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Exam 15: Chi-Squared Tests Optional

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Question 71

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The rule of five states that in order to conduct the chi-squared goodness-of-fit test, the ____________________ value for each cell must be five or more.

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Question 72

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Mathematical statisticians have established that if we square the value of z, the test statistic for the test of one proportion p, we produce the $\chi$ ² statistic. That is, z² = $\chi$ ².

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Question 73

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The values of a chi-squared distribution are always ____________________ zero.

A small chi-squared test statistic in a goodness-of-fit test supports the null hypothesis.

There are two critical factors in identifying the technique used when the data are nominal. The first is the problem objective. The second is the number of ____________________ that nominal variable can assume.

A Deli proposes to serve 4 main Sandwiches. For planning purposes, the manager expects that the proportions of each that will be selected by her customers will be: $\begin{array} { | l | c | } \hline \text { Selection } & \text { Proportion } \\\hline \text { Turkey } & 0 .50 \\\text { Roast Beef } & 0 .20 \\\text { Pastrami } &0 .10 \\\text { Tuna } &0 .20 \\\hline\end{array}$ Of a random sample of 100 customers, 44 selected chicken, 24 selected roast beef, 13 selected Pastrami, and 10 selected tuna. Should the manager revise her estimates? Use $\alpha$ = 0.01.

The rule of ____________________ states that in order to conduct the chi-squared goodness-of-fit test, the expected value for each cell must be ____________________ or more.

To test for normality, the ____________________ hypothesis is that at least two proportions differ from their specified values.

In a goodness-of-fit test, H₀ lists specific values for proportions and the test of a contingency table does not.

The number of degrees of freedom in a chi-squared test for normality, where the number of standardized intervals is 5 and there are 2 population parameters to be estimated from the data, is equal to:

A chi-squared test statistic in a test of a contingency table that is equal to zero means:

NARRBEGIN: Students Absenteeism Student Absenteeism Consider a multinomial experiment involving n = 200 students of a large high school. The attendance department recorded the number of students who were absent during the weekdays. The null hypothesis to be tested is: H₀: p₁ = 0.10, p₂ = 0.25, p₃ = 0.30, p₄ = 0.20, p₅ = 0.15.NARREND -{Student Absenteeism Narrative} Test the hypothesis at the 5% level of significance with the following frequencies: $\begin{array} { | l | c c c c c | } \hline \text { Day of the Week } & \text { Mon. } & \text { Tues. } & \text { Wed. } & \text { Thurs. } & \text { Fri } \\\hline \text { Number Absent } & 16 & 44 & 56 & 48 & 36 \\\hline\end{array}$

To test for normality, the ____________________ hypothesis specifies probabilities of certain intervals within the normal distribution.

A chi-squared distribution is symmetric.

If there are only two categories, the chi-squared goodness-of-fit test is the same as the z-test for p, the population proportion (as long as the sample/cell sizes meet the conditions).

A test statistic that lies in the far right tail of the chi-squared distribution indicates you will ____________________ H₀.

NARRBEGIN: Students Absenteeism Student Absenteeism Consider a multinomial experiment involving n = 200 students of a large high school. The attendance department recorded the number of students who were absent during the weekdays. The null hypothesis to be tested is: H₀: p₁ = 0.10, p₂ = 0.25, p₃ = 0.30, p₄ = 0.20, p₅ = 0.15.NARREND -{Student Absenteeism Narrative} Test the hypothesis at the 5% level of significance with the following frequencies: $\begin{array} { | l | c c c c c | } \hline \text { Day of the Week } & \text { Mon. } & \text { Tues. } & \text { Wed. } & \text { Thurs. } & \text { Fri. } \\\hline \text { Number Absent } & 4 & 11 & 14 & 12 & 9 \\\hline\end{array}$

The human resources manager of a consumer product company asked a random sample of employees how they felt about the work they were doing. The following table gives a breakdown of their responses by whether the employee is part time or full time (aka work status). Do the data provide sufficient evidence to conclude that the level of job satisfaction is related to their work status? Use $\alpha$ = 0.10. $\begin{array}{l}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text { Response }\\\begin{array} { | l | c c c | } \hline \text { Gender } & \text { Very Interesting } & \text { Fairly Interesting } & \text { Not Interesting } \\\hline \text { Full time } & 70 & 41 & 9 \\\text { Part time } & 35 & 34 & 11 \\\hline\end{array}\end{array}$

The chi-squared goodness-of-fit test compares the ____________________ frequencies in the table to the ____________________ frequencies based on the null hypothesis.

Conduct a test to determine whether the two classifications A and B are independent, using the data in the accompanying table and $\alpha$ = 0.05. $\begin{array} { | l | l l l | } \hline & \boldsymbol { B } _ { 1 } & \boldsymbol { B } _ { \mathbf { 7 } } & \boldsymbol { B } _ { \mathbf { 3 } } \\\hline A _ { 1 } & 35 & 25 & 20 \\A _ { 7 } & 25 & 20 & 25 \\\hline\end{array}$