. (e) 0.99.

Packaging \(\operatorname{CDs}(6.2,5.3)\) A manufacturer of compact discs (CDs) wants to be sure that their CDs will fit inside the plastic cases they have bought for packaging. Both the CDs and the cases are circular. According to the supplier, the plastic cases vary Normally with mean diameter \(\mu=4.2\) inches and standard deviation \(\sigma=0.05\) inches. The CD manufacturer decides to produce CDs with mean diameter \(\mu=4\) inches. Their diameters follow a Normal distribution with \(\sigma=0.1\) inches. (a) Let X the diameter of a randomly selected CD and Y the diameter of a randomly selected case. Describe the shape, center, and spread of the distribution of the random variable X Y. What is the importance of this random variable to the CD manufacturer? (b) Compute the probability that a randomly selected CD will fit inside a randomly selected case. (c) The production process actually runs in batches of 100 CDs. If each of these CDs is paired with a randomly chosen plastic case, find the probability that all the CDs fit in their cases.

Reporting cheating What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: 鈥淵ou witness two students cheating on a quiz. Do you go to the professor?鈥 The Minitab output below shows the results of a significance test and a 95% confidence interval based on the survey data. (a) Define the parameter of interest. (b) Check that the conditions for performing the significance test are met in this case. (c) Interpret the P-value in context. (d) Do these data give convincing evidence that the actual population proportion differs from 0.15? Justify your answer with appropriate evidence.

Losing weight A Gallup Poll found that 59% of the people in its sample said 鈥淵es鈥 when asked, 鈥淲ould you like to lose weight?鈥 Gallup announced: 鈥淔or results based on the total sample of national adults, one can say with 95% confidence that the margin of (sampling) error is \(\pm 3\) percentage points." \(^{\prime \prime 6}\) Can we use this interval to conclude that the actual proportion of U.S. adults who would say they want to lose weight differs from 0.55\(?\) Justify your answer.

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