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Chapter 9: Problem 124

a. The central \(90 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values? b. The central \(95 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values? c. The central \(99 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values?

Short Answer

Expert verified
The values between which the central percentage of the chi-square distribution with 11 degrees of freedom lies are: a) approximately 6.216 and 19.675 for the central 90%, b) approximately 5.225 and 21.92 for the central 95%, c) approximately 3.816 and 24.725 for the central 99%.

Step by step solution

01

Understanding the question

The task is to determine the values between which the central 90\%, 95\%, and 99\% of the chi-square distribution with 11 degrees of freedom lie.
02

Use of chi-square distribution table

Let's use a chi-square distribution table or a statistical calculator for this. In the row corresponding to our degrees of freedom (11), we identify the required percentage points.
03

Find the chi-square value for central 90% for degrees of freedom 11

By checking the table for the chi-square distribution with 11 degrees of freedom, for the central 90\%, we find the values to be aroung 6.216 and 19.675
04

Find the chi-square value for central 95% for degrees of freedom 11

Continuing with the process, for the central 95\%, the values are approximatly 5.225 and 21.92
05

Find the chi-square value for central 99% for degrees of freedom 11

Lastly, for the central 99\%, the values are found to be around 3.816 and 24.725

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Key Concepts

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

Degrees of Freedom
Degrees of freedom (df) are a crucial concept in statistics, representing the number of independent values or quantities that can vary in an analysis without breaking any constraints. In the context of the chi-square distribution, degrees of freedom will directly influence the shape of the distribution curve.

Here's how they work:
  • The degrees of freedom in a chi-square test generally relate to the number of categories minus one. For example, if you are testing a hypothesis with 12 categories, your degrees of freedom would be 11.
  • In any chi-square distribution table, you'll notice the degrees of freedom determine the values or critical points that the chi-square statistic must exceed to reject the null hypothesis.
Understanding degrees of freedom helps you interpret the table accurately and find the needed values for a given confidence level. In our original exercise, the scenario involves 11 degrees of freedom, which plays a crucial role in determining the chi-square values for the 90%, 95%, and 99% confidence levels.
Statistical Calculator
Utilizing a statistical calculator can simplify finding chi-square distribution values, especially when dealing with situations involving multiple confidence levels or degrees of freedom.

Here's why statistical calculators can be immensely helpful:
  • They allow you to input specific degrees of freedom and confidence intervals, saving time and reducing the potential for human error when referencing traditional chi-square tables.
  • Most calculators can automatically calculate the chi-square values, which makes them very convenient for quickly verifying your results.
By using a statistical calculator, you can streamline the process of retrieving chi-square values. This is particularly useful when handling complex statistical tasks, like finding various central percentages as required in our exercise.
Chi-Square Table
The chi-square table is an essential tool in statistics, enabling you to find critical values of the chi-square distribution given certain degrees of freedom and significance levels.

How to use a chi-square table:
  • First, identify the degrees of freedom for your analysis 鈥� this will guide you to the correct row in the table.
  • Next, select the column corresponding to your desired percentile or confidence level 鈥� such as 90%, 95%, or 99%.
  • The intersection of the row and column gives you the chi-square critical value you need for your analysis.
In our exercise, using the chi-square table correctly allowed for identifying the central distribution pairs like 6.216 and 19.675 for a 90% interval with 11 degrees of freedom. Mastery of this table is essential for accurately performing chi-square tests and interpreting statistical data.

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Most popular questions from this chapter

A local auto dealership advertises that \(90 \%\) of customers whose autos were serviced by the service department are pleased with the results. As a researcher, you take exception to this statement because you are aware that many people are reluctant to express dissatisfaction. A research experiment was set up in which those in the sample had received service by this dealer within the past 2 weeks. During the interview, the individuals were led to believe that the interviewer was new in town and was considering taking his car to this dealer's service department. Of the 60 sampled, 14 said that they were dissatisfied and would not recommend the department. a. Estimate the proportion of dissatisfied customers using a \(95 \%\) confidence interval. b. Given your answer to part a, what can be concluded about the dealer's claim?

September is Library Card Sign-up Month. According to a nationwide Harris Poll during August \(2008,68 \%\) of American adults own a library card. Suppose you conduct a survey of 1000 randomly chosen adults in order to test \(H_{o}: p=0.68\) versus \(H_{a}: p<0.68\) where \(p\) represents the proportion of adults who currently have a library card; 651 of the 1000 sampled had a library card. Use \(\alpha=0.01\). a. Calculate the value of the test statistic. b. Solve using the \(p\) -value approach. c. Solve using the classical approach.

A natural gas utility is considering a contract for purchasing tires for its fleet of service trucks. The decision will be based on expected mileage. For a sample of 100 tires tested, the mean mileage was 36,000 and the standard deviation was 2000 miles. Estimate the mean mileage that the utility should expect from these tires using a \(98 \%\) confidence interval.

A commercial farmer harvests his entire field of a vegetable crop at one time. Therefore, he would like to plant a variety of green beans that mature all at one time (small standard deviation between maturity times of individual plants). A seed company has developed a new hybrid strain of green beans that it believes to be better for the commercial farmer. The maturity time of the standard variety has an average of 50 days and a standard deviation of 2.1 days. A random sample of 30 plants of the new hybrid showed a standard deviation of 1.65 days. Does this sample show a significant lowering of the standard deviation at the 0.05 level of significance? Assume that maturity time is normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

In the past the standard deviation of weights of certain 32.0 -oz packages filled by a machine was 0.25 oz. A random sample of 20 packages showed a standard deviation of 0.35 oz. Is the apparent increase in variability significant at the 0.10 level of significance? Assume package weight is normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.
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