/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Calculate the value of \(t \star... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Calculate the value of \(t \star\) for the hypothesis test: \(H_{o}: \mu=32, H_{a}: \mu>32, n=16, \bar{x}=32.93, s=3.1.\)

Short Answer

Expert verified
The t-value (t*) is calculated using the given information and the standard formula for a right-tail t-test. fter conducting the above operations, the t-value (t*) can be obtained, which would be the final step in this procedure. The calculated t-value will then be used to decide whether to accept or reject the null hypothesis in statistical analysis.

Step by step solution

01

Identify the Sample Mean and Population Mean

From the hypotheses, the population mean (\mu) is 32. The sample mean (\bar{x}) is 32.93.
02

Identify the sample size and sample standard deviation

From the given exercise, the sample size (n) is 16 and the sample standard deviation (s) is 3.1.
03

Calculate the t-score (t*)

In a right-tail t-test, t* is calculated using the formula \[t* = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\].Substituting the values \(\bar{x}=32.93\), \(\mu=32\), \(s=3.1\) and \(n=16\), the formula results in \[t* = \frac{32.93 - 32}{\frac{3.1}{\sqrt{16}}}\]. Hence, the t-score t* is computed to obtain the test statistic. This value will then be compared with the critical t-value for a specified level of significance to decide whether to reject or not reject null hypothesis H_{o}.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

t-test
A t-test is a statistical method used to determine whether there is a significant difference between the means of two groups, which may be related in certain features. It's widely used when the test statistic follows a normal distribution if the value of a scaling term in the test statistic is known. When the scaling term is unknown and is replaced by an estimate based on the sample data, the test statistic (under certain conditions) follows a t distribution.

For the exercise provided, a one-sample t-test is used because we're comparing the sample mean to a known population mean to see if the sample provides enough evidence to conclude that the population mean is actually different from the value claimed under the null hypothesis. The calculation will demonstrate how the test helps to make this determination.
sample mean
The sample mean, often represented by \( \bar{x} \), is simply the average of all the data points in a sample. It is computed by summing up all the observations and dividing by the number of observations. For the exercise at hand, if we were to add up all the individual observations in the sample and divide by the sample size, the result would be 32.93. This value serves as the point estimate of the population mean and is key to conducting the hypothesis test.
sample standard deviation
The sample standard deviation (s) is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.

In the context of our exercise, the sample standard deviation is given as 3.1, which tells us about the variability of students' test scores around the sample mean. When conducting the t-test, this measure of variability is crucial for calculating the standard error, which then helps to determine the test statistic.
test statistic
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to decide whether to reject the null hypothesis. The formula typically depends on the type of test being performed, but all test statistics are a function of the difference between the sample data and the null hypothesis, adjusted for the amount of sample data variability.

In this case, the test statistic is a t-score represented by \( t* \). It's calculated as the difference between the sample mean and the population mean, divided by the standard error of the mean (which is the sample standard deviation divided by the square root of the sample size). The larger the absolute value of the t-score, the more likely it is that the null hypothesis will be rejected, which means there is evidence that the sample provides sufficient proof of a difference between the sample mean and the population mean.
population mean
The population mean, denoted as \( \mu \), is the average of all the values in a population. In hypothesis testing, the population mean is what you compare your sample mean (\( \bar{x} \)) against to determine if there are differences that are statistically significant, rather than due to random chance.

In our exercise, the null hypothesis (\( H_{0} \)) states that the population mean is 32. This is the assumed value based on the statement of the hypothesis unless evidence suggests otherwise. It's this stated population mean that we are testing against our sample mean to either provide evidence against the null hypothesis or fail to do so.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the test statistic for the hypothesis test: a. \(H_{o}: \sigma^{2}=532\) versus \(H_{a}: \sigma^{2}>532\) using sample information \(n=18\) and \(s^{2}=785\) b. \(H_{o}: \sigma^{2}=52\) versus \(H_{a}: \sigma^{2} \neq 52\) using sample information \(n=41\) and \(s^{2}=78.2\)

Find \(n\) for a \(90 \%\) confidence interval for \(p\) with \(E=0.02\) using an estimate of \(p=0.25\).

a. What is the relationship between \(p=P(\text { success })\) and \(q=P(\text { failure }) ?\) Explain. b. Explain why the relationship between \(p\) and \(q\) can be expressed by the formula \(q=1-p\) c. If \(p=0.6,\) what is the value of \(q ?\) d. If the value of \(q^{\prime}=0.273,\) what is the value of \(p^{\prime} ?\)

It is important that the force required to extract a cork from a wine bottle not have a large standard deviation. Years of production and testing indicate that the no.9 corks in Applied Example 6.13 (p. 285 ) have an extraction force that is normally distributed with a standard deviation of 36 Newtons. Recent changes in the manufacturing process are thought to have reduced the standard deviation. a. What would be the problem with the standard deviation being relatively large? What would be the advantage of a smaller standard deviation? A sample of 20 randomly selected bottles is used for testing. Extraction Force in Newtons $$\begin{array}{llllllllll}\hline 296 & 338 & 341 & 261 & 250 & 347 & 336 & 297 & 279 & 297 \\\259 & 334 & 281 & 284 & 279 & 266 & 300 & 305 & 310 & 253 \\\\\hline\end{array}$$ b. Is the preceding sample sufficient to show that the standard deviation of extraction force is less than 36.0 Newtons, at the 0.02 level of significance? During a different testing, a sample of eight bottles is randomly selected and tested. Extraction Force in Newtons $$\begin{array}{rrrrrr}331.9 & 312.0 & 289.4 & 303.6 & 346.9 & 308.1 & 346.9 & 276.0\end{array}$$ c. Is the preceding sample sufficient to show that the standard deviation of extraction force is less than 36.0 Newtons, at the 0.02 level of significance? d. What effect did the two different sample sizes have on the calculated test statistic in parts b and c? What effect did they have on the \(p\) -value or critical value? Explain. e. What effect did the two different sample standard deviations have on the answers in parts b and c? What effect did they have on the \(p\) -value or critical value? Explain.

State the null hypothesis, \(H_{o},\) and the alternative hypothesis, \(H_{a},\) that would be used to test these claims: a. The probability of our team winning tonight is less than 0.50. b. At least \(50 \%\) of all parents believe in spanking their children when appropriate. c. At most, \(80 \%\) of the invited guests will attend the wedding. d. The single-digit numbers generated by the computer do not seem to be equally likely with regard to being odd or even. e. Less than half of the customers like the new pizza.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.