/*! 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 424 A manufacturer of kitchen clocks... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A manufacturer of kitchen clocks clalms that a certam model will last 5 years on the average with a standard deviation of \(1.2\) years. A random sample of six of the clocks lasted \(6,5.5\), \(4,5.2,5\), and \(4.3\) years. Compute $$ \left\\{(n-1) S^{2}\right\\} / \sigma^{2} $$ and use the chi-square tables to find the probability of a \(\mathrm{X}^{2}\) value this high.

Short Answer

Expert verified
The calculated test statistic \(\frac{(n-1) S^{2}}{\sigma^{2}}\) is 1.2847. Using the chi-square distribution table with 5 degrees of freedom, we find that the probability of getting a chi-square value equal to or higher than 1.2847 lies between 0.8 and 0.9.

Step by step solution

01

Sample mean calculation

Let's first compute the sample mean, which is the average of the six observed lifetimes: \[ \overline{X} = \frac{6+5.5+4+5.2+5+4.3}{6} = 5\] years.
02

Sample variance calculation

Next, we need to compute the sample variance. The formula for the variance is: \[ S^2=\frac{\sum_{i=1}^n (X_i-\overline{X})^2}{n-1} \] Substitute the given values and compute the sample variance: \[S^2= \frac{(6-5)^2+(5.5-5)^2+(4-5)^2+(5.2-5)^2+(5-5)^2+(4.3-5)^2}{6-1}= 0.37\] **Step 2: Calculate the test statistic \(\left\\{(n-1) S^{2}\right\\} / \sigma^{2}\)**
03

Test statistic calculation

The test statistic is given by the following formula: \[ \frac{(n-1)S^2}{\sigma^2} \] Using the sample variance we calculated in step 1 (0.37), the given standard deviation (1.2), and sample size (6), we can compute the test statistic: \[ \frac{(6-1)(0.37)}{1.2^2}=\frac{5\cdot 0.37}{1.44} = 1.2847 \] **Step 3: Use the chi-square distribution table to find the probability**
04

Finding the probability using the chi-square table

Now, we need to use the chi-square distribution table to find the probability corresponding to the calculated test statistic and degrees of freedom. The degrees of freedom are (n-1), which is (6-1) = 5. Looking up in the table with 5 degrees of freedom, we find that the calculated test statistic 1.2847 lies between the chi-square values 1.145 and 2.675 with respective probabilities 0.9 and 0.8. Hence, the probability of getting a chi-square value equal to or higher than 1.2847 for a random sample of 6 clocks with 5 degrees of freedom lies between 0.8 and 0.9.

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.

Sample Variance
Sample variance is a key concept in statistics that helps us measure the dispersion or spread of numbers in a data set. It tells us how different the data points are from the mean, i.e., their average value. When you have a group of numbers, the sample variance calculates the extent of variance or how each number differs from the average of the group.

To compute sample variance, you use the formula:
  • S^2 = \(\frac{\sum_{i=1}^n (X_i-\overline{X})^2}{n-1}\)
  • \(X_i\) represents each individual data point.
  • \(\overline{X}\) is the sample mean (average of the data set).
  • n is the total number of data points.
The formula subtracts the mean from each individual data point, squares the result, and averages these squared differences, adjusting for all but one observation. This adjustment, dividing by \(n-1\) instead of \(n\), compensates for having a sample rather than a full population, providing an unbiased estimate of the population variance.
Standard Deviation
Standard deviation is another measure of data variability that complements the concept of sample variance. While variance gives us a general sense of dispersion by focusing on squared differences, standard deviation provides a more intuitive measure since it calculates dispersion in the same units as the original data. In essence, standard deviation is simply the square root of the variance.

If you have a sample variance \(S^2\), you compute the standard deviation \(S\) using:
  • \(S = \sqrt{S^2}\)
Using standard deviation offers a clearer idea of data spread since it portrays deviation in the same dimension you are working with. For instance, if the sample variance of clock lifespan is years squared, the standard deviation provides a deviation measurement in years, which is more direct and interpretable.

Standard deviation is especially useful when comparing variability between different data sets or when you need to understand the likelihood of data falling within certain ranges around the mean.
Degrees of Freedom
Degrees of freedom is a concept refering to the number of independent values or quantities which can be assigned to a statistical distribution. In simple terms, it is how many values in the final calculation of a statistic are free to vary. In many statistical computations, like calculating variance or conducting a chi-square test, this concept is pivotal.

When calculating the sample variance, you compute it based on a sample mean rather than the true population mean, which consumes one degree of freedom. That's why, in variance calculations, you usually divide by \(n-1\) rather than \(n\). With a total of \(n\) observations, \(n-1\) represents the number of values that can independently vary while ensuring the mean remains constant.

In chi-square tests, the degrees of freedom often determine the shape of the chi-square distribution. The choice of degrees of freedom precisely reflects how many "free" variations are feasible within your data sample relative to the statistical model being used. Knowing the degrees of freedom is crucial when interpreting chi-square statistics against a chi-square table, as it informs you of how probable or rare your test statistic is.

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

Let \(\mathrm{Y}_{1} \leq \mathrm{Y}_{2} \leq \ldots \ldots \leq \mathrm{Y}_{\mathrm{n}}\) represent the order statistics of the sample \(\left\\{\mathrm{X}_{1}, \ldots, \mathrm{X}_{n}\right\\}\) from a cumulative distribution function \(\mathrm{F}\). Find the marginal cumulative distribution function of \(\mathrm{Y}_{\alpha}\). \(\alpha=1,2, \ldots, n\)

Suppose that light bulbs made by a standard process have an average life of 2000 hours, with a standard deviation of 250 hours, and suppose that it is worthwhile to change the process if the mean life can be increased by at least 10 percent. An engineer wishes to test a proposed new process, and he is willing to assume that the standard deviation of the distribution of lives is about the same as for the standard process. How large a sample should he examine if he wishes the probability to be about \(.01\) that he will fail to adopt the new process if in fact it produces bulbs with a mean life of 2250 hours?

Each of the following sampling procedures is to be classified as producing a random sample or as producing a biased sample. Consider each case and decide whether the procedure is random or biased. a.) The population about which inference is to be made is a population of scores is dart throwing, and the question to be decide is whether men achieve higher scores than women. The investigator selects 10 men and 10 women at random; he then obtain five score from each, making a sample of 100 scores . Is this a random sample? b.) In attacking the same problem, the investigator takes 100 averages of five scores, each average coming from a different randomly selected person. Is this sample of average scores random sample? c.) The population of interest is population of attitude scores at Alpha College; assume that it is an infinite population. The experimenter whishes to obtain a sample of about 100 scores. He administers the attitude test to four existing groups. In this case they are four classes selected at random whose total enrollment is 100 . Is this a random sample? d.) The same investigator interested in attitude scores sends out the attitude test by mail to a sample of 100 students whose names he has selected by taking every seventy-fifth name in the student directory after randomly choosing a starting point. He receives 71 completed tests; the other 29 students fail to respond. Is the original list a random sample? Is the sample of 71 tests random?

Random samples of size 100 are drawn, with replacement, from two populations, \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\), and their means, \(\mathrm{X}_{1}\) and \(\mathrm{X}_{2}\) computed. If \(\mu_{1}=10, \sigma_{1}=2, \mu_{2}=8\), and \(\sigma_{2}=1\), find (a) \(\mathrm{E}\left(\underline{\mathrm{X}}_{1}-\underline{\mathrm{X}}_{2}\right)\); (b) \(\sigma_{\\{\underline{X}) 1-(\underline{x}) 2\\}}\) (c) the probability that the difference between a given pair of sample means is less than 1.5; (d) the probability that the difference between a given pair of sample means is greater than \(1.75\) but less than \(2.5\)

Suppose that the life of a certain light bulb is exponentially distributed with mean 100 hours. If 10 such light bulbs are installed simultaneously, what is the distribution of the life of the light bulb that fails first, and what is its expected life? Let \(\mathrm{X}_{1}\) denote the life of the ith light bulb; then \(\mathrm{Y}_{1}=\min \left[\mathrm{X}_{1}, \ldots, \mathrm{X}_{10}\right]\) is the life of the light bulb that fails first. Assume that the \(\mathrm{X}_{1}\) 's are independent.

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.