/*! 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 57 A manufacturer of industrial lig... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 57

A manufacturer of industrial light bulbs likes its bulbs to have a mean length of life that is acceptable to its customers and a variation in length of life that is relatively small. A sample of 20 bulbs tested produced the following lengths of life (in hours): $$ \begin{array}{llllllllll} 2100 & 2302 & 1951 & 2067 & 2415 & 1883 & 2101 & 2146 & 2278 & 2019 \\ 1924 & 2183 & 2077 & 2392 & 2286 & 2501 & 1946 & 2161 & 2253 & 1827 \end{array} $$ The manufacturer wishes to control the variability in length of life so that \(\sigma\) is less than 150 hours. Do the data provide sufficient evidence to indicate that the manufacturer is achieving this goal? Test using \(\alpha=.01\).

Short Answer

Expert verified
There is sufficient evidence to indicate that the manufacturer is not achieving their goal of controlling the variability in the length of life to less than 150 hours. This conclusion is based on a hypothesis test where the null hypothesis was rejected as the test statistic (44.692) was greater than the critical value (36.191) with a significance level of 0.01.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis \((H_0)\) is that the population standard deviation is less than or equal to 150 hours, and the alternative hypothesis \((H_1)\) is that it is more than 150 hours. Mathematically, this is represented as: $$ H_0: \sigma \leq 150 \\ H_1: \sigma > 150 $$
02

Calculate sample mean and sample standard deviation

Using the given data, we need to calculate the sample mean \((\bar{x})\) and sample standard deviation \((s)\). To do this, we first find the sum of all the length of life values, then divide by the total number of bulbs to get the mean. Next, we find the variance by calculating the sum of squared differences divided by \((n-1)\), and take the square root to get the standard deviation. Sample mean: $$ \bar{x} = \frac{\sum_{i=1}^n x_i}{n} = \frac{2100 + 2302 + \cdots + 2253 + 1827}{20} $$ Sample standard deviation: $$ s = \sqrt{\frac{\sum_{i=1}^{20}(x_i - \bar{x})^2}{n-1}} $$ After calculating, we get: $$ \bar{x} \approx 2141.85 \ \text{hours} \quad \text{and} \quad s \approx 183.88 \ \text{hours} $$
03

Determine the critical value

We need to find the critical value from a chi-square distribution table using the degrees of freedom \((df)\) and the significance level \((\alpha)\). $$ df = n - 1 = 20 - 1 = 19 $$ For a right-tailed test with \(\alpha=0.01\), we get the critical value: $$ \chi^2_{19, 0.01} = 36.191 $$
04

Calculate the test statistic

Now, we will calculate the test statistic using the formula: $$ \chi^2 = \frac{(n - 1)s^2}{\sigma_0^2} $$ where \(\sigma_0\) is the value we are testing (150 hours). Plug in the values: $$ \chi^2 = \frac{(20 - 1)(183.88)^2}{(150)^2} \approx 44.692 $$
05

Make a decision

Now, we compare the test statistic with the critical value to make a decision: $$ \chi^2 = 44.692 > \chi^2_{19, 0.01} = 36.191 $$ Since the test statistic is greater than the critical value, we reject the null hypothesis \((H_0)\). Based on the analysis, there is sufficient evidence to indicate that the manufacturer is not achieving the goal of controlling the variability in the length of life to less than 150 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.

Understanding Standard Deviation
Standard deviation is a way to measure the amount of variation or dispersion in a set of values. In simpler terms, it tells us how spread out the numbers in our data set are. A low standard deviation indicates that the values are close to the mean, while a high standard deviation suggests that they are more spread out.

To calculate the standard deviation of a sample, follow these steps:
  • Find the mean (average) of your data set.
  • Subtract the mean from each data point to get the differences.
  • Square each of the differences to make them positive.
  • Sum up all the squared differences.
  • Divide this sum by one less than the number of data points (this value is known as the degrees of freedom).
  • Finally, take the square root of this value.
In the exercise, the calculated sample standard deviation was approximately 183.88 hours, which reflects how the lifespan of the light bulbs differs from the mean. Understanding standard deviation helps us grasp how consistent the lifespans of the bulbs are.
Chi-Square Distribution
The chi-square distribution is a mathematical tool used in statistics, often applied when assessing the goodness of fit or testing hypotheses concerning variance. It is particularly useful for categorical data analysis and works for variables measured on a nominal or ordinal scale.

For this type of statistic, we deal with the 'degrees of freedom,' which is determined by the sample size minus one (n-1). The larger the sample size, the more data points we have to help estimate the distribution. The chi-square distribution is skewed to the right, meaning it only takes on positive values.

When testing hypotheses involving standard deviation, like in our exercise, we compare a calculated chi-square test statistic against a critical value from a chi-square distribution table for a given significance level (such as \(\alpha=0.01\)). If the test statistic is larger than the critical value, we reject the null hypothesis.

In our exercise, a critical value of 36.191 was identified using a 19-degree freedom and a 0.01 significance level. The calculated test statistic was 44.692, leading us to reject the null hypothesis. This result indicates that the variability exceeds what the manufacturer aims to achieve.
Exploring Sample Statistics
Sample statistics are numerical characteristics of a sample, which is a subset of the population. Calculating sample statistics allows us to make inferences or conclusions about a population without surveying every individual. Common sample statistics include the mean, median, variance, and standard deviation, among others.

Firstly, the sample mean is calculated by adding up all sample values and dividing by the number of samples. This forms the basis for many statistical analyses, providing insights into the typical value found within the data set.

Another critical aspect is the sample standard deviation, which gives us an idea of how much variation or spread exists within the sample data, as seen earlier.
  • In hypothesis testing, sample statistics are used to calculate test statistics, such as the chi-square statistic.
  • These tests help determine whether observed data can be explained by a particular model or if there are differences under the evaluated conditions.
In the context of this exercise, with a calculated sample mean of approximately 2141.85 hours and standard deviation of about 183.88 hours, these sample statistics serve as the foundation for testing the manufacturer's variability control goal. Understanding and analyzing sample statistics is essential for making informed decisions based on observed data.

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

Use Table 4 in Appendix I to find the following critical values: a. An upper one-tailed rejection region with \(\alpha=.05\) and \(11 d f\). b. A two-tailed rejection region with \(\alpha=.05\) and \(7 d f\). c. A lower one-tailed rejection region with \(\alpha=.01\) and \(15 d f\).

How much sleep do you get on a typical school night? A group of 10 college students were asked to report the number of hours that they slept on the previous night with the following results: $$ \begin{array}{llllllllll} 7, & 6, & 7.25, & 7, & 8.5, & 5, & 8, & 7, & 6.75, & 6 \end{array} $$ a. Find a \(99 \%\) confidence interval for the average number of hours that college students sleep. b. What assumptions are required in order for this confidence interval to be valid?

A manufacturer of hard safety hats for construction workers is concerned about the mean and the variation of the forces helmets transmit to wearers when subjected to a standard external force. The manufacturer desires the mean force transmitted by helmets to be 800 pounds (or less), well under the legal 1000 -pound limit, and \(\sigma\) to be less than \(40 .\) A random sample of \(n=40\) helmets was tested, and the sample mean and variance were found to be equal to 825 pounds and 2350 pounds \(^{2}\), respectively. a. If \(\mu=800\) and \(\sigma=40,\) is it likely that any helmet, subjected to the standard external force, will transmit a force to a wearer in excess of 1000 pounds? Explain. b. Do the data provide sufficient evidence to indicate that when the helmets are subjected to the standard external force, the mean force transmitted by the helmets exceeds 800 pounds?

To compare the demand for two different entrees, the manager of a cafeteria recorded the number of purchases of each entree on seven consecutive days. The data are shown in the table. Do the data provide sufficient evidence to indicate a greater mean demand for one of the entrees? Use the Excel printout. $$ \begin{array}{lcc} \text { Day } & \mathrm{A} & \mathrm{B} \\ \hline \text { Monday } & 420 & 391 \\ \text { Tuesday } & 374 & 343 \\ \text { Wednesday } & 434 & 469 \\ \text { Thursday } & 395 & 412 \\ \text { Friday } & 637 & 538 \\ \text { Saturday } & 594 & 521 \\ \text { Sunday } & 679 & 625 \end{array} $$

A paired-difference experiment was conducted using \(n=10\) pairs of observations. a. Test the null hypothesis \(H_{0}:\left(\mu_{1}-\mu_{2}\right)=0\) against \(H_{\mathrm{a}}:\left(\mu_{1}-\mu_{2}\right) \neq 0\) for \(\alpha=.05, \bar{d}=.3,\) and \(s_{d}^{2}=\) .16. Give the approximate \(p\) -value for the test. b. Find a \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right)\). c. How many pairs of observations do you need if you want to estimate \(\left(\mu_{1}-\mu_{2}\right)\) correct to within .1 with probability equal to \(.95 ?\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

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.