Exam 14: Additional Tests for Nominal Data: Chi-Squared Tests

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

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 500 observations. In addition, the researcher used 6 standardised intervals to test for normality. Using a 5% level of significance, the critical value for this test is:

Of the values for a chi-squared test statistic listed below, which one is likely to lead to rejection of the null hypothesis in a goodness-of-fit test?

Consider a multinomial experiment involving n = 200 trials and k = 5 cells. The observed frequencies resulting from the experiment are shown in the following table: Cell 1 2 3 4 5 Frequency 8 22 28 24 18 The null hypothesis to be tested is as follows. $H _ { 0 } : p _ { 1 } = 0.10$ , $p _ { 2 } = 0.25,$ $p _ { 3 } = 0.30,$ $p _ { 4 } = 0.20,$ $p _ { 5 } = 0.15$ $p _ { 5 } = 0.15$ .

The chi-squared test of independence is a Chi-squared test of a contingency table.

Consider a multinomial experiment involving n = 200 trials and k = 5 cells. The observed frequencies resulting from the experiment are shown in the following table. Cell 1 2 3 4 5 Frequency 4 11 14 12 9 The null hypothesis to be tested is as follows. $H _ { 0 } : p _ { 1 } = 0.10$ , $p _ { 2 } = 0.25,$ $p _ { 3 } = 0.30,$ $p _ { 4 } = 0.20,$ $p _ { 5 } = 0.15$ $p _ { 5 } = 0.15$ .

Consumer panel preferences for three proposed fast food restaurants are as follows: Restaurant A Restaurant B Restaurant C 48 62 40 Using the 0.05 level of significance, test to see if there is a preference among the three restaurants.

Which statistical technique is appropriate when we compare two populations of nominal data with exactly two categories?

Which of the following best describes a Chi-squared goodness-of-fit test?

A chi-squared test for independence with 10 degrees of freedom results in a test statistic of 17.894. Using the chi-squared table, the most accurate statement that can be made about the p-value for this test is that 0.05 < p-value < 0.10.

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.

A Chi-squared test can be used to test the equality of two population means.

A biology professor claimed that the proportions of grades in his classes are the same. A sample of 100 students showed the following frequencies. Grade A B C D F Frequency 14 23 27 26 10 Compute the value of the test statistic.