Exam 15: Chi-Squared Tests

When we describe a population of nominal data, with exactly two categories, the multinomial experiment is actually a binomial experiment with one of the categorical outcomes labeled "success" and the other labeled "failure".

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 300 observations.In addition, the researcher used 6 standardized intervals to test for normality.Using a 2.5% level of significance, the critical value for this test is 14.4494.

A cable company prepared four versions of a set of instructions for hooking up a TV.The company asked a sample of 1,600 people which one of the four forms was easiest to understand.In the sample, 425 people preferred Form A, 385 preferred Form B, 375 preferred Form C, and 415 preferred Form D.At the 5% level of significance, can one conclude that in the population there is a preferred form?

Sales Volumes A telemarketer makes five calls per day.A sample of 200 days gives the frequencies of sales volumes listed below: Assume the population is binomial distribution with a probability of purchase p equal to .50. -{Sales Volumes Narrative} Compute the expected frequencies for x = 0, 1, 2, 3, 4, and 5 by using the binomial probability function or the binomial tables.Combine categories if necessary to satisfy the rule of five.

In a goodness-of-fit test, all of the proportions specified in the null hypothesis must be equal to each other.

If we want to perform a one-tail test of a population proportion p, we can employ either the z-test of p, or the chi-squared goodness-of-fit test.

For a chi-squared distributed random variable with 12 degrees of freedom and a level of significance of .05, the test statistics is 25.168.The chi-squared value from the table is 21.0261.These results will lead us to reject the null hypothesis.

In the test of a contingency table, the observed cell frequencies must satisfy the rule of 5.

For a chi-squared distributed random variable with 10 degrees of freedom and a level of significance of .025, the chi-squared table value is 20.4831.Suppose the value of your test statistic is 16.857.This will lead you to reject the null hypothesis.

The null hypothesis states that the sample data came from a normally distributed population.The researcher calculates the sample mean and the sample standard deviation from the data.The data arrangement consisted of five categories.Using $\alpha$ = .05, the appropriate critical value for this chi-squared test for normality is 5.99147.

What are the two critical factors in identifying the technique used when the data are nominal?

A chi-squared goodness-of-fit test is always conducted as a(n):

If we use the chi-squared method of analysis we must first check that there are at least 5 observations in each cell of the contingency table.

If you want to compare two populations that each have more than two categories, which test(s) can you use?

Which of the following conditions indicate that H₀ should be rejected in a goodness-of-fit test?

The following data are believed to have come from a normal probability distribution. 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 whether the data indeed come from a normal distribution.

A chi-squared test of a contingency table is applied to a contingency table with 4 rows and 4 columns for two qualitative variables.The degrees of freedom for this test must be 9.

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 $\alpha$ = .10.

Which test(s) can you use when you want to describe a population with two categories?

If the observed frequencies are all smaller than the expected frequencies in a goodness-of-fit test, the chi-squared test statistic will be negative.