Chi Square Tests Review

An appropriate null hypothesis is to test whether the food choices are equally popular is

• $H_0:\mu=$ where mu is the mean number of students that prefer each type of food.
  

• $H_0:p=0.25$ where p is the proportion of all students who prefer Asian food.

• $H_0:n_A=n_M=n_P=n_H=25$ where nA is the number of students in the school who would choose Asian food, and so on.

• $H_0:p_A=p_B=p_P=p_H=0.25$ where pA is the proportion of students in the school who would choose Asian food, and so on.

• $H_0:\overline{p_A}=\overline{p_M}=\overline{p_P}=\overline{p_H}$ where p-hatA is the proportion of students in the sample who chose Asian food, and so on.

The chi square statistic is

• $\frac{\left(18-25\right)^2}{25}+...+\frac{\left(21-25\right)^2}{25}$

• $\frac{\left(25-18\right)^2}{18}+...+\frac{\left(25-21\right)^2}{21}$

• $\frac{\left(18-25\right)}{25}+...+\frac{\left(21-25\right)}{25}$

• $\frac{\left(18-25\right)^2}{100}+...+\frac{\left(21-25\right)^2}{100}$

• $\frac{\left(0.18-0.25\right)^2}{0.25}+...+\frac{\left(0.39-0.25\right)^2}{0.25}$

The p-value for a chi-square test for goodness of fit is 0.0129. Which of the following is the most appropriate conclusion?

• Because 0.0129 is less than 0.05, reject the null hypothesis. There is not convincing evidence that the food choices are equally popular.

• Because 0.0129 is less than 0.05, reject the null hypothesis. There is not convincing evidence that the food choices are equally popular.

• Because 0.0129 is less than 0.05, reject the null hypothesis. There is convincing evidence that the food choices are not equally popular.

• Because 0.0129 is less than 0.05, we fail to reject the null hypothesis. There is not convincing evidence that the food choices are equally popular.

• Because 0.0129 is less than 0.05, fail to reject the null hypothesis. There is convincing evidence that the food choices are equally popular.

Which of the following is false?

• A chi-square distribution with k degrees of freedom is more right skewed than a chi-square distribution with k+1 degrees of freedom.

• A chi-square distribution never takes negative values.

• The degrees of freedom for a chi-square test is determined by sample size.

• $P\left(\chi^2>10\right)$ is greater than df=k+1 than when df=k.

• The area under a chi-square density curve is always equal to 1.

Which of the following would be most appropriate type of graph for these data?

• A bar chart showing the marginal distribution of opinion about marriage.

• A bar chart showing the marginal distribution of gender.

• A bar chart showing the conditional distribution of gender for each opinion about marriage.

• A bar chart showing the conditional distribution of opinion about marriage for each gender.

• Dotplots that display the number in each opinion category for each gender.

The appropriate null hypothesis for performing a chi-square test is that

• equal proportions of female and male teenagers are almost certain they will be married in 10 years.

• there is no difference between the distributions of female and male teenagers' opinions about marriage in this sample.

• there is no difference between the distributions of female and male teenagers' opinions about marriage in the population.

• there is no association between gender and opinion about marriage in the sample.

• there is no association between gender and opinion about marriage in the population.

The expected count of females who respond "almost certain" is

• 487.7

• 525

• 965

• 1038.8

• 1174

Tags

CCSS.HSS.MD.A.2

The degrees of freedom for the chi-square test for this two-way table are

• 4

• 8

• 10

• 20

• 4876

For these data, the chi-square test statistic is 69.8 with a p-value of approximately 0. Assuming that the researchers used a significance level of 0.05, which of the following is true?

• A type I error is possible.

• A type II error is possible.

• Both a type I and a type II error are possible.

• There is no chance of making a type I or type II error because the p-value is approximately zero.

• There is no chance of making a type I or type II error because the calculations are correct.

When analyzing results from a two-way table, the main distinction between a test for independence and homogeneity is

• how the degrees of freedom are calculated.

• how the expected counts are calculated.

• the number of samples obtained.

• the number of rows in the two-way table.

• the number of columns in the two-way table.