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

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A chi-squared test for independence with 6 degrees of freedom results in a test statistic of 13.25. Using the chi-squared table, the most accurate statement that can be made about the p-value for this test is that 0.025 < p-value < 0.05.

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The personnel 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 gender. \multicolumn 3 |c| Response Gender Very interesting Fairly interesting Not interesting Male 70 41 9 Female 35 34 11 Do the data provide sufficient evidence to conclude that the level of job satisfaction is related to gender? Use α=\alpha = 0.10.

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If we want to test for differences between two populations of nominal data with exactly two categories, we can employ either the z-test of p1p2p _ { 1 } - p _ { 2 } , or the chi-squared test of a contingency table. (Squaring the value of the z-statistic yields the value of the x2x ^ { 2 } -statistic.)

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Which of the following may be used to determine whether data were drawn a particular distribution?

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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.

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For a chi-squared distributed random variable with 12 degrees of freedom and a level of significance of 0.05, the chi-squared value from the table is 21.0261. The computed value of the test statistics is 25.1687. This will lead us to reject the null hypothesis.

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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.

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In 2015, the student body of a University in NSW consisted of 30% first-years, 25% second-years, 27% third-years, and 18% fourth-years. A sample of 400 students taken from the 2004 student body showed that there are 138 first-years, 88 second-years, 94 third-years, and 80 fourth-years. Test with 5% significance level to determine whether the student body proportions have changed.

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In a chi-squared test of independence, the value of the test statistic was x2=x ^ { 2 } = 15.652, and the critical value at α=0.025\alpha = 0.025 was 11.1433. Thus we must reject the null hypothesis at α=0.025\alpha = 0.025 .

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A psychologist claims that people who exercise regularly are more content with their body image. A random sample of 102 people was taken, where they were asked if they regularly exercise and if they were content with their body image. Not content Content Exercised regularly 20 30 Did not exercise regularly 28 24 Is there significant evidence to support this claim? Test at the 1% significance level.

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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 Do the data provide enough evidence to support the professor's claim?

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A left-tailed area in the chi-squared distribution equals 0.975. For df = 11, the table value equals:

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An Australian tyre manufacturer operates a plant in Melbourne and another plant in Adelaide. Employees at each plant have been evenly divided among three issues (wages, working conditions and super benefits) in terms of which one they feel should be the primary issue in the upcoming contract negotiations. The secretary of the union has recently circulated pamphlets among the employees, attempting to convince them that super benefits should be the primary issue. A subsequent survey revealed the following breakdown of the employees according to the plant at which they worked and the issue that they felt should be supported as the primary one. Issues Plant Location Very interesting Fairly interesting Not interesting Melbourne 60 62 78 Adelaide 70 56 74 Can you infer at the 5% significance level that the proportional support by the Adelaide employees for the three issues has changed since the pamphlet was circulated?

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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.

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An Australian tyre manufacturer operates a plant in Melbourne and another plant in Adelaide. Employees at each plant have been evenly divided among three issues (wages, working conditions and super benefits) in terms of which one they feel should be the primary issue in the upcoming contract negotiations. The secretary of the union has recently circulated pamphlets among the employees, attempting to convince them that super benefits should be the primary issue. A subsequent survey revealed the following breakdown of the employees according to the plant at which they worked and the issue that they felt should be supported as the primary one. Issues Plant Location Very interesting Fairly interesting Not interesting Melbourne 60 62 78 Adelaide 70 56 74 Do the data indicate at the 5% significance level that there are differences between the two plants regarding which issue should be the primary one?

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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 do not reject Ho.

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Which of the following best describes the sampling distribution of the test statistic for a goodness-of-fit test with k categories?

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

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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.

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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.

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