Question 1

In chi-squared tests, the conventional and conservative rule - known as the rule of five - is to require that difference between the observed and expected frequency for each cell be at least 5.

Accepted Answer

The rule of five in chi-squared tests actually states that no more than 20% of the cells should have expected frequencies less than 5, and all individual expected frequencies should be 1 or greater.

Question 2

An Australian firm has been accused of engaging in prejudicial hiring practices against women who have young children, where a young child is defined to be a child less than 12 years of age. According to the most recent census, the percentages of men, women without young children and women with young children in a certain community are 72%, 10% and 18%, respectively. A random sample of 200 employees of the firm revealed that 165 were men, 14 were women without young children and 21 were women with young children. Do the data provide sufficient evidence to conclude at the 10% level of significance that the firm has been engaged in prejudicial hiring practices?

Accepted Answer

The answer of An Australian firm has been accused of...

Question 3

If we want to test for differences between two populations of nominal data with exactly two categories, we can employ either the z-test of $p _ { 1 } - p _ { 2 }$ , or the chi-squared test of a contingency table. (Squaring the value of the z-statistic yields the value of the $\chi ^ { 2 }$ -statistic.)

Accepted Answer

The z-test for two proportions and the chi-squared test for a 2x2 contingency table are related; squaring the z-statistic from the z-test gives the chi-squared statistic for the same data. Both tests assess the difference between two proportions.

Question 4

The area to the right of a chi-squared value is 0.01. For 8 degrees of freedom, the table value is 20.0902.

Accepted Answer

For a chi-squared distribution with 8 degrees of freedom, a table or calculator will show that the critical value corresponding to an area (or p-value) of 0.01 to the right (in the upper tail) is indeed 20.0902. This means that if the chi-squared statistic is greater than 20.0902, the result is in the top 1% of the distribution, indicating a very low probability (less than 0.01) that the observed distribution matches the expected distribution under the null hypothesis, assuming the null hypothesis is true.

Question 5

If we want to perform a one-tail test for differences between two populations of nominal data with exactly two categories, we must employ the z-test of $p _ { 1 } - p _ { 2 }$ .

Accepted Answer

The z-test for two proportions is appropriate for comparing differences between two populations with nominal data having two categories, especially when sample sizes are large enough to approximate normal distribution.

Question 6

The rule of five is used to ensure that the discrete distribution of the test statistic can be approximated by the continuous Chi-squared distribution.

Accepted Answer

The rule of 5 states that in order to use the chi-squared distribution to approximate the distribution of the test statistic, each expected frequency should be at least 5. This helps ensure that the discrete distribution of the test statistic can be accurately approximated by the continuous chi-squared distribution.

Question 7

The number of degrees of freedom associated with the chi-squared test for normality is the number of intervals used minus the number of parameters estimated from the data.

Accepted Answer

The answer of The number of degrees of freedom associated...

Question 8

At present in Australia, voting in a Federal election is compulsory for all citizens 18 years or over.
There has been some discussion to have this changed from being compulsory to voluntary.
A study was done to investigate whether there was a relationship between gender and attitude towards changing voting in Australia from compulsory to voluntary. Participants were asked "Do you agree to change voting in Australian Federal Elections from compulsory to voluntary? A table of results from the sample is given below.
At the 1% level of significance, is there a relationship between gender and attitude towards voluntary voting? $$\begin{array} { | l | l | l | l | } 
\hline & \text { Agreed } & \text { Disagreed } & \text { Undecided } \
\hline \text { Female } & 15 & 25 & 30 \
\hline \text { Male } & 25 & 20 & 25 \
\hline
\end{array}$$

Accepted Answer

The answer of At present in Australia, voting in a...

Question 9

The personnel 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 gender. $$\begin{array} { | l | c | c | c | } 
\hline & & { \text { Response } } \
\hline \text { Gender } & \text { Very interesting } & \text { Fairly interesting } & \text { Not interesting } \
\hline \text { Male } & 70 & 41 & 9 \
\hline \text { Female } & 35 & 34 & 11 \
\hline
\end{array}$$ Do the data provide sufficient evidence to conclude that the level of job satisfaction is related to gender? Use $\alpha =$ 0.10.

Accepted Answer

The answer of The personnel manager of a consumer product...

Question 10

A sport preference poll showed the following data for men and women: $$\begin{array} { | l | c | c | c | c | c | } 
\hline && { \text { Fanourite Sport } } \
\hline \text { Gender } & \text { Rugby } & \text { B asketball } & \text { Football } & \text { Golf } & \text { Tennis } \
\hline \text { Male } & 24 & 17 & 30 & 18 & 22 \
\hline \text { Female } & 21 & 20 & 22 & 12 & 28 \
\hline
\end{array}$$ Using the 5% level of significance, test to determine whether sport preferences depend on gender.

Accepted Answer

The answer of A sport preference poll showed the following...

Question 11

The following data are believed to have come from a normal probability distribution. 
$$\begin{array} { | r | r | r | r | r | r | r | r | r | r | } 
\hline 26 & 21 & 25 & 20 & 21 & 29 & 26 & 23 & 22 & 24 \
\hline 24 & 30 & 23 & 32 & 26 & 24 & 32 & 16 & 36 & 26 \
\hline 21 & 31 & 26 & 23 & 32 & 35 & 40 & 30 & 14 & 26 \
\hline 46 & 27 & 33 & 25 & 27 & 21 & 26 & 18 & 29 & 36 \
\hline
\end{array}$$ The mean of this sample equals 26.80, and the standard deviation equals 6.378. Use the goodness-of-fit test at the 5% significance level to test this claim.

Accepted Answer

The answer of The following data are believed to have...

Question 12

Like that of the Student t-distribution, the shape of the chi-squared distribution depends on its number of degrees of freedom.

Accepted Answer

The shape of the chi-squared distribution depends on its degrees of freedom. As the degrees of freedom increase, the distribution becomes more symmetric and bell-shaped.

Question 13

The number of degrees of freedom for a contingency table with r rows and c columns is (r -1)(c - 1), provided that both r and c follow the rule of five.

Accepted Answer

The answer of The number of degrees of freedom for...

Question 14

When doing a Chi-squared goodness of fit test, we use Ho: The data are not normally distributed.

Accepted Answer

When doing a Chi-squared goodness of fit test, we assume that the data are normally distributed and use Ho: The data follow a normal distribution.

Question 15

If we want to perform a two-tail test for differences between two populations of nominal data with exactly two categories, we can employ either the z-test of $p _ { 1 } - p _ { 2 }$ , or the chi-squared test of a contingency table. (Squaring the value of the z-statistic yields the value of the $\chi ^ { 2 }$ -statistic.)

Accepted Answer

The z-test for two proportions and the chi-squared test for a 2x2 contingency table are equivalent; squaring the z-statistic gives the chi-squared statistic.

Question 16

In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 200 observations. In addition, the researcher used 5 standardised intervals to test for normality. Using a 10% level of significance, the critical value for this test is 4.60517.

Accepted Answer

This statement is true as the critical value for a goodness-of-fit test with 5 intervals and a 10% level of significance is indeed 4.60517.

Question 17

The chi-squared goodness-of-fit test is usually used as a test of multinomial parameters, but it can also be used to determine whether data were drawn from any distribution.

Accepted Answer

The chi-squared goodness-of-fit test can be used to test whether observed data follow a given distribution or not, not just for testing multinomial parameters.

Question 18

The chi-squared test of a contingency table is used to determine if there is enough evidence to infer that two nominal variables are related, and to infer that differences exist among two or more populations of nominal variables.

Accepted Answer

This statement accurately describes the purpose of the chi-squared test for a contingency table.

Question 19

A multinomial experiment, where the outcome of each trial can be classified into one of two categories, is identical to the binomial experiment.

Accepted Answer

A binomial experiment has exactly two possible outcomes (success or failure) while a multinomial experiment can have more than two possible outcomes. However, if each trial of a multinomial experiment has only two possible outcomes, then the experiment is identical to a binomial experiment.

Question 20

Five brands of orange juice are displayed side by side in several supermarkets in a large city. It was noted that in one day, 180 customers purchased orange juice. Of these, 30 picked Brand A, 40 picked Brand B, 25 picked Brand C, 35 picked Brand D, and 50 picked brand
E. In this city, can you conclude at the 5% significance level that there is a preferred brand of orange juice?

Accepted Answer

The answer of Five brands of orange juice are displayed...

Quiz 17: Additional Tests for Nominal Data: Chi-Squared Tests