122 Section 13.7: Problem 7 (10 points) Consider the following sequence of 1 and 6: 611166111661161166611611116661166 With a 0.01 significance level, we wish to test the claim that the above sequence was produced in a random manner. Answer each of the following questions: (a) The null hypothesis H0 is given by A. G=0 B. ρ=0 C. β=0 D. r=0 E. The data are in an order that is not random F. The data are in a random order G. n1=n2 H. Median=0 I. None of the above. The null hypothesis H1 is given by A. r!=0 B. G!=0 C. The data are in a random order D. Median!=0 E. n1!=n2 F. ρ!=0 G. β!=0 H. The data are in an order that is not random (c) The number of runs G is A. 21 B. 27 C. 15 D. 30 E. 17 F. 19 G. None of the above. (d) What kind of test should you conduct A. Runs test for randomness and the test statistic is a z-score B. Both sign and goodness of fit tests C. Independence Test D. Sign test E. Runs test for randomness and the test statistic is G F. Goodness of fit test G. None of the above. (e) What is/are the critical value(s) A. The only critical value is 11 B. The only critical value is -2.575 C. The smallest one is -1.96 and the largest one is 1.96 D. The only critical value is 23 E. The only critical value is 24 F. The negative one is -1.645 and the positive one is 1.645 G. The smallest one is -2.575 and the largest one is 2.575 H. The smallest one is 11 and the largest one is 24 None of the above. (f) The test statistic is A. z=1.56 with n1=18 and n2=15. B. z=1.56 with n1=15 and n2=18. C. G=15 with n1=15 and n2=18. D. z=-.84 with μ(G)=17.36 and σ(G)=2.8. E. z=1.49 with μ(G)=19.37 and σ(G)=4.80. F. G=15 with n1=18 and n2=15. G. z=1.65 with n1=15 and n2=18. H. z=-.49 with μ(G)=17.36 and σ(G)=4.80. I. z=-.49 with μ(G)=19.37 and σ(G)=2.8. J. None of the above. (g) The conclusion: A. We fail to reject H0 and then there is enough evidence to reject the claim that the above sequence was in a random order B. We reject H0 and then there is enough evidence to reject the claim that the above sequence was in a random order C. We reject H0 and then there is enough evidence to support the claim that the above sequence was in a random order D. We fail to reject H0 and then there isn't enough evidence to reject the claim that the above sequence was in a random order. E. We reject H0 and then there isn't enough evidence to support the claim that the above sequence was in a random order F. We reject H0 and then there isn't enough evidence to reject the claim that the above sequence was in a random order G. We fail to reject H0 and then there isn't enough evidence to support the claim that the above sequence was in a random order H. We fail to reject H0 and then there is enough evidence to support the claim that the above sequence was in a random order None of the above.
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In this case, the null hypothesis is that the data are in an order that is not random. Therefore, the correct answer is E. The data are in an order that is not random. Show more…
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The results were analyzed in R via a 2-sample t-test with a 2-sided alternative hypothesis, and some of the output is summarized below (assume ̑ = 0.05 where necessary). mean sd n Alumania 3.67 3.34 35 Beautinium 4.01 3.74 35 Two Sample t-test data: impurity by supplier t = -0.413, df = 68, p-value = 0.6895 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -2.03 1.35 Levene's Test for Homogeneity of Variance (center = "median") Df F value Pr(>F) group 1 0.1977 0.658 68 Let ̑1 and ̑2 denote the mean amount of impurities from the suppliers “Alumania‐ and “Beautinium" respectively, and ̑1^2 and ̑2^2 be the respective variances. If you were asked to use this analysis to answer the question of which supplier of aluminium to use, what is your conclusion? a. Beautinium has a higher sample mean of impurities, so we should use Beautinium. b. p = 0.6895> 0.05 so we cannot reject H0, ie there is no significant evidence that there is any difference in the amount of impurities produced by the two suppliers. c. Aluminia has a lower sample mean of impurities, so we should use Aluminia. d. p = 0.658 > 0.05 so we cannot reject H0, ie there is no significant evidence that there is any difference in the amount of impurities produced by the two suppliers.
Kari H.
a) It is reported that scores on a national accounting certification exam are approximately normally distributed with a mean of 78. Professor Rose believes that students from her three sections of her preparatory course will score significantly higher than the national average on this test. After they write the exam, her students have scored an average of 84. After conducting the appropriate statistical test, Professor Rose finds that the p-value is 0.0022 Which of the following is the best interpretation of the p-value? A p-value of 0.0022 indicates that there is a very small chance that Professor Rose's class outperformed the other exam writers across the nation. A p-value of 0.0022 provides strong evidence that Professor Rose's class outperformed the other exam writers across the nation. A p-value of 0.0022 provides evidence that Professor Rose is an exceptional teacher who was able to prepare her students well for this exam. None of the above are reasonable interpretations of this result. b) Food inspectors inspect samples of Alberta beef to see if they are safe. This can be thought of as a hypothesis test where H0: the beef is safe H1: the beef is not safe. At one plant, the inspector concludes the beef is safe when it actually is not safe. Which of the following statements is correct? The inspector has made a Type I error. The inspector has made a Type II error. It is impossible to know which of the two types of errors has been made without knowing the level of significance. It is impossible to know which of the two types of errors has been made without knowing the p-value. c) Prairie Craft Brewery has an automated filling machine that is set to fill each bottle with 500 millilitres of the its new Redhead Rye Ale. They have just completed the testing on a sample of 180 bottles to to see if the machine is filling the bottles with an average of 500 millilitres of ale. The tester has obtained a mean of 510 ml and p-value of 0.05 for a right-tail test. Which definition of the p-value is the most accurate? It is the probability that the observed mean of 510 ml will occur again. It is the probability that the null hypothesis is true. It is the probability of observing a mean of 510 ml or more under the assumption that the null hypothesis is true. It is the value that any sample mean must reach in order to be considered significant under the null hypothesis. d) An entrepreneurial scientist has invented a supplement that she believes can enhance short-term memory. She conducts an experiment to test the effect of the supplement on 23 volunteers, 8 of whom are given a placebo and 15 of whom are given the supplement. In analyzing in her study, she obtains a p-value of 0.24. Which of the following is a reasonable interpretation of her results? She should reject the null hypothesis. She should fail to reject the null hypothesis. This proves that at least 24% of the time, the experimental treatment has no effect on memory. This proves that at most 24% of the time, the experimental treatment has no effect on memory. e) A human resources study claims that the average age of Canadians who retire is 66 years. You believe that the average age is less than that. You take a random sample of 15 Canadians who have retired, and find their average age to be 65.4 years, with a standard deviation of 0.80 years. Using the resulting p-value of 0.0017, you conclude that this is significantly lower than the age of 66 stated in the study. What would be the most appropriate interpretation of this result? An error must have been made. This difference is too small to be statistically significant. Without knowing the level of significance, it is impossible to know if the results are statistically significant. Although the result is statistically significant, the difference in age is of little practical importance. The statistically significant result indicates that the majority of people who retire are younger than 66.
Jon S.
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Sri K.
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