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

For a chi-squared distributed random variable with 10 degrees of freedom and a level of significance of 0.025, the chi-squared value from the table is 20.5. The computed value of the test statistic is 16.857. The decision is to retain Ho.

Which of the following best describes the number of degrees of freedom used in a Chi-squared test of independence if there is a contingency table with 5 rows and 3 columns?

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

Last year, Brand A mobile android phones had 45% of the market, Brand B had 35% and Brand C had 20%. This year the makers of Brand C launched a heavy advertising campaign. A random sample of mobile phone stores shows that of the 10 000 android mobile phones sold, 4350 were Brand A, 3450 were Brand B, and 2200 were Brand C. Has the market changed? Test at $\alpha =$ 0.05.

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:

Which of the following are the degrees of freedom used in a Chi-squared test of independence between gender and mode of transport to university, if the mode of transport choices are public transport, car, bicycle or other.

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.

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? Agreed Disagreed Undecided Female 15 25 30 Male 25 20 25

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.

Which of the following is the correct decision in a goodness-of-fit test, if the calculated value of the test statistic is 20.08 and there are 10 categories, testing at α of 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.

If each element in a population is classified into one and only one of several categories, the population is a:

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 100 observations. In addition, the researcher used 6 standardised intervals to test for normality. Using a 2.5% level of significance, the critical value for this test is 9.3484.

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

A large carpet store wishes to determine if the brand of carpet purchased is related to the purchaser's family income. As a sampling frame, they mailed a survey to people who have a store credit card. Five hundred customers returned the survey and the results follow: Brand of Carpet Family Income Brand Brand B Brand C High income 65 32 32 Middle income 80 68 104 Low income 25 35 59 At the 5% level of significance, can you conclude that the brand of carpet purchased is related to the purchaser's family income?

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?

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. $H _ { 0 } : p _ { 1 } = 0.10$ , $p _ { 2 } = 0.25$ $p _ { 3 } = 0.30$ $p _ { 4 } = 0.20$ $p _ { 5 } = 0.15$ $p _ { 5 } = 0.15$ .

Consider a multinomial experiment involving 100 trials and 4 categories (cells). The observed frequencies resulting from the experiment are shown in the accompanying table. Category 1 2 3 4 Frequency 18 30 25 27 Use the 5% significance level to test the hypotheses. H₀ : p₁ = 0.25, p₂ = 0.30, p₃ = 0.20, p₄ = 0.25. H₁ ;At least two proportions differ from their specified values.

A test for independence is applied to a contingency table with 5 rows and 2 columns for two nominal variables. The number of degrees of freedom for this chi-squared test must be 4.

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 $x ^ { 2 }$ -statistic.)