Exam 14: Additional Tests for Nominal Data: Chi-Squared Tests
Exam 1: What Is Statistics17 Questions
Exam 2: Types of Data, Data Collection and Sampling18 Questions
Exam 3: Graphical Descriptive Techniques Nominal Data17 Questions
Exam 4: Graphical Descriptive Techniques Numerical Data65 Questions
Exam 5: Numerical Descriptive Measures149 Questions
Exam 6: Probability113 Questions
Exam 7: Random Variables and Discrete Probability Distributions50 Questions
Exam 8: Continuous Probability Distributions113 Questions
Exam 9: Statistical Inference and Sampling Distributions69 Questions
Exam 10: Estimation: Describing a Single Population125 Questions
Exam 11: Estimation: Comparing Two Populations36 Questions
Exam 12: Hypothesis Testing: Describing a Single Population124 Questions
Exam 13: Hypothesis Testing: Comparing Two Populations69 Questions
Exam 14: Additional Tests for Nominal Data: Chi-Squared Tests113 Questions
Exam 15: Simple Linear Regression and Correlation213 Questions
Exam 16: Multiple Regression122 Questions
Exam 17: Time-Series Analysis and Forecasting147 Questions
Exam 18: Index Numbers27 Questions
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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.
(True/False)
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In a goodness-of-fit test, suppose that a sample showed that the observed frequency and expected frequency were equal for each cell i. Which of the following best describes the decision for the hypothesis test?
(Multiple Choice)
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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 200 observations. In addition, the researcher used 5 standardised intervals to test for normality. Using a 10% level of significance, the critical value for this test is 4.60517.
(True/False)
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Whenever the expected frequency of a cell is less than 5, one possible remedy for this condition is to combine it with one or more other cells.
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Suppose that a random sample of 150 observations was drawn from a population. After calculating the mean and standard deviation, each observation was standardised and the number of observations in each of the intervals below was counted. Can we infer at the 5% significance level that the data were drawn from a normal population? Intervals Frequency Z\leq-1.5 15 -1.51.5 13
(Essay)
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A telephone company prepared four versions of a set of instructions for placing collect calls. The company asked a sample of 1600 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?
(Essay)
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Consider a multinomial experiment involving 160 trials and 4 categories (cells). The observed frequencies resulting from the experiment are shown in the following table. Category 1 2 3 4 Frequency 53 35 30 42 Use the 10% significance level to test the hypotheses. .
HA: At least two proportions differ from their specified values.
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Which of the following would be used to analyse the relationship between two categorical variables?
(Multiple Choice)
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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 seven categories. Using a 0.05 significance level, the appropriate critical value for this chi-squared test for normality is 11.0705.
(True/False)
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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.
HA:At least two proportions differ from their specified values.
(Essay)
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A simple random sample of 150 people was taken to investigate whether there is a link between smoking status and having had a tertiary education. Test at the 5% level of significance. Non-smoker Smoker None-tertiary educated 42 28 Tertiary educated 23 57
(Essay)
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Which of the following best describes the degrees of freedom needed in a Chi-squared test for normality?
(Multiple Choice)
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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 Use statistical software to compute the p-value for this test.
(Short Answer)
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In a chi-squared goodness-of-fit test, if the expected frequencies and the observed frequencies were quite different, which of the following is the best conclusion?
(Multiple Choice)
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A left-tailed area in the chi-squared distribution equals 0.90. For 10 degrees of freedom, the table value equals 15.9871.
(True/False)
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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?
(Essay)
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An Australian firm has been accused of engaging in prejudicial hiring practices against women who have young children, where a young child is defined to be a child less than 12 years of age. According to the most recent census, the percentages of men, women without young children and women with young children in a certain community are 72%, 10% and 18%, respectively. A random sample of 200 employees of the firm revealed that 165 were men, 14 were women without young children and 21 were women with young children. Do the data provide sufficient evidence to conclude at the 1% level of significance that the firm has been engaged in prejudicial hiring practices?
(Essay)
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The number of degrees of freedom in a test of a contingency table with 7 rows and 5 columns is:
(Multiple Choice)
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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.
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