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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

Tests and CIs The P-value for a two-sided test of the null hypothesis \(H_{0} : \mu=10\) is \(0.06 .\) (a) Does the 95\(\%\) confidence interval for \(\mu\) include 10 ? Why or why not? (b) Does the 90\(\%\) confidence interval for \(\mu\) include 10? Why or why not?

Bullies in middle school A University of Illinois study on aggressive behavior surveyed a random sample of 558 middle school students. When asked to describe their behavior in the last 30 days, 445 students said their behavior included physical aggression, social ridicule, teasing, name-calling, and issuing threats. This behavior was not defined as bullying in the questionnaire. Is this evidence that more than three-quarters of the students at that middle school engage in bullying behavior? To find out, Maurice decides to perform a significance test. Unfortunately, he made a few errors along the way. Your job is to spot the mistakes and correct them. $$ \begin{array}{l}{H_{0} : p=0.75} \\ {H_{a} : \hat{p}>0.797}\end{array} $$ where \(p=\) the true mean proportion of middle school students who engaged in bullying. A random sample of 558 middle school students was surveyed. \(558(0.797)=444.73\) is at least 10 $$ z=\frac{0.75-0.797}{\sqrt{\frac{0.797(0.203)}{445}}}=-2.46 ; P \text { -value }=2(0.0069)=0.0138 $$ The probability that the null hypothesis is true is only \(0.0138,\) so we reject \(H_{0} .\) This proves that more than three-quarters of the school engaged in bullying behavior.

Slow response times by paramedics, firefighters, and policemen can have serious consequences for accident victims. In the case of life-threatening injuries, victims generally need medical attention within 8 minutes of the accident. Several cities have begun to monitor emergency response times. In one such city, the mean response time to all accidents involving life- threatening injuries last year was \(\mu=6.7\) minutes. Emergency personnel arrived within 8 minutes after 78\(\%\) of all calls involving life-threatening injuries last year. The city manager shares this information and encourages these first responders to 鈥渄o better.鈥 At the end of the year, the city manager selects an SRS of 400 calls involving life-threatening injuries and examines the response times. Awful accidents (a) State hypotheses for a significance test to determine whether the average response time has decreased. Be sure to define the parameter of interest. (b) Describe a Type I error and a Type II error in this setting, and explain the consequences of each. (c) Which is more serious in this setting: a Type I error or a Type II error? Justify your answer.

Teen drivers A state鈥檚 Division of Motor Vehicles (DMV) claims that 60% of teens pass their driving test on the first attempt. An investigative reporter examines an SRS of the DMV records for 125 teens; 86 of them passed the test on their first try. Is this good evidence that the DMV鈥檚 claim is incorrect? Carry out a test at the A 0.05 significance level to help answer this question.

The \(z\) statistic for a test of \(H_{0} : p=0.4\) versus \(H_{a} :\) \(p>0.4\) is \(z=2.43\) . This test is (a) not significant at either \(\alpha=0.05\) or \(\alpha=0.01\) (b) significant at \(\alpha=0.05\) but not at \(\alpha=0.01\) (c) significant at \(\alpha=0.01\) but not at \(\alpha=0.05\) (d) significant at both \(\alpha=0.05\) and \(\alpha=0.01\) (e) inconclusive because we don't know the value of \(\hat{p} .\)

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