/*! 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 574 Suppose it is required that the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 574

Suppose it is required that the mean operating life of size "D" batteries be 22 hours. Suppose also that the operating life of the batteries is normally distributed. It is known that the standard deviation of the operating life of all such batteries produced is \(3.0\) hours. If a sample of 9 batteries has a mean operating life of 20 hours, can we conclude that the mean operating life of size "D" batteries is not 22 hours? Then suppose the standard deviation of the operating life of all such batteries is not known but that for the sample of 9 batteries the standard deviation is \(3.0 .\) What conclusion would we then reach?

Short Answer

Expert verified
In the first part, using a z-test with a known population standard deviation, we find a p-value of 0.0456, which is less than the 0.05 significance level. Therefore, we reject the null hypothesis and conclude that the mean operating life of size "D" batteries is not 22 hours. In the second part, using a t-test with the given sample standard deviation, we find a p-value of 0.0818, which is greater than the 0.05 significance level. Therefore, we fail to reject the null hypothesis, and we do not have enough evidence to conclude that the mean operating life of size "D" batteries is not 22 hours.

Step by step solution

01

State the null and alternative hypotheses

We need to set up null and alternative hypotheses for the problem. The null hypothesis is that the mean operating life of size "D" batteries is 22 hours, and the alternative hypothesis is that it is not 22 hours. \(H_0: \mu = 22\) \(H_1: \mu \neq 22\) #Step 2: Calculate the z-score of the sample#
02

Calculate the z-score of the sample

Next, we will calculate the z-score of the sample using the formula: \(z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\) Where: - \(z\) is the z-score - \(\bar{x}\) is the sample mean (20 hours) - \(\mu\) is the population mean (22 hours) - \(\sigma\) is the population standard deviation (3.0 hours) - \(n\) is the sample size (9 batteries) \(z = \frac{20 - 22}{\frac{3}{\sqrt{9}}} = -2\) #Step 3: Find the p-value associated with the z-score#
03

Find the p-value associated with the z-score

Let's determine the p-value associated with the z-score. Since our alternative hypothesis is two-tailed (i.e., we are testing for a not equal to relationship), we need to find the probability that a z-score is more extreme than -2 in both the left and right tails of the normal distribution. P(Z < -2) = 0.0228 (from the standard normal probability table or using a calculator) P(Z > 2) = 0.0228 Two-tailed P(Z < -2 鈭� Z > 2) = 2 脳 0.0228 = 0.0456 #Step 4: Compare the p-value with the significance level#
04

Compare the p-value with the significance level

Let's assume a significance level of 0.05. Since the p-value (0.0456) is less than the significance level (0.05), we reject the null hypothesis. Thus, there is enough evidence to conclude that the mean operating life of size "D" batteries is not 22 hours. #Part 2: Perform a t-test with the given sample standard deviation# #Step 5: Calculate the t-score of the sample#
05

Calculate the t-score of the sample

Now, we will calculate the t-score of the sample using the formula: \(t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\) Where: - \(t\) is the t-score - \(\bar{x}\) is the sample mean (20 hours) - \(\mu\) is the population mean (22 hours) - \(s\) is the sample standard deviation (3.0 hours) - \(n\) is the sample size (9 batteries) \(t = \frac{20 - 22}{\frac{3}{\sqrt{9}}} = -2\) #Step 6: Find the p-value associated with the t-score#
06

Find the p-value associated with the t-score

Since our alternative hypothesis is two-tailed, we need to find the probability that a t-score is more extreme than -2 in both the left and right tails of the Student's t-distribution with 8 degrees of freedom (df = 9 - 1 = 8). P(T < -2) = 0.0409 (from the t-probability table or using a calculator) P(T > 2) = 0.0409 Two-tailed P(T < -2 鈭� T > 2) = 2 脳 0.0409 = 0.0818 #Step 7: Compare the p-value with the significance level#
07

Compare the p-value with the significance level

Let's assume the same significance level of 0.05. Since the p-value (0.0818) is greater than the significance level (0.05), we fail to reject the null hypothesis. With the given sample standard deviation, we do not have enough evidence to conclude that the mean operating life of size "D" batteries is not 22 hours.

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 91影视!

Key Concepts

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

z-score
In statistics, a z-score, also known as a standard score, is a way to describe the position of a raw score in relation to the mean of a group of scores. It is a measure of how many standard deviations an element is from the mean. For this reason, it's a powerful tool for comparing results across different data sets.
The formula for calculating a z-score is:
  • \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \)
Where:
  • \( \bar{x} \) is the sample mean,
  • \( \mu \) is the population mean,
  • \( \sigma \) is the population standard deviation,
  • \( n \) is the sample size.
In our exercise, we calculate a z-score to assess whether the sample mean significantly differs from the population mean. A z-score of \(-2\) indicates the sample mean is 2 standard deviations away from the population mean, suggesting a potential difference.
t-test
The t-test is a statistical test that is used to compare the means of two groups. It's particularly useful when the population standard deviation is unknown, which often happens in real-life scenarios. Instead of using the population standard deviation, it uses the sample standard deviation and is applied using the t-distribution, which accounts for the smaller sample size by having 'fatter' tails than the normal distribution.
The formula for calculating the t-score is:
  • \( t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \)
Where:
  • \( \bar{x} \) is the sample mean,
  • \( \mu \) is the population mean,
  • \( s \) is the sample standard deviation,
  • \( n \) is the sample size.
In our battery life example, a t-score of \(-2\) was calculated. However, the evidence wasn't strong enough to reject the null hypothesis when the sample standard deviation was used. This showcases the importance of considering sample size and variance when drawing conclusions from data.
standard deviation
Standard deviation is a measure that indicates the amount of variation or dispersion in a set of values. Simply put, it tells us how much the individual data points differ from the mean of the data set. A small standard deviation means the data points are close to the mean, while a large one indicates they are spread out over a wider range of values.
In statistical testing, like in our battery life test, standard deviation plays a crucial role. It allows us to understand how much our sample results can vary. In cases where the population standard deviation is known, we use it directly in z-score calculations. When unknown, we rely on the sample standard deviation for t-tests. This transformation ensures our tests accommodate variation in data, providing more accurate results.

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

An electrical company claimed that at least \(95 \%\) of the parts which they supplied on a government contract conformed to specifications. A sample of 400 parts was tested, and 45 did not meet specifications. Can we accept the company's claim at a \(.05\) level of significance?

\(\mathrm{H}_{0}: \mu=\mu_{0}\) \(\mathrm{H}_{1}: \mu=\mu_{1}\) and \(\alpha\) and \(\beta\) are probabilities of making type I and type II errors respectively, show that for a two-tailed test the required sample size \(\mathrm{n}\) is given approximately by $$ \mathrm{n}=\left[\left\\{\left(\mathrm{Z}_{\alpha / 2}+\mathrm{Z}_{\mathrm{B}}\right)^{2} \sigma^{2}\right\\} /\left(\mu_{1}-\mu_{0}\right)^{2}\right] \text { , } $$ provided that $$ \operatorname{Pr}\left[Z<-Z_{\alpha / 2}-\left\\{\left(\sqrt{n}\left|\mu_{1}-\mu_{0}\right|\right) / \sigma\right\\}\right] $$ is small when \(\mu=\mu_{1}\).

Past experience has shown that the scores of students who take a certain mathematics test are normally distributed with mean 75 and variance \(36 .\) The Mathematics Department members would like to know whether this year's group of 16 students is typical. They decide to test the hypothesis that this year's students are typical versus the alternative that they are not typical. When the students take the test the average score is 82 . What conclusion should be drawn?

Out of 57 men at a weekly college dance, 36 had been at the dance the week before, and of these, 23 had brought the same date on both occasions, and 13 had brought a different date or had come alone. Test whether the number of men who came both weeks with the same date is significantly different from the number who came both weeks but not with the same date. Use a \(10 \%\) level of significance.

For the following samples of data, compute \(\mathrm{t}\) and determine whether \(\mu_{1}\) is significantly less than \(\mu_{2}\). For your test use a level of significance of \(10 .\) Sample \(1: \mathrm{n}=10, \underline{\mathrm{X}}=10.0\) \(\mathrm{S}=5.2\); sample \(2: \mathrm{n}=10, \underline{\mathrm{X}}=13.3, \mathrm{~S}=5.7\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation
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.