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Chapter 12: Problem 10

Determine \((a)\) the \(\chi^{2}\) test statistic, \((b)\) the degrees of freedom, (c) the critical value using \(\alpha=0.05,\) and (d) test the hypothesis at the \(\alpha=0.05\) level of significance. \(H_{0}:\) The random variable \(X\) is binomial with \(n=4, p=0.3\) \(H_{1}:\) The random variable \(X\) is not binomial with \(n=4\) \(p=0.3\) $$ \begin{array}{llllll} \boldsymbol{X} & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\ \hline \text { Observed } & 260 & 400 & 280 & 50 & 10 \\ \hline \text { Expected } & 240.1 & 411.6 & 264.6 & 75.6 & 8.1 \\ \hline \end{array} $$

Short Answer

Expert verified
The chi-square test statistic is 11.645. Degrees of freedom are 4. Critical value at \( \alpha = 0.05 \) is 9.488. Reject \( H_0 \).

Step by step solution

01

- Calculate the Chi-Square Test Statistic

The Chi-Square test statistic \(n \chi^{2} \) is calculated using the formula \[ \chi^{2} = \sum \frac{(O_i - E_i)^2}{E_i} \] where \( O_i \) are the observed frequencies and \( E_i \) are the expected frequencies. Compute this for each value in the table: \(\chi^{2} = \frac{(260 - 240.1)^2}{240.1} + \frac{(400 - 411.6)^2}{411.6} + \frac{(280 - 264.6)^2}{264.6} + \frac{(50 - 75.6)^2}{75.6} + \frac{(10 - 8.1)^2}{8.1}\) Calculate each term separately: \[ \chi^{2}_0 = \frac{(260 - 240.1)^2}{240.1} \approx 1.656 \] \[ \chi^{2}_1 = \frac{(400 - 411.6)^2}{411.6} \approx 0.328 \] \[ \chi^{2}_2 = \frac{(280 - 264.6)^2}{264.6} \approx 0.854 \] \[ \chi^{2}_3 = \frac{(50 - 75.6)^2}{75.6} \approx 8.379 \] \[ \chi^{2}_4 = \frac{(10 - 8.1)^2}{8.1} \approx 0.428 \] Sum these up: \[ \chi^{2} \approx 11.645 \]
02

- Determine the Degrees of Freedom

Degrees of freedom \( df \) are calculated as the number of categories minus 1. Here, there are 5 categories (0, 1, 2, 3, 4): \[ df = 5 - 1 = 4 \]
03

- Find the Critical Value

Using a chi-square distribution table, find the critical value for \( df = 4 \) at \(n \alpha = 0.05 \). The critical value is \[ \chi^{2}_{critical} = 9.488 \]
04

- Test the Hypothesis

Compare the calculated chi-square test statistic (Step 1) with the critical value (Step 3). \[ \chi^{2}_{calculated} = 11.645 \]\[ \chi^{2}_{critical} = 9.488 \] Since \( 11.645 > 9.488 \), reject the null hypothesis \( H_0 \). There is sufficient evidence to conclude that the random variable \( X \) is not binomial with \( n=4 \) and \( p=0.3 \).

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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 key concept in various statistical tests, including the Chi-Square test. Degrees of freedom generally refer to the number of independent values or quantities which can be assigned to a statistical distribution. In the context of the Chi-Square test, degrees of freedom are calculated as the number of categories minus 1. This accounts for the fact that if you know the total number of observations, and you know the counts in all but one category, then the count in the remaining category is fixed. For example, in our given table with 5 categories (0, 1, 2, 3, 4), the degrees of freedom are:
  • df = 5 - 1
Therefore, we get:
  • df = 4
These 4 degrees of freedom mean that 4 categories can vary independently, providing the flexibility needed for the chi-square test calculations. Understanding degrees of freedom helps in grasping the underlying mechanics of the Chi-Square test and other related statistical methods.
Critical Value
The critical value in hypothesis testing is a threshold that distinguishes whether your test statistic falls in the rejection region (leading to rejecting the null hypothesis) or not. For the Chi-Square test, the critical value is found using a Chi-Square distribution table at a predetermined significance level ( (alpha) ). In this exercise, our significance level is (alpha = 0.05). To find the critical value:
  • Determine the degrees of freedom.
  • Locate the critical value in a Chi-Square distribution table corresponding to (degr=f = 4) and (degree = 0.05).
In our case, the critical value at (df = 4) and (alpha = 0.05) is:
  • ( ( Chi^2_{critical} = 9.488)
This means that if our calculated Chi-Square test statistic is greater than 9.488, we reject the null hypothesis.
Hypothesis Testing
Hypothesis testing is a method used to determine if there is enough statistical evidence in favor of a specific hypothesis. In this exercise, we are testing the claim (null hypothesis (H_0)) that the random variable (X) follows a binomial distribution with parameters ( = 4) and (p=0.3). The steps for hypothesis testing include:
  • State the null and alternate hypotheses: (H_0) and (H_1)).
  • Choose a significance level, (alpha = 0.05) in this case.
  • Compute the test statistic (Using Chi-Square in this case).
  • Determine the critical value.
  • Compare the calculated test statistic to the critical value.
  • Make a decision: reject or fail to reject the null hypothesis ((H_0)).
For our exercise, we calculated the Chi-Square test statistic to be approximately 11.645. Our critical value was found to be 9.488. Since 11.645 is greater than 9.488, we reject (H_0), supporting the alternate hypothesis (H_1) that (X) is not binomial with (n = 4) and (p = 0.3).
Statistical Significance
Statistical significance is a determination of whether the result of a hypothesis test is extreme enough to reject the null hypothesis. It relates to the (alpha) level, which is the threshold for deciding significance. In this exercise, the (degree = 0.05) level indicates that there's a 5% risk of concluding that a difference exists when there is no actual difference.
  • If the test statistic exceeds the critical value derived from the Chi-Square distribution for given degrees of freedom, it indicates statistical significance.
Our test statistic was 11.645, which exceeded our critical value of 9.488 at <(alpha = 0.05).) This means our result is statistically significant. Hence, we reject the null hypothesis (H_0) and accept that the data does not fit a binomial distribution with the provided parameters. Understanding statistical significance helps in making informed decisions in hypothesis testing and interpreting test results.

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Most popular questions from this chapter

Our number system consists of the digits \(0,1,2,3,4,5,6,7,8,\) and \(9 .\) The first significant digit in any number must be \(1,2,3,4,5,6,7,8,\) or 9 because we do not write numbers such as 12 as \(012 .\) Although we may think that each first digit appears with equal frequency so that each digit has a \(\frac{1}{9}\) probability of being the first significant digit, this is not true. In 1881 , Simon Newcomb discovered that first digits do not occur with equal frequency. This same result was discovered again in 1938 by physicist Frank Benford. After studying much data, he was able to assign probabilities of occurrence to the first digit in a number as shown. $$ \begin{array}{lccccc} \text { Digit } & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Probability } & 0.301 & 0.176 & 0.125 & 0.097 & 0.079 \\ \hline \text { Digit } & 6 & 7 & 8 & 9 & \\ \hline \text { Probability } & 0.067 & 0.058 & 0.051 & 0.046 & \\ \hline \end{array} $$ The probability distribution is now known as Benford's Law and plays a major role in identifying fraudulent data on tax returns and accounting books. For example, the following distribution represents the first digits in 200 allegedly fraudulent checks written to a bogus company by an employee attempting to embezzle funds from his employer. $$ \begin{array}{lrrrrrrrrr} \hline \text { First digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \text { Frequency } & 36 & 32 & 28 & 26 & 23 & 17 & 15 & 16 & 7 \\ \hline \end{array} $$ (a) Because these data are meant to prove that someone is guilty of fraud, what would be an appropriate level of significance when performing a goodness- of-fit test? (b) Using the level of significance chosen in part (a), test whether the first digits in the allegedly fraudulent checks obey Benford's Law. (c) Based on the results of part (b), do you think that the employee is guilty of embezzlement?

Determine \((a)\) the \(\chi^{2}\) test statistic, \((b)\) the degrees of freedom, (c) the critical value using \(\alpha=0.05,\) and (d) test the hypothesis at the \(\alpha=0.05\) level of significance. \(H_{0}: p_{\mathrm{A}}=p_{\mathrm{B}}=p_{\mathrm{C}}=p_{\mathrm{D}}=p_{\mathrm{E}}=\frac{1}{5}\) \(H_{1}\) : At least one of the proportions is different from the others. $$ \begin{array}{llllll} \hline\text { Outcome } & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} & \mathbf{E} \\ \hline \text { Observed } & 38 & 45 & 41 & 33 & 43 \\ \hline \text { Expected } & 40 & 40 & 40 & 40 & 40 \\ \hline \end{array} $$

In Section 10.2, we tested hypotheses regarding a population proportion using a z-test. However, we can also use the chi-square goodness-of-fit test to test hypotheses with \(k=2\) possible outcomes. In Problems 25 and \(26,\) we test hypotheses with the use of both methods. In \(2000,25.8 \%\) of Americans 15 years of age or older lived alone, according to the Census Bureau. A sociologist, who believes that this percentage is greater today, conducts a random sample of 400 Americans 15 years of age or older and finds that 164 are living alone. (a) If the proportion of Americans aged 15 years or older living alone is 0.258 , compute the following expected numbers: Americans 15 years of age or older who live alone; Americans 15 years of age or older who do not live alone. (b) Test the sociologist's belief at the \(\alpha=0.05\) level of significance using the goodness-of-fit test. (c) Test the belief by using the approach presented in Section 10.2

A researcher wanted to determine whether bicycle deaths were uniformly distributed over the days of the week. She randomly selected 200 deaths that involved a bicycle, recorded the day of the week on which the death occurred, and obtained the following results (the data are based on information obtained from the Insurance Institute for Highway Safety). $$ \begin{array}{lc|lc} \begin{array}{l} \text { Day of } \\ \text { the Week } \end{array} & \text { Frequency } & \begin{array}{l} \text { Day of } \\ \text { the Week } \end{array} & \text { Frequency } \\ \hline \text { Sunday } & 16 & \text { Thursday } & 34 \\ \hline \text { Monday } & 35 & \text { Friday } & 41 \\ \hline \text { Tuesday } & 16 & \text { Saturday } & 30 \\ \hline \text { Wednesday } & 28 & & \\ \hline \end{array} $$ Is there reason to believe that bicycle fatalities occur with equal frequency with respect to day of the week at the \(\alpha=0.05\) level of significance?

In a famous study by the Physicians Health Study Group from Harvard University from the late 1980 s, 22,000 healthy male physicians were randomly divided into two groups; half the physicians took aspirin every other day, and the others were given a placebo. Of the physicians in the aspirin group, 104 heart attacks occurred; of the physicians in the placebo group, 189 heart attacks occurred. The results were statistically significant, which led to the advice that males should take an aspirin every other day in the interest of reducing the chance of having a heart attack. Does the same advice apply to women? In a randomized, placebo-controlled study, 39,876 healthy women 45 years of age or older were randomly divided into two groups. The women in group 1 received \(100 \mathrm{mg}\) of aspirin every other day; the women in group 2 received a placebo every other day. The women were monitored for 10 years to determine if they experienced a cardiovascular event (such as heart attack or stroke). Of the 19,934 in the aspirin group, 477 experienced a heart attack. Of the 19,942 women in the placebo group, 522 experienced a heart attack. Source: Paul M. Ridker et al. "A Randomized Trial of Low-Dose Aspirin in the Primary Prevention of Cardiovascular Disease in Women." New England Journal of Medicine \(352: 1293-1304 .\) (a) What is the population being studied? What is the sample? (b) What is the response variable? Is it qualitative or quantitative? (c) What are the treatments? (d) What type of experimental design is this? (e) How does randomization deal with the explanatory variables that were not controlled in the study? (f) Determine whether the proportion of cardiovascular events in each treatment group is different using a two-sample \(Z\) -test for comparing two proportions. Use the \(\alpha=0.05\) level of significance. What is the test statistic? (g) Determine whether the proportion of cardiovascular events in each treatment group is different using a chi-square test for homogeneity of proportions. Use the \(\alpha=0.05\) level of significance. What is the test statistic? (h) Square the test statistic from part (f) and compare it to the test statistic from part \((\mathrm{g}) .\) What do you conclude?
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