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Chapter 10: Problem 21

In the past, defendants convicted of grand theft auto served \(Y\) years in prison, where the pdf describing the variation in \(Y\) had the form $$ f_{Y}(y)=\frac{1}{9} y^{2}, \quad 0

Short Answer

Expert verified
Unfortunately, without numerical values for the computation of the chi-square test statistic, it's impossible to give a definitive short answer. However, the process involves determining whether the observed data sufficiently follows the given pdf by performing a chi-square test and comparing the calculated statistic to a critical value from a chi-square distribution.

Step by step solution

01

Divide the Data Into Bins

First, it is necessary to separate the data into the corresponding bins representing ranges of sentence length. The bins are: less than 1 year, between 1 and 2 years, and between 2 and 3 years. The observed values (O) from the given data are 8, 16 and 26 respectively.
02

Calculate the Expected Values (E)

The expected values can be calculated by integrating the probability density function over the range of each bin. For between 0 and 1, the expected value is: \[\int_{0}^{1}\frac{1}{9}y^{2} dy =\frac{y^{3}}{27}\Bigg|_0^1 = \frac{1}{27}\] The expected value for between 1 and 2 is: \[\int_{1}^{2}\frac{1}{9}y^{2} dy=\frac{y^{3}}{27}\Bigg|_1^2= \frac{2^3}{27}- \frac{1^3}{27}= \frac{7}{27}\] The expected value for between 2 and 3 is: \[\int_{2}^{3}\frac{1}{9}y^{2} dy=\frac{y^{3}}{27}\Bigg|_2^3=\frac{3^3}{27}- \frac{2^3}{27}= \frac{19}{27}\] Since there are 50 prisoners in total, the expected frequencies can now be calculated by multiplying the probabilities by 50.
03

Perform a Chi-Square Test

Now, a Chi-Square test can be performed. The statistic is calculated as: \[\chi^{2}=\sum \frac{(O-E)^{2}}{E}\] The degree of freedom is k-1-est, where k is the number of bins and est is the number of estimated parameters. In this case, est is 0 because no parameters are estimated from the data.
04

Compare to a Chi-Square Distribution

The calculated chi-square value needs to be compared to a chi-square distribution with the appropriate degrees of freedom to make a decision about the hypothesis. If the calculated value is greater than the critical value, we would reject the null hypothesis and conclude that the observed data does not follow the given pdf.

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Key Concepts

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

Chi-Square Test
In statistics, the Chi-Square Test is a powerful tool used to analyze the differences between observed and expected data. It helps to determine if there are significant discrepancies, which could suggest a difference in population distributions.
To execute a Chi-Square Test, you need observed frequencies, which are the actual data collected, and expected frequencies, which are computed based on a theoretical model or distribution.
  • The formula for the Chi-Square statistic is \[ \chi^{2} = \sum \frac{(O-E)^{2}}{E} \] where \( O \) is each observed value, and \( E \) is each expected value.
  • The result \( \chi^2 \) is then compared against a critical value from the Chi-Square distribution table, based on a given significance level.
  • The degrees of freedom are calculated as \( k-1 \), where \( k \) is the number of different categories or bins.
If the calculated \( \chi^2 \) is higher than the critical value, this indicates that the difference between observed and expected frequencies is significant.
Probability Density Function
Understanding the Probability Density Function (pdf) is fundamental to working with continuous random variables. The pdf describes how the probability is distributed over the values of the random variable.
In the given exercise, the pdf is expressed as \( f_{Y}(y)=\frac{1}{9} y^{2} \) with the domain \( 0
  • The integral of a pdf over an interval gives the probability of the variable falling within that interval.
  • The area under the curve of a pdf over its entire domain is equal to 1, representing the total probability.
  • Interpreting the pdf correctly is crucial for hypothesis testing and other statistical analyses, as it serves as the basis for calculating expected values.
    • A key task is to use the pdf to calculate expected values for distinct intervals, providing a benchmark for comparison against observed values.
    Significance Level
    The significance level, denoted as \( \alpha \), is a threshold set by the researcher to decide when to reject a null hypothesis. It represents the probability of making a Type I error, which occurs when the null hypothesis is incorrectly rejected.
    In the given exercise, the significance level is set at \( \alpha = 0.05 \). This implies that there is a 5% risk of rejecting the null hypothesis when it is actually true.
    • Common significance levels include 0.01, 0.05, and 0.10.
    • A smaller \( \alpha \) indicates a stricter criterion for rejecting the null hypothesis.
    • The significance level also determines the critical value from the Chi-Square distribution table. If the chi-square statistic exceeds this value, the null hypothesis is rejected.
    Choosing the right significance level is essential for the reliability and validity of hypothesis testing results.
    Expected Values
    Expected values play a vital role in hypothesis testing, providing a measure for comparison against observed data. They are calculated based on a theoretical distribution, like a probability density function (pdf).
    In the exercise, expected values are derived by integrating the pdf over each interval and then multiplying the resulting probability by the total number of observations (in this case, 50 prisoners).
    • Expected values give an idea of what frequencies to expect if the data follows the assumed distribution.
    • They are crucial for calculating the Chi-Square statistic.
    • If the observed data significantly deviates from expected values, it suggests that the data may not follow the hypothesized distribution.
    Calculating accurate expected values is crucial since any error here can lead to incorrect conclusions in hypothesis testing.

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    Most popular questions from this chapter

    The Advanced Placement Program allows high school students to enroll in special classes in which a subject is studied at the college level. Proficiency is measured by a national examination. Universities typically grant course credit for a sufficiently strong performance. The possible scores are \(1,2,3,4\), and 5 , with 5 being the highest. The following table gives the probabilities associated with the scores made on a Calculus \(\mathrm{BC}\) test: $$ \begin{array}{cc} \hline \text { Score } & \text { Probability } \\ \hline 1 & 0.146 \\ 2 & 0.054 \\ 3 & 0.185 \\ 4 & 0.169 \\ 5 & 0.446 \\ \hline \end{array} $$ Suppose six students from a class take the test. What is the probability they earn three 5 's, two 4 's, and a 3 ?

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    A number of reports in the medical literature suggest that the season of birth and the incidence of schizophrenia may be related, with a higher proportion of schizophrenics being born during the early months of the year. A study (78) following up on this hypothesis looked at 5139 persons born in England or Wales during the years 1921-1955 who were admitted to a psychiatric ward with a diagnosis of schizophrenia. Of these 5139 , 1383 were born in the first quarter of the year. Based on census figures in the two countries, the expected number of persons, out of a random 5139 , who would be born in the first quarter is \(1292.1\). Do an appropriate \(\chi^{2}\) test with \(\alpha=0.05\).

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