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

In a chi-squared goodness-of-fit test, if the expected frequencies $e _ { i }$ and the observed frequencies $f _ { i }$ were quite different, which of the following is the best conclusion? A. Null hypothesis is false, and reject B. Null hypothesis is true, and retain . C. Alternative hypothesis is false, reject D. Chi-squared distribution is invalid, use t -distribution instead.

Which of the following statements is not correct? A. The chi-squared distribution is symmetrical. B. The chi-squared distribution is skewed to the right. C. All values of the chi-squared distribution are positive. D. The critical region for a goodness-of-fit test with k categories is >.

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.

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 16 44 56 48 36 The null hypothesis to be tested is as follows. H₀ : p₁ = 0.05, p₂ = 0.25, p₃ = 0.35, p₄ = 0.20, p₅ = 0.15. Test the hypothesis at the 5% level of significance.

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.

Which of the following is not a characteristic of a multinomial experiment? A. The experiment consists of a fixed number, $n$ , of trials. B. The outcome of each trial can be classified into one of two categories called success and failure. C. The probability $p_{i}$ that the outcome will fall into cell $i$ remain constant for each trial. D. Each trial of the experiment is independent of the other trials.

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

A biology professor claimed that the proportions of grades in his classes are the same Grade A B C D F Frequency 14 23 27 26 10 State the null and alternative hypotheses to be tested.

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 Use statistical software to compute the p-value for this test.

A major insurance firm interviewed a random sample of 1500 college students to find out the type of life insurance preferred, if any. The results follow: Insurance Preference Gender Term Whole life No insurance Female 170 110 470 Male 195 75 480 Is there evidence that the life insurance preference of male students is different to that of female students? Test using the 5% level of significance.

Whenever the expected frequency of a cell is less than 5, we must increase the significance level.

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.

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.

Which statistical technique is appropriate when we compare two populations of nominal data with exactly two categories? A. The z -test of a population proportion. B. The z -test of the difference between two proportions. C. The chi-squared test of a contingency table. D. The z -test of a population proportion and the z -test of the differ ence between two proportions. E. The z -test of the difference between two proportions and the chi-squared test of a contingency table.

Which of the following may be used to determine whether data were drawn a particular distribution? A. Chi-squared goodness-of-fit test. B. Chi-squared test of a contingency table. C. Z test. D. All of these choices are correct.

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.

In a number of universities in Australia, lecture recordings online are provided to all students. There is growing concern amongst teaching staff that lecture recordings are used as a substitute for lecture attendance and that this is leading to a greater fail rate. At one particular university, a random sample of 224 students is taken, where the students are asked if they attended the majority of lectures for the last exam they took, or if they viewed the majority of lecture recordings online and then the student's records are followed up to see if the student passed that course. Attended lectures Viewed lecture recordings online Fail 40 80 Pass 56 48 Conduct a test to determine if enough evidence exists to infer that lecture attendance and passing a university course are related. Test at the 10% level of significance.

A chi-squared goodness-of-fit test can be conducted either as a two-tail test or as a one-tail test.

The following data are believed to have come from a normal probability distribution. 26 21 25 20 21 29 26 23 22 24 24 30 23 32 26 24 32 16 36 26 21 31 26 23 32 35 40 30 14 26 46 27 33 25 27 21 26 18 29 36 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.

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.