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A quality control specialist tests samples of the seed being packaged and uses a chi-square goodness-of-fit test to see whether the proportions in the samples match what is claimed by the company. Which of the following best describes the null hypothesis and the alternative hypothesis for the test? A) H0: p=0.20 Ha: p≠0.20 B) H0: pF=0.20, pB=0.20, pBG=0.20, pI=0.20, pNG=0.20 Ha: At least one of the proportions is different. C) H0: pF=0.55, pB=0.22, pBG=0.10, pI=0.07, pNG=0.06 Ha: At least one of the proportions is different. D) H0: All proportions are equally likely. Ha: All of the proportions are different. E) H0: There is no association between the grass-seed types. Ha: There is an association between the grass-seed types.

A quality control specialist tests samples of the seed being packaged and uses a chi-square goodness-of-fit test to see whether the proportions in the samples match what is claimed by the company. Which of the following best describes the null hypothesis and the alternative hypothesis for the test? A) H0: p=0.20 Ha: p≠0.20 B) H0: … Read more

Jimmy believes that the shuffle feature on his music player is malfunctioning by not playing songs that meet this distribution of music types. To test this, he listens to 100 songs randomly chosen when his player is in shuffle mode and records the number of songs in each category. Which inference procedure should he use to test whether or not the shuffle feature is working correctly? A) A one-sample z-test for a population proportion B) A two-sample z-test for a difference between population proportions C) A chi-square goodness-of-fit test D) A chi-square test for homogeneity E) A matched pairs t-test for a mean difference

Jimmy believes that the shuffle feature on his music player is malfunctioning by not playing songs that meet this distribution of music types. To test this, he listens to 100 songs randomly chosen when his player is in shuffle mode and records the number of songs in each category. Which inference procedure should he use … Read more

Ms. Harper knows that her students in a computing course can choose from one of three operating systems for the semester: Doors, Banana, or Duix. Ms. Harper wants to test the hypothesis that her students will select the operating systems in the same proportion as students in other computing courses at the university. She conducts a χ2 goodness-of-fit test and calculates χ2=3.79 with a corresponding p-value of 0.15. Which of the following is correct at a 5-percent level of significance? A) Reject the null hypothesis, since 3.79>2. B) Fail to reject the null hypothesis, since 3.79>2. C) Reject the null hypothesis, since 0.15>0.05. D) Fail to reject the null hypothesis, since 0.15>0.05. E) Reject the null hypothesis, since 0.15<3.790.

Ms. Harper knows that her students in a computing course can choose from one of three operating systems for the semester: Doors, Banana, or Duix. Ms. Harper wants to test the hypothesis that her students will select the operating systems in the same proportion as students in other computing courses at the university. She conducts … Read more

A spinner made for a game of chance has 8 equally likely spaces. Alfonso records the result of a sample of 400 spins. Alfonso decides to calculate a chi-square test statistic for a goodness-of-fit test to see whether the spinner is fair. Which of the following is the appropriate null hypothesis? A) H0 =0.125,p2=0.125,p3=0.125,p4=0.125,p5=0.125,p6=0.125,p7=0.125,p8=0.12 B) H0:H0: At least one proportion is different. C) H0 =0.125 D) H0 ≠0.125,p2≠0.125,p3≠0.125,p4≠0.125,p5≠0.125,p6≠0.125,p7≠0.125,p8≠0.125 E) H0 =0.08,p2=0.08,p3=0.08,p4=0.08,p5=0.08,p6=0.08,p7=0.08,p8=0.08

A spinner made for a game of chance has 8 equally likely spaces. Alfonso records the result of a sample of 400 spins. Alfonso decides to calculate a chi-square test statistic for a goodness-of-fit test to see whether the spinner is fair. Which of the following is the appropriate null hypothesis? A) H0 =0.125,p2=0.125,p3=0.125,p4=0.125,p5=0.125,p6=0.125,p7=0.125,p8=0.12 B) … Read more

A certain type of legal proceeding has three possible outcomes: in favor of party A, in favor of party B, or not in favor of either party. The outcomes are expected to be 40 percent, 20 percent, and 40 percent, respectively. A random sample of 40 cases is selected from a certain judge to investigate whether the judge’s outcomes are consistent with the expected outcomes. A chi-square goodness-of-fit test is conducted, and the value of the chi-square test statistic is χ2=9.19 with a corresponding p-value of 0.01. Assuming the conditions for inference were met, which of the following is the correct interpretation of the p-value? A) There is a 1 percent chance that the company’s claim is correct. B) If the null hypothesis is true, there is a 1 percent chance that the company’s claim is correct. C) If the null hypothesis is true, there is a 1 percent chance of obtaining a chi-square value of 9.19. D) If the null hypothesis is true, there is a 1 percent chance of obtaining a chi-square value of at least 9.19. E) There is a 1 percent chance of obtaining a chi-square value of at least 9.19.

A certain type of legal proceeding has three possible outcomes: in favor of party A, in favor of party B, or not in favor of either party. The outcomes are expected to be 40 percent, 20 percent, and 40 percent, respectively. A random sample of 40 cases is selected from a certain judge to investigate … Read more

Each student also identified which extracurricular activity (out of a total of 5 extra curricular activities) they were involved in. The calculated chi-square test statistic was 7.53 with a corresponding p-value of 0.4807. Based on this p-value, which of the following is the correct decision for the appropriate hypothesis test at the α=0.05 significance level? A) Reject the null hypothesis. The test is statistically significant because a p-value of 0.4807 is greater than a significance level of 0.05. B) Reject the null hypothesis. The test is statistically significant because a p-value of 0.4807 is less than the test statistic of 7.53. C) Fail to reject the null hypothesis. The test is not statistically significant because a p-value of 0.4807 is greater than a significance level of 0.05. D) Fail to reject the null hypothesis. The test is not statistically significant because a p-value of 0.4807 is less than the test statistic of 7.53. E) Accept the null hypothesis. The test is not statistically significant because a p-value of 0.4807 is greater than a significance level of 0.05.

Each student also identified which extracurricular activity (out of a total of 5 extra curricular activities) they were involved in. The calculated chi-square test statistic was 7.53 with a corresponding p-value of 0.4807. Based on this p-value, which of the following is the correct decision for the appropriate hypothesis test at the α=0.05 significance level? … Read more

A regional highway uses 8 tollbooths that are open to all vehicles. A chi-square goodness-of-fit test using a significance level of α=0.05 was conducted to determine whether the tollbooths are all used in equal proportions. A chi-square value of χ2=19.1 was calculated with a corresponding p-value of 0.008. Which of the following is correct? A) There is insufficient evidence to suggest that the tollbooths are not used in equal proportions. B) There is insufficient evidence to suggest that the tollbooths are used in equal proportions. C) There is sufficient evidence to suggest that the tollbooths are not used in equal proportions. D) There is sufficient evidence to suggest that the tollbooths are used in equal proportions. E) The tollbooths are used in equal proportion.

A regional highway uses 8 tollbooths that are open to all vehicles. A chi-square goodness-of-fit test using a significance level of α=0.05 was conducted to determine whether the tollbooths are all used in equal proportions. A chi-square value of χ2=19.1 was calculated with a corresponding p-value of 0.008. Which of the following is correct? A) … Read more

A newspaper article indicated that 43 percent of cars with black seats are white, 46 percent of cars with black seats are blue, 7 percent of cars with black seats are red, and 4 percent of cars with black seats are black. A test was conducted to investigate whether the color of cars with black seats was consistent with the newspaper article. A random sample of cars of these colors was selected, and the value of the chi-square test statistic was χ2=8.2. Which of the following represents the p-value for the test? A) P(χ2≥8.2)=0.08 B) P(χ2≥8.2)=0.04 C) P(χ2≤8.2)=0.96 D) P(χ2=8.2)=0.00 E) The p-value cannot be calculated because the sample size is not given.

A newspaper article indicated that 43 percent of cars with black seats are white, 46 percent of cars with black seats are blue, 7 percent of cars with black seats are red, and 4 percent of cars with black seats are black. A test was conducted to investigate whether the color of cars with black … Read more

A large factory that builds machines has three shifts, one that starts at 4:00 A.M., one that starts at noon, and one that starts at 8:00 P.M. The manager of the factory wanted to know whether there is an association between an employee’s work experience (less than five years with the company, between five and twenty years with the company, over twenty years with the company) and the time of the employee’s shift. The manager selected a random sample of 125 employees and classified employees by their shift time and work experience. Which of the following is an appropriate pair of hypotheses for the manager to use? A) H0 : Work experience is independent of the time of an employee’s shift. Ha : Work experience is dependent on the time of an employee’s shift. B) H0 : Employees with over 20 years of experience are just as likely to start work at 8 P.M. Ha : Employees with over 20 years of experience are less likely to start work at 8 P.M. C) H0 : There is an association between work experience and an employee’s shift. Ha : There is no association between work experience and an employee’s shift. D) H0 : For each shift, there is no difference between the proportions of work experience level. Ha : For each shift, there is a difference between the proportions of work experience level. E) H0 : Among those in the sample, the proportion of employees for each work experience level did not differ by shift. Ha : Among those in the sample, the proportion of employees for each work experience level did differ by shift.

A large factory that builds machines has three shifts, one that starts at 4:00 A.M., one that starts at noon, and one that starts at 8:00 P.M. The manager of the factory wanted to know whether there is an association between an employee’s work experience (less than five years with the company, between five and … Read more

The logging company would like to use its sample to provide convincing statistical evidence that over 50 percent of the trees in the forest are spruce trees. The logging company has decided to use a chi-square goodness-of-fit test to justify its claim. Why is the chi-square goodness-of-fit test not an appropriate procedure for the logging company to use? A) A chi-square goodness-of-fit test would be used to show that the entire distribution of trees in the forest is different than what the forester reported, not necessarily the individual proportion representing the spruce trees. B) The logging company should find the average number of spruce trees using several samples and then construct a confidence interval for a difference in population means to show that there are more spruce trees in the forest than reported. C) The logging company does not need to complete an inference procedure; there are more than 50 percent spruce trees in the sample. D) In order to perform a chi-square test, the logging company needs expected counts, not percentages. The logging company should declare its current sample as expected values and then generate a new sample of observed values to compute the test statistic. E) The sample does not meet the minimum requirements needed for a chi-square goodness-of-fit test.

The logging company would like to use its sample to provide convincing statistical evidence that over 50 percent of the trees in the forest are spruce trees. The logging company has decided to use a chi-square goodness-of-fit test to justify its claim. Why is the chi-square goodness-of-fit test not an appropriate procedure for the logging … Read more