For each of the following rejection regions, sketch the sampling distribution for z and indicate the location of the rejection reg Type I error will be made n, and determine the probability that a 2 *> 2.575 2* - 2.33 d. z x 1.28 b. zx 1.96 6. 2 2.33 a. Which sketch below shows the sampling distribution for the rejection region z 2.575? O A O B O c. 2.573 Reject Ho Reject 14." -2.575 Reject le A survey found that 55% of college professors believe that their online education courses are as good as or superior to courses that use traditional face-to-face instruction. a. Give the null hypothesis for testing the claim made by the survey. b. Give the rejection region for a two-tailed test conducted at a = 0.10. a. What is the null hypothesis? Ho: P (Type an integer or a decimal.) A biologist and a zoologist at a university are testing the effectiveness of a high-tech handheld device designed to instantly identify the DNA of an animal species. They the DNA-reading device on tissue samples collected from mollusks with brightly colored shells. The scientists discover that the error rate of this device is less than 4.1 percent. Set up the null and alternative hypotheses as if you want to support the findings. Complete the null and alternative hypotheses. (Type integers or decimals, Do not round.) Ho: Ha: A certain national lunch program mandates that for a high school to receive reimbursement for school lunches, the number of calories served at lunch must be no more than 940 calories. Suppose a nutritionist believes that the true mean number of calories served at lunch at all schools in the country is not 940 calories. e parameter of interest. potheses for testing this claim. b Specify the null and alternative hypotheses for featin message a Type I error in the words of the problem. d. Describe a Type II error in the words of the problem a. The parameter of interest is According to a university wildlife ecology and conservation researcher, the average level of mercury uptake in wading birds in a certain region has increased over the past several years. Ten years ago, the average level was 16.6 parts per million. Complete parts a through c a. Give the null and alternative hypotheses for testing whether the average level today is greater than 16.6 ppm. (Type integers or decimals. Do not round.) Ho: HVAn intrusion detection system (IDS) is designed to provide an alarm whenever unauthorized access to a computer system occurs. The probability of the system giving a false alarm is defined by the symbol a, while the probability of a missed detection is defined by the symbol B. These symbols are used to represent Type I and Type II r rates, respectively, in a hypothesis testing soona a. What is the null hypothesis, Ho? b. What is the alternative hypothesis, H,? c. According to a college research laboratory, only 19 in 1,900 computer sessions with no intrusions resulted in a false alarm. For the same system, the laboratory found that only 700 of 2,800 intrusions were actually detected. Use this information to estimate the values of a and p. a. The null hypothesis is Ho: There is an intrusion There is no intrusion If a hypothesis test were conducted using a = 0.05, for which of the following p-values would the null hypothesis be rejected? a. 0.06 b. 0.036 a. What is the conclusion for a p-value of 0.06? O A. Reject the null hypothesis since the p-value is less than the value of a. O B. Do not reject the null hypothesis since the p-value is less than the value of a. O C. Do not reject the null hypothesis since the p-value is not less than the value of o. O D. Reject the null hypothesis since the p-value is not less than the value of a. For the a and observed significance level (p-value) pair, indicate whether the null hypothesis would be rejected. o = 0.05, p-value =0.10 Choose the correct conclusion below. O A. Do not reject the null hypothesis since the p-value is not less than the value of a. O B. Reject the null hypothesis since the p-value is less than the value of a. O C. Do not reject the null hypothesis since the p-value is less than the value of a. O D. Reject the null hypothesis since the p-value is not less than the value of o. A random sample of 100 observations from a population with standard deviation 82 yielded a sample mean of 115. Complete parts a through c below. a. Test the null hypothesis that u = 100 against the alternative hypothesis that u > 100, using a = 0.05. Interpret the results of the test. What is the value of the test statistic? z= (Round to two decimal places as needed.) Consider a test of Ho: u = 75 performed with the computer. The software reports a two-tailed p-value of 0.1032. Make the appropriate conclusion for each of the following situations. a. Ha: H 75, z = 1.63, of = 0.07 d. Ha: H # 75, z= - 1.63, a = 0.01 a. Choose the correct answer below. There is sufficient evidence to reject Ho There is insufficient evidence to reject Ho.In a test of Ho: H = 80 against Ha: H > 80, the sample data yielded the test statistic z = 2.26. Find and interpret the p-value for the test. Click the icon to view a table of standard normal values The p-value is p =]. (Round to three decimal places as needed.) .00 .01 .02 03 .04 -0000 .05 .0040 -06 -0080 .0438 -0478 0120 .08 .0160 -0199 09 0596 -0239 -0871 0279 -1179 .0832 -0319 0359 -1255 -1293 -1331 0987 -0636 -1026 0675 :1915 .1950 1628 -1985 -1654 -1368 2010 1700 -1736 -1406 -2257 -2054 2321 2058 2580 -2357 2380 .2123 2157 -2642 -2422 2454 -2190 2224 3159 -2734 -2764 2517 2540 3413 7264 3289 3051 2794 2823 3643 -3461 3485 3508 -3531 3554 3365 -3686 -3517 3509 -3858 3621 -39017 3749 -4060 -3925 -3944 -37 70 3790 -398 -3810 3830 332 4343 4272 -4236 4251 4265 4131 4279 4147 -3907 4357 4370 4382 -4394 4292 4306 -4177 4463 4406 -4484 -4418 4495 4429 -4441 4582 -4505 -4599 -4525 4535 -4616 -4545 tobub hawk bobb WANE bbbyb WINE bbbub WAwNib -4732 .4738 -4678 -4744 4750 -4693 -4633 4793 -4706 4821 4830 -4708 4803 4761 4808 4767 -4834 4838 4812 .4817 4868 4871 -4846 4918 4881 4854 185T -4906 4884 4938 -4922 4927 4909 -4911 890 4953 -4943 4979 4946 4931 -4932 4956 4948 40 19 496 495 4951 -4936 974 4967 4959 4952 -4968 4960 -4976 -4969 4961 -4970 4962 -4971 4963 4987 4973 4987 4982 -4977 -4978 -4972 -4987 498S 40903 40906 4988 4989 4985 49934 49910 4989 49936 49913 49916 -4989 4990 49952 49918 49953 49955 409 38 49924 49066 49068 49057 499-40 49942 -49960 49944 49061 49946 49977 49969 49958 49078 40970 .49962 49978 49971 49064 40970 40972 40980 49073 49981 49974 -40965 49984 49075 40080 49085 49985 49081 49456 49982 49976 490R3 40900 40986 49057 40043 40991 49087 .49088 49038 49995 49002 49095 49996 49994 -40902 40996 49906 49096 49096 -49996 49097 19097 A study analyzed the sustainability behaviors of CPA corporations. The level of support for corporate sustainability (measured on a quantitative scale ranging from 0 to 160 points) was obtained for each in a sample of 973 senior managers at CPA firms. The CEO of a CPA firm claims that the true mean level of support for sustainability is 68. Complete parts a through e below. i Click the icon to view a printout of the analysis. a. Specify the null and alternative hypotheses for testing this claim. O A. Ho: H=68 Ha: H 68 OD. Ho: H #68 Hai H = 68