/*! 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 9 The response time of a computer ... [FREE SOLUTION] | 91Ó°ÊÓ

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The response time of a computer system to a request for a certain type of information is hypothesized to have an exponential distribution with parameter \(\lambda=1\) [so if \(X=\) response time, the pdf of \(X\) under \(H_{0}\) is \(f_{0}(x)=e^{-x}\) for \(x \geq 0\) ]. a. If you had observed \(X_{1}, X_{2}, \ldots, X_{n}\) and wanted to use the chi-squared test with five class intervals having equal probability under \(H_{0}\), what would be the resulting class intervals? b. Carry out the chi-squared test using the following data resulting from a random sample of 40 response times: \(\begin{array}{rrrrrrrr}.10 & .99 & 1.14 & 1.26 & 3.24 & .12 & .26 & .80 \\\ .79 & 1.16 & 1.76 & .41 & .59 & .27 & 2.22 & .66 \\ .71 & 2.21 & .68 & .43 & .11 & .46 & .69 & .38 \\ .91 & .55 & .81 & 2.51 & 2.77 & .16 & 1.11 & .02 \\\ 2.13 & .19 & 1.21 & 1.13 & 2.93 & 2.14 & .34 & .44\end{array}\)

Short Answer

Expert verified
Class intervals: \([0, 0.223), [0.223, 0.511), [0.511, 0.916), [0.916, 1.609), [1.609, \infty)\). \( \chi^2 \approx 3.25, \) do not reject \( H_0 \).

Step by step solution

01

Determine Class Intervals

For a chi-squared test with five class intervals having equal probability under \( H_0 \), we will divide the integral from 0 to \( \infty \) into five equal probability intervals. For an exponential distribution with \( \lambda = 1 \), the cumulative distribution function (CDF) is \( F(x) = 1 - e^{-x} \). Each interval should have a probability of 0.2, so we will find the points where the CDF equals 0.2, 0.4, 0.6, and 0.8. We solve for \( x \) in \( F(x) = q \) for \( q = 0.2, 0.4, 0.6, 0.8 \):
02

Solve for Class Boundaries

Using the equation \( 1 - e^{-x} = q \), solve for \( x \) at each probability level:- For \( q = 0.2 \): \[ 1 - e^{-x} = 0.2 \] \[ e^{-x} = 0.8 \] \[ x = -\ln(0.8) \approx 0.223 \]- For \( q = 0.4 \): \[ e^{-x} = 0.6 \] \[ x = -\ln(0.6) \approx 0.511 \]- For \( q = 0.6 \): \[ e^{-x} = 0.4 \] \[ x = -\ln(0.4) \approx 0.916 \]- For \( q = 0.8 \): \[ e^{-x} = 0.2 \] \[ x = -\ln(0.2) \approx 1.609 \]Thus, we have class intervals: \([0, 0.223), [0.223, 0.511), [0.511, 0.916), [0.916, 1.609), [1.609, \infty)\).
03

Calculate Observed Frequencies

Sort the provided data into the determined class intervals and calculate the observed frequencies:- \([0, 0.223)\): 5 observations- \([0.223, 0.511)\): 10 observations- \([0.511, 0.916)\): 9 observations- \([0.916, 1.609)\): 10 observations- \([1.609, \infty)\): 6 observations
04

Calculate Expected Frequencies

Under \( H_0 \), each class interval should have an expected frequency calculated by dividing the total sample size, 40, by 5 (the number of intervals) because each class has an equal probability:- Expected frequency for each class = 40/5 = 8
05

Perform Chi-squared Test

Calculate the chi-squared statistic:\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]Where \(O_i\) are the observed frequencies, and \(E_i\) are the expected frequencies:\[\chi^2 = \frac{(5-8)^2}{8} + \frac{(10-8)^2}{8} + \frac{(9-8)^2}{8} + \frac{(10-8)^2}{8} + \frac{(6-8)^2}{8} \approx 3.25\]
06

Decision Rule

The critical value for \( \chi^2 \) with 4 degrees of freedom (5-1) at a 0.05 significance level is 9.488. Since our \( \chi^2 = 3.25 \) is less than 9.488, we do not reject the null hypothesis \( H_0 \).

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Key Concepts

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

Exponential Distribution
The exponential distribution is a type of probability distribution that is often used to model the time between independent events that happen at a constant rate. In this context, it is used to describe the response time of a computer system. The key parameter of an exponential distribution is \( \lambda \), which is the rate parameter. In this exercise, the rate parameter \( \lambda \) is 1, simplifying the calculations, as we are given the probability density function (pdf) of \( X \) under the hypothesis \( H_0 \) as \( f_0(x) = e^{-x} \) for \( x \geq 0 \).
  • This distribution has a mean of \( 1/\lambda \) and a variance of \( 1/\lambda^2 \).
  • It is memoryless, meaning the probability of an event occurring in the future is independent of the past events.
The cumulative distribution function (CDF), \( F(x) = 1 - e^{-\lambda x} \), helps in determining the probability that a random variable \( X \) is less than or equal to \( x \). Here, it simplifies to \( F(x) = 1 - e^{-x} \) due to \( \lambda = 1 \). This feature is crucial in determining class intervals with equal probability in this chi-squared test scenario.
Class Intervals
In hypothesis testing, class intervals are ranges into which observations are sorted for the purpose of comparing observed and expected frequencies. For the chi-squared test, these intervals must have equal probability under the null hypothesis.

In the context of this exercise with an exponential distribution, the cumulative distribution function (CDF) \( F(x) = 1 - e^{-x} \) is used to determine these intervals. Since we want five intervals, and each should cover 20% probability (0.2) under \( H_0 \), we solve \( F(x) = q \) for \( q = 0.2, 0.4, 0.6, 0.8 \):
  • For \( q = 0.2 \), solve \( 1 - e^{-x} = 0.2 \) which gives \( x \approx 0.223 \).
  • For \( q = 0.4 \), solve \( 1 - e^{-x} = 0.4 \) which gives \( x \approx 0.511 \).
  • For \( q = 0.6 \), solve \( 1 - e^{-x} = 0.6 \) which results in \( x \approx 0.916 \).
  • For \( q = 0.8 \), solve \( 1 - e^{-x} = 0.8 \) leading to \( x \approx 1.609 \).
Thus, the class intervals in this exercise are: - \([0, 0.223)\)- \([0.223, 0.511)\)- \([0.511, 0.916)\)- \([0.916, 1.609)\)- \([1.609, \infty)\)
Hypothesis Testing
Hypothesis testing is a statistical method used to make inferences or draw conclusions about a population based on sample data. In this instance, we are testing the hypothesis that the response times of a computer system follow an exponential distribution. The null hypothesis \( H_0 \) assumes that the distribution is exponential with \( \lambda = 1 \).Steps in Hypothesis Testing:
  • Formulate the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \).
  • Select a significance level (\( \alpha \)), commonly 0.05, to determine the threshold for rejecting \( H_0 \).
  • Calculate a test statistic from the sample data.
  • Compare the test statistic to a critical value to decide whether to reject \( H_0 \).
The chi-squared test is used here to compare observed frequencies, obtained by sorting sample data into determined class intervals, against expected frequencies. A chi-squared statistic \( \chi^2 \) is computed and compared with a critical value to decide the fate of \( H_0 \). If \( \chi^2 \) is less than the critical value, we fail to reject \( H_0 \). In this exercise, \( \chi^2 = 3.25 \), which is less than the critical value 9.488 for 4 degrees of freedom at a 0.05 significance level, thereby failing to reject the null hypothesis.
Degrees of Freedom
Degrees of freedom (df) in statistical tests refer to the number of values or quantities that can vary in the final calculation of a statistic. For a chi-squared test, the degrees of freedom are calculated as the number of categories minus 1.

In this exercise, where the data is divided into five class intervals, the degrees of freedom are calculated as 5 (the number of categories) minus 1, which results in 4. Degrees of freedom are vital because they determine the critical value from the \( \chi^2 \) distribution tables. They reflect the number of independent comparisons that can be made between the observed and expected frequencies.
  • The more class intervals, or more complex the model, the higher the degrees of freedom.
  • Each degree of freedom represents an independent direction in which data can vary.
In our chi-squared test, with 4 degrees of freedom, we use this to find the critical value (9.488 with \( \alpha = 0.05 \)) against which the test statistic (\( \chi^2 = 3.25 \)) is compared. The degrees of freedom ensure that the test correctly assesses the hypothesis given the available data.

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Most popular questions from this chapter

A certain type of flashlight is sold with the four batteries included. A random sample of 150 flashlights is obtained, and the number of defective batteries in each is determined, resulting in the following data: \begin{tabular}{l|ccccc} Number Defective & 0 & 1 & 2 & 3 & 4 \\ \hline Frequency & 26 & 51 & 47 & 16 & 10 \end{tabular} Let \(X\) be the number of defective batteries in a randomly selected flashlight. Test the null hypothesis that the distribution of \(X\) is \(\operatorname{Bin}(4, \theta)\). That is, with \(p_{i}=P(i\) defectives), test \(H_{0}: p_{i}=\left(\begin{array}{l}4 \\ i\end{array}\right) \theta^{i}(1-\theta)^{4-i} \quad i=0,1,2,3,4\) [Hint: To obtain the mle of \(\theta\), write the likelihood (the function to be maximized) as \(\theta^{u}(1-\theta)^{v}\), where the exponents \(u\) and \(v\) are linear functions of the cell counts. Then take the natural \(\log\), differentiate with respect to \(\theta\), equate the result to 0 , and solve for \(\hat{\theta}\).]

A statistics department at a large university maintains a tutoring center for students in its introductory service courses. The center has been staffed with the expectation that \(40 \%\) of its clients would be from the business statistics course, \(30 \%\) from engineering statistics, \(20 \%\) from the statistics course for social science students, and the other \(10 \%\) from the course for agriculture students. A random sample of \(n=120\) clients revealed \(52,38,21\), and 9 from the four courses. Does this data suggest that the percentages on which staffing was based are not correct? State and test the relevant hypotheses using \(\alpha=.05\).

The authors of the article "Predicting Professional Sports Game Outcomes from Intermediate Game Scores" (Chance, 1992: 18-22) used a chisquared test to determine whether there was any merit to the idea that basketball games are not settled until the last quarter, whereas baseball games are over by the seventh inning. They also considered football and hockey. Data was collected for 189 basketball games, 92 baseball games, 80 hockey games, and 93 football games. The games analyzed were sampled randomly from all games played during the 1990 season for baseball and football and for the 1990-1991 season for basketball and hockey. For each game, the late-game leader was determined, and then it was noted whether the late-game leader actually ended up winning the game. The resulting data is summarized in the accompanying table. $$ \begin{array}{lcc} \hline \text { Sport } & \begin{array}{c} \text { Late-Game } \\ \text { Leader Wins } \end{array} & \begin{array}{l} \text { Late-Game } \\ \text { Leader Loses } \end{array} \\ \hline \text { Basketball } & 150 & 39 \\ \text { Baseball } & 86 & 6 \\ \text { Hockey } & 65 & 15 \\ \text { Football } & 72 & 21 \end{array} $$ The authors state, "Late-game leader is defined as the team that is ahead after three quarters in basketball and football, two periods in hockey, and seven innings in baseball. The chi-square value on three degrees of freedom is \(10.52\) \((P<.015) . "\) a. State the relevant hypotheses and reach a conclusion using \(\alpha=.05\). b. Do you think that your conclusion in part (a) can be attributed to a single sport being an anomaly?

The article "Psychiatric and Alcoholic Admissions Do Not Occur Disproportionately Close to Patients' Birthdays"' (Psych. Rep., 1992: 944–946) focuses on the existence of any relationship between date of patient admission for treatment of alcoholism and patient's birthday. Assuming a 365day year (i.e., excluding leap year), in the absence of any relation, a patient's admission date is equally likely to be any one of the 365 possible days. The investigators established four different admission categories: (1) within 7 days of birthday, (2) between 8 and 30 days, inclusive, from the birthday, (3) between 31 and 90 days, inclusive, from the birthday, and (4) more than 90 days from the birthday. A sample of 200 patients gave observed frequencies of \(11,24,69\), and 96 for categories 1,2 , 3 , and 4, respectively. State and test the relevant hypotheses using a significance level of \(.01\).

The article "Compatibility of Outer and Fusible Interlining Fabrics in Tailored Garments (Textile Res. J., 1997: 137-142) gave the following observations on bending rigidity \((\mu \mathrm{N} \cdot \mathrm{m})\) for medium-quality fabric specimens, from which the accompanying MINITAB output was obtained: \(\begin{array}{rrrrrrrr}24.6 & 12.7 & 14.4 & 30.6 & 16.1 & 9.5 & 31.5 & 17.2 \\\ 46.9 & 68.3 & 30.8 & 116.7 & 39.5 & 73.8 & 80.6 & 20.3 \\ 25.8 & 30.9 & 39.2 & 36.8 & 46.6 & 15.6 & 32.3 & \end{array}\) Would you use a one-sample \(t\) confidence interval to estimate true average bending rigidity? Explain your reasoning.

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