/*! 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 7 Conduct the appropriate test. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Conduct the appropriate test. A simple random sample of size \(n=15\) is drawn from a population that is normally distributed. The sample mean is found to be \(23.8,\) and the sample standard deviation is found to be \(6.3 .\) Test whether the population mean is different from 25 at the \(\alpha=0.01\) level of significance.

Short Answer

Expert verified
Fail to reject \(H_0\). No significant difference from 25.

Step by step solution

01

- State the hypotheses

Formulate the null hypothesis (ullhypothesis) and the alternative hypothesis (alternativehypothesis). For this test: \(H_0: \text{\mu} = 25\) \(H_1: \text{\mu} eq 25\)
02

- Determine the test statistic

Since the sample size is small and the population standard deviation is unknown, use the t-test. The test statistic is calculated as follows: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] where: \(\bar{x} = 23.8\) \(\text{(sample mean)}\)\(\mu = 25\) \(\text{(population mean under } H_0 \text{)}\)\(s = 6.3\) \(\text{(sample standard deviation)}\)\(n = 15\) \(\text{(sample size)}\)
03

- Calculate the test statistic

Substitute the known values into the test statistic formula: \[ t = \frac{23.8 - 25}{\frac{6.3}{\sqrt{15}}} = \frac{-1.2}{\frac{6.3}{3.872}} = \frac{-1.2}{1.626} \approx -0.738 \]
04

- Determine the degrees of freedom

The degrees of freedom for this test is \(df = n - 1 = 15 - 1 = 14\).
05

- Find the critical value

Using a t-distribution table or calculator, find the critical t-value for a two-tailed test at the \(\alpha = 0.01\) level of significance with 14 degrees of freedom. The critical value is approximately \(\pm 2.977\).
06

- Compare the test statistic to the critical value

Compare the test statistic \(t \approx -0.738\) to the critical values \(\pm 2.977\). Since \(-0.738\) lies within the range \([-2.977, 2.977]\), we fail to reject the null hypothesis.
07

- Conclusion

Based on the test, there is not enough evidence to reject the null hypothesis at the \(\alpha = 0.01\) level of significance. Therefore, we conclude that the population mean is not significantly different from 25.

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 type of inferential statistic used to determine if there is a significant difference between the means of two groups. This can apply to a single group compared to a known mean (one-sample t-test), or two groups compared to each other (two-sample t-test). In this scenario, we use the one-sample t-test because we are comparing our sample mean to a known population mean.

The formula for the t-test statistic is \( t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \). Here, the symbol \( \bar{x} \) represents the sample mean, \( \mu \) is the population mean under the null hypothesis, \( s \) stands for the sample standard deviation, and \( n \) is the sample size. This formula helps us understand how far our sample mean is from the population mean in terms of the standard error.
null hypothesis
In hypothesis testing, the null hypothesis (\text{H}_0) is a general statement or default position that there is no difference or relationship between the measured phenomena. The null hypothesis is what we are testing against. In this exercise, the null hypothesis is \( H_0: \mu = 25 \). This means that we assume the population mean is 25 until we have enough evidence to prove otherwise.

Rejecting the null hypothesis would mean we have enough statistical evidence to support that the population mean is different from 25. However, if we fail to reject it, we do not have sufficient evidence to say the population mean is different from 25.
levels of significance
The level of significance, denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. Common levels of significance are 0.05, 0.01, and 0.10, but in this exercise, \( \alpha = 0.01 \) is used. This strict criterion implies that we are willing to accept only a 1% chance of mistakenly rejecting the null hypothesis.

Using a lower level of significance makes the test more conservative, reducing the risk of Type I errors (false positives), but increasing the risk of Type II errors (false negatives). Therefore, choosing an appropriate \( \alpha \) level is crucial to balancing these risks.
degrees of freedom
Degrees of freedom (df) refer to the number of values in a calculation that are free to vary. For the t-test, the degrees of freedom are calculated as \( df = n - 1 \), where \( n \) is the sample size. In this problem, with \( n = 15 \), the degrees of freedom is \( df = 15 - 1 = 14 \).

The degrees of freedom play a crucial role in determining the critical value from the t-distribution table. The more degrees of freedom you have, the closer the t-distribution gets to a normal distribution. With fewer degrees of freedom, the t-distribution becomes more spread out.
sample mean
The sample mean (\text{\bar{x}}) is the average value obtained from the sample data. It is calculated by summing all the sample values and then dividing by the number of samples. In this exercise, the sample mean is 23.8. The formula for the sample mean is \( \bar{x} = \frac{\sum x_i}{n} \), where \( \sum x_i \) is the sum of all sample values, and \( n \) is the number of samples.

The sample mean is an important statistic used to estimate the population mean. By comparing the sample mean to the population mean under the null hypothesis using the t-test statistic, we can draw inferences about the population based on our sample.

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

Talk to the Animals In a survey conducted by the American Animal Hospital Association, \(37 \%\) of respondents stated that they talk to their pets on the answering machine or telephone. A veterinarian found this result hard to believe, so he randomly selected 150 pet owners and discovered that 54 of them spoke to their pet on the answering machine or telephone. Does the veterinarian have a right to be skeptical? Use the \(\alpha=0.05\) level of significance.

A null and alternative hypothesis is given. Determine whether the hypothesis test is left-tailed, right-tailed, or two tailed. What parameter is being tested? $$\begin{aligned}&H_{0}: \mu=120\\\&H_{1}: \mu<120\end{aligned}$$

Test the hypothesis, using (a) the classical approach and then (b) the P-value approach. Be sure to verify the requirements of the test. $$\begin{aligned}&H_{0}: p=0.25 \text { versus } H_{1}: p<0.25\\\&n=400 ; x=96 ; \alpha=0.1\end{aligned}$$

Carl Reinhold August Wunderlich said that the mean temperature of humans is \(98.6^{\circ} \mathrm{F}\). Researchers Philip Mackowiak, Steven Wasserman, \(\begin{array}{lllllll}\text { and } & \text { Myron } & \text { Levine } & {[\mathrm{JAMA},} & \text { Sept. } & 23-30 & 1992\end{array}\) \(268(12): 1578-80]\) thought that the mean temperature of humans is less than \(98.6^{\circ}\) F. They measured the temperature of 148 healthy adults 1 to 4 times daily for 3 days, obtaining 700 measurements. The sample data resulted in a sample mean of \(98.2^{\circ} \mathrm{F}\) and a sample standard deviation of \(0.7^{\circ} \mathrm{F}\) (a) Test whether the mean temperature of humans is less than \(98.6^{\circ} \mathrm{F}\) at the \(\alpha=0.01\) level of significance using the classical approach. (b) Determine and interpret the \(P\) -value.

Suppose that we are testing the hypotheses \(H_{0}: \mu=\mu_{0}\) versus \(H_{1}: \mu \neq \mu_{0}\) and we find the \(P\) -value to be 0.02 Explain what this means. Would you reject \(H_{0} ?\) Why?

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.