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91Ó°ÊÓ

Use a computer or calculator to find the area (a) to the left, and (b) to the right of \(\chi^{2} \star=20.2\) with df \(=15\).

Short Answer

Expert verified
The area to the left of \(\chi^{2} \star=20.2\) and to the right of \(\chi^{2} \star=20.2\) can be calculated using the Chi-Square cumulative distribution function (CDF) and its complement respectively, with a degrees of freedom of 15. The actual numerical results depend on the specific calculation with a computer or calculator.

Step by step solution

01

Compute Area to the Left

To compute the area to the left of \(\chi^{2} \star=20.2\), one needs to use the cumulative distribution function (CDF) of the Chi-Square distribution. This gives the probability that a random variable drawn from the Chi-Square distribution is less than or equal to the given value. In many calculators or statistical software, this is done with the chisq.cdf function, like: chisq.cdf(20.2, df=15). Input the chi-square value (20.2) and the degrees of freedom (15) into the first and the second argument of the function respectively.
02

Compute Area to the Right

To compute the area to the right of \(\chi^{2} \star=20.2\), one needs to use the complement of the cumulative distribution function (CDF) of the Chi-Square distribution. This gives the probability that a random variable drawn from the Chi-Square distribution is greater than the given value. This is usually computed as 1- chisq.cdf(20.2, df=15). Input the chi-square value (20.2) and the degrees of freedom (15) into the first and the second argument of the function respectively.

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Key Concepts

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

Cumulative Distribution Function
Understanding the cumulative distribution function (CDF) is crucial when analyzing statistical data. Think of the CDF as a way of adding up probabilities, piece by piece, up to a certain point. In the context of the Chi-Square distribution, the CDF represents the probability that a randomly chosen observation from the distribution will have a value less than or equal to a specific cut-off point, in this case, \(\chi^{2} = 20.2\).

For any continuous probability distribution, the CDF is a curve that increases from 0 to 1 as you move along the x-axis. The value of the CDF at any given point can be interpreted as the area under the probability density function (PDF) curve to the left of that point. In layman's terms, the CDF tells us 'how much' of the distribution lies to the left of the value we are interested in.

Using statistical software or a calculator, as mentioned in the example problem, it's easy to find this area. You enter your Chi-Square value and degrees of freedom into the CDF function, and it churns out a probability, effectively answering the question: what is the chance that an observation is equal to or less than 20.2?
Degrees of Freedom
The concept of degrees of freedom (df) can be a bit abstract, but it's fundamental in statistical analysis, especially when working with Chi-Square distributions. In simple terms, df refers to the number of independent values or quantities which can vary in an analysis without breaking any constraints. It's a way of accounting for the size of the sample or the number of categories you are testing.

The number of degrees of freedom plays a pivotal role in shaping the Chi-Square distribution. To visualize it, you could think of df as 'settings' that alter the Chi-Square curve. The more degrees of freedom there are, the more the distribution curve looks like a normal distribution. In the textbook exercise, when we specify df = 15, we are tailoring the Chi-Square curve to reflect 15 independent pieces of information in our dataset.

Remember that the degrees of freedom are not just a technicality; they are a key part of ensuring that any statistical conclusions drawn are based on the appropriate distribution, which considers the size and constraints of the data we're analyzing.
Probability
Probability is a measure of how likely it is that an event will occur. In statistical terms, it's a value between 0 (the event never occurs) and 1 (the event always occurs). Probabilities are the foundation of inferential statistics and are used to make predictions and decisions based on data.

In the context of the Chi-Square distribution, when we talk about finding the area to the left or right of a certain Chi-Square value, we are referring to the probability of observing a value less than or greater than that point. The value given by the CDF function for a specific point on a Chi-Square curve tells us the probability of observing a value less than or equal to that point.

Meanwhile, the complement of the CDF, which is 1 minus the CDF value, gives us the probability of observing a value greater than the point in question (the area to the right). In our exercise, when we computed 1 - CDF(20.2, df=15), we found the probability of a Chi-Square value being greater than 20.2 with 15 degrees of freedom. Knowing how to interpret these probabilities is key to making sense of statistical data and arriving at accurate conclusions.

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Most popular questions from this chapter

How important is the assumption "The sampled population is normally distributed" to the use of Student's \(t\) -distribution? Using a computer, simulate drawing 100 samples of size 10 from each of three different types of population distributions, namely, a normal, a uniform, and an exponential. First generate 1000 data values from the population and construct a histogram to see what the population looks like. Then generate 100 samples of size 10 from the same population; each row represents a sample. Calculate the mean and standard deviation for each of the 100 samples. Calculate \(t \neq\) for each of the 100 samples. Construct histograms of the 100 sample means and the 100 t \(\star\) values. (Additional details can be found in the Student Solutions Manual.) For the samples from the normal population: a. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. b. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. For the samples from the rectangular or uniform population: c. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. d. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. For the samples from the skewed (exponential) population: e. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. f. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. In summary: g. In each of the preceding three situations, the sampling distribution for \(\bar{x}\) appears to be slightly different from the distribution of \(t \star .\) Explain why. h. Does the normality condition appear to be necessary in order for the calculated test statistic \(t \star\) to have a Student's \(t\) -distribution? Explain.

a. What value of chi-square for 5 degrees of freedom subdivides the area under the distribution curve such that \(5 \%\) is to the right and \(95 \%\) is to the left? b. What is the value of the 95 th percentile for the chi-square distribution with 5 degrees of freedom? c. What is the value of the 90th percentile for the chi-square distribution with 5 degrees of freedom?

Determine the \(p\) -value for each of the following hypothesis-testing situations. a. \(H_{o}: p=0.5, H_{a}: p \neq 0.5, z \star=1.48\) b. \(H_{o}: p=0.7, H_{a}: p \neq 0.7, z \star=-2.26\) c. \(H_{o}: p=0.4, H_{a}: p>0.4, z \star=0.98\) d. \(H_{o}: p=0.2, H_{a}: p<0.2, z \star=-1.59\)

You are interested in comparing the null hypothesis \(p=0.8\) against the alternative hypothesis \(p<0.8 .\) In 100 trials you observe 73 successes. Calculate the \(p\) -value associated with this result.

Find: a. \(\chi^{2}(10,0.01)\) b. \(\chi^{2}(12,0.025)\) c. \(\chi^{2}(10,0.95)\) d. \(\chi^{2}(22,0.995)\)

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