FiltersDone

Question type

Essay

Multiple Choice

Short Answer

True False

Matching

Exam 11: Chi-Square and Analysis of Variance

Show Answer

Share

---

All types

Filters

Study Flashcards

Question 1

Essay

Review LaterAdd Note

Review Later

According to Benford's Law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. Test for goodness-of-fit with Benford's Law. $\begin{array} { | l | c | c | c | c | c | c | c | c | c | } \hline \text { Leading Digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\hline \begin{array} { l } \text { Benford's law: } \\\text { distribution of } \\\text { leading digits }\end{array} & 30.1 \% & 17.6 \% & 12.5 \% & 9.7 \% & 7.9 \% & 6.7 \% & 5.8 \% & 5.1 \% & 4.6 \% \\\hline\end{array}$ -When working for the Brooklyn District Attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 9, 0, 70, 485, 189, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford's Law, the check amounts appear to result from fraud. Use a 0.05 significance level to test for goodness-of-fit with Benford's Law. Does it appear that the checks are the result of fraud?

Correct Answer

verified

Test statistic . Critical value: -value ...

View Answer

Show Answer

Question 2

Essay

Review LaterAdd Note

Review Later

Describe the null and alternative hypotheses for one-way ANOVA. Give an example.

Correct Answer

verified

The null hypothesis for one-way ANOVA is...

View Answer

Show Answer

Question 3

Essay

Review LaterAdd Note

An observed frequency distribution is as follows: $\begin{array}{l|ccc}\text { Number of successes } & 0 & 1 & 2 \\\hline \text { Frequency } & 47 & 98 & 55\end{array}$ i) Assuming a binomial distribution with $\mathrm { n } = 2$ and $\mathrm { p } = 1 / 2$ , use the binomial formula to find the probability corresponding to each category of the table. ii) Using the probabilities found in part (i), find the expected frequency for each category. iii) Use a $0.05$ level of significance to test the claim that the observed frequencies fit a binomial distribution for which $\mathrm { n } = 2$ and $\mathrm { p } = 1 / 2$ .

Identify the p-value. $\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\\text { Factor } & 3 & 30 & 10.00 & 1.6 & 0.264 \\\text { Error } & 8 & 50 & 6.25 & & \\\text { Total } & 11 & 80 & & &\end{array}$

Use a $\chi ^ { 2 }$ test to test the claim that in the given contingency table, the row variable and the column variable are independent. -Responses to a survey question are broken down according to employment status and the sample results are given below. At the 0.10 significance level, test the claim that response and employment status are independent. $\begin{array} { r | r r r } & \text { Yes } & \text { No } & \text { Undecided } \\\hline \text { Employed } & 30 & 15 & 5 \\\text { Unemployed } & 20 & 25 & 10\end{array}$

According to Benford's Law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. Test for goodness-of-fit with Benford's Law. $\begin{array} { | l | c | c | c | c | c | c | c | c | c | } \hline \text { Leading Digit } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\hline \begin{array} { l } \text { Benford's law: } \\\text { distribution of } \\\text { leading digits }\end{array} & 30.1 \% & 17.6 \% & 12.5 \% & 9.7 \% & 7.9 \% & 6.7 \% & 5.8 \% & 5.1 \% & 4.6 \% \\\hline\end{array}$ -When working for the Brooklyn District Attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 12, 0, 73, 482, 186, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford's Law, the check amounts appear to result from fraud. Use a 0.05 significance level to test for goodness-of-fit with Benford's Law. Does it appear that the checks are the result of fraud?

A company manager wishes to test a union leader's claim that absences occur on the different week days with the same frequencies. Test this claim at the 0.05 level of significance if the following sample data have been compiled. $\begin{array} { l | } \text { Day } & \text { Mon} & \text {Tue} & \text { Wed} & \text { Thur} & \text { Fri }\\\hline \text {absences }&37&15&12&23&43\\\end{array}$

Use the data given below to verify that the t test for independent samples and the ANOVA method are equivalent. $\begin{array} { c | c } \mathrm { A } & \mathrm { B } \\\hline 85 & 74 \\81 & 72 \\73 & 65 \\91 & 83 \\64 & 59\end{array}$ i) Use a t test with a 0.05 significance level to test the claim that the two samples come from populations with the same means. ii) Use the ANOVA method with a 0.05 significance level to test the same claim. iii) Verify that the squares of the t test statistic and the critical value are equal to the F test statistic and critical value.

In the chi-square test of independence, the formula used is $\chi ^ { 2 } = \frac { \Sigma ( \mathrm { O } - \mathrm { E } ) ^ { 2 } } { \mathrm { E } }$ . Discuss the meaning of $\mathrm { O }$ and $\mathrm { E }$ and explain the circumstances under which the $\chi ^ { 2 }$ values will be smaller or larger. What is the relationship between a significant $\chi ^ { 2 }$ value and the values of O and E?

Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance. -Random samples of four different models of cars were selected and the gas mileage of each car was measured. The results are shown below. $\begin{array} { l } \text { Model A}&\text { Model B}&\text { Model C}&\text { Model D }\\\hline23&28&30&25\\25&26&28&26\\24&29&32&25\\26&30&27&28\end{array}$ Test the claim that the four different models have the same population mean. Use a significance level of 0.05.

Use a $\chi ^ { 2 }$ test to test the claim that in the given contingency table, the row variable and the column variable are independent. -Responses to a survey question are broken down according to gender and the sample results are given below. At the 0.05 significance level, test the claim that response and gender are independent. $\begin{array} { r | r r r } & \text { Yes } & \text { No } & \text { Undecided } \\\hline \text { Male } & 25 & 50 & 15 \\\text { Female } & 20 & 30 & 10\end{array}$

Fill in the missing entries in the following partially completed one-way ANOVA table. $\begin{array} { | l | l | l | l | l | } \hline \text { Source } & \text { df } & \text { SS } & \text { MS=SS/df } & \text { F-statistic } \\\hline \text { Treatment } & & 28.9 & & \\\hline \text { Error } & 30 & & 3.5 & \\\hline \text { Total } & 33 & & & \\\hline\end{array}$

Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance. -Given the sample data below, test the claim that the populations have the same mean. Use a significance level of 0.05. $\begin{array} { l } \text { Brand A }&\text { Brand B }&\text { Brand C }&\text { Brand D }\\\hline\mathrm{n}=16 & \mathrm{n}=16 & \mathrm{n}=16 & \mathrm{n}=16 \\\overline{\mathrm{x}}=2.09 & \overline{\mathrm{x}}=3.48 & \overline{\mathrm{x}}=1.86 & \overline{\mathrm{x}}=2.84 \\\mathrm{~s}=0.37 & \mathrm{~s}=0.61 & \mathrm{~s}=0.45 & \mathrm{~s}=0.53\end{array}$

At the 0.025 significance level, test the claim that the four brands have the same mean if the following sample results have been obtained. $\begin{array} { l } \text { Brand A}&\text { Brand B}&\text { Brand C }&\text { Brand D }\\\hline17&18&21&22\\20&18&24&25\\21&23&25&27\\22&25&26&29\\21&26&29&35\\&&29&36\\&&&37\end{array}$

You roll a die 48 times with the following results. $\begin{array} { c | l | l | r | r | r | r } \text { Number } & 1 & 2 & 3 & 4 & 5 & 6 \\\hline \text { Frequency } & 2 & 4 & 12 & 13 & 14 & 3\end{array}$ Use a significance level of 0.05 to test the claim that the die is fair.

Draw an example of an F distribution and list the characteristics of the F distribution.

Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance. -At the 0.025 significance level, test the claim that the three brands have the same mean if the following sample results have been obtained. $\begin{array} { l } \text { Brand A}& \text {Brand B }& \text {Brand C }\\\hline32&27&22\\34&24&25\\37&33&32\\33&30&22\\36&&21\\39\end{array}$

An observed frequency distribution of exam scores is as follows: $\begin{array} {| l|ccccc| } \hline \text {Exam Score }&\text {under 60 }&60-69&70-79&80-89&90-100\\\hline \text { Frequency}&30&30&140&60&40\\\hline\end{array}$ i) Assuming a normal distribution with $\mu = 75$ and $\sigma = 15$ , find the probability of a randomly selected subject belonging to each class. (Use boundaries of $59.5,69.5,79.5,89.5$ , 100.) ii) Using the probabilities found in part (i), find the expected frequency for each category. iii) Use a $0.05$ significance level to test the claim that the exam scores were randomly selected from a normally distributed population with $\mu = 75$ and $\sigma = 15$ .

What can you conclude about the equality of the population means? $\begin{array} { l r c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \\ \text { Error } & 16 & 13.925 & 0.870 & & \\ \text { Total } & 19 & 27.425 & & & \end{array}$

At a high school debate tournament, half of the teams were asked to wear suits and ties and the rest were asked to wear jeans and t-shirts. The results are given in the table below. Test the hypothesis at the 0.05 level that the proportion of wins is the same for teams wearing suits as for teams wearing jeans. $\begin{array} { r | r r } & \text { Win } & \text { Loss } \\\hline \text { Suit } & 22 & 28 \\\text { T-shirt } & 28 & 22\end{array}$