If, in a (two-tail) hypothesis test, the p-value is 0.1116, what is your statistical decision if you test the null hypothesis at the 0.06 level of significance? . . . Choose the correct answer below? O A. Since the p-value is less than a, do not reject Ho. O B. Since the p-value is greater than a, do not reject Ho. O C. Since the p-value is less than a, reject Ho. O D. Since the p-value is greater than a, reject Ho.If a hypothesis test were conducted using a = 0.10, for which of the following p-values would the null hypothesis be rejected? a. 0.002 b. 0.11 E) a. What is the conclusion for a p-value of 0.002? O A. Reject the null hypothesis since the p-value is less than the value of 0t. 0 B. Do not reject the null hypothesis since the p-value is not less than the value of 0L. 0 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 a. b. What is the conclusion for a p-value of 0.1 1? O A. Do not reject the null hypothesis since the p-value is not less than the value of a. O B. Do not reject the null hypothesis since the p-value is less than the value of 0:. O C. Reject the null hypothesis since the p-value is less than the value of 0t. 0 D. Reject the null hypothesis since the p-value is not less than the value of a. The manager of a paint supply store wants to determine whether the mean amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer is actually 1 gallon. You know from the manufacturer's specications that the standard deviation of the amount of paint is 0.009 gallon. You select a random sample of 55 cans, and the mean amount of paint per 1-gallon can is 0.995 gallon. Complete parts (a) through (e) below. Click here to view p_age 1 of the cumulative standardized normal distribution table. Click here to view [gs 2 of the cumulative standardized normal distribution table. a. Is there evidence that the mean amount is different from 1.0 gallon? (Use a = 0.05.) Let |.1 be the population mean. Determine the null hypothesis, H0, and the alternative hypothesis, H1. H0:p. 1.0 H1:p. 1.0 Cumulative probabilities for negative z-scores Entry represents area under the cumulative standardized normal distribution from - co to Z. Cumulative Probabilities 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 -6.0 0.000000001 -5.5 0.000000019 -5.0 0.000000287 -4.5 0.000003398 -4.0 0.000031671 -3.9 0.00005 0.00005 0.00004 0.00004 0.00004 0.00004 0.00004 0.00004 0.00003 0.00003 -3.8 0.00007 0.00007 0.00007 0.00006 0.00006 0.00006 0.00006 0.00005 0.00005 0.00005 -3.7 0.00011 0.00010 0.00010 0.00010 0.00009 0.00009 0.00008 0.00008 0.00008 0.00008 -3.6 0.00016 0.00015 0.00015 0.00014 0.00014 0.00013 0.00013 0.00012 0.00012 0.00011 -3.5 0.00023 0.00022 0.00022 0.00021 0.00020 0.00019 0.00019 0.00018 0.00017 0.00017 -3.4 0.00034 0.00032 0.00031 0.00030 0.00029 0.00028 0.00027 0.00026 0.00025 0.00024 -3.3 0.00048 0.00047 0.00045 0.00043 0.00042 0.00040 0.00039 0.00038 0.00036 0.00035 -3.2 0.00069 0.00066 0.00064 0.00062 0.00060 0.00058 0.00056 0.00054 0.00052 0.00050 -3.1 0.00097 0.00094 0.00090 0.00087 0.00084 0.00082 0.00079 0.00076 0.00074 0.00071 -3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00103 0.00100 -2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 -2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 -2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026 -2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 -2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048 -2.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 -2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084 -2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110 -2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143 -2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 -1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 -1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294\fCumulative probabilities for positive z-scores Entry represents area under the cumulative standardized normal distribution from - co to Z. \fAt a recent meeting, the manager of a national call center for a major Internet bank made the statement that the average past-due amount for customers who have been called previously about their bills is now no larger than $33.00. Other bank managers at the meeting suggested that this statement may be in error and that it might be worthwhile to conduct a test to see if there is statistical support for the call center manager's statement. The accompanying table contains data for a random sample of 67 customers from the call center population. Assuming that the population standard deviation for past due amounts is known to be $58.00, what should be concluded based on the sample data? Test using or = 0.05. E Click the icon to view the data table. State the appropriate null and alternative hypotheses. Choose the correct answer below. A. H0:p>33.00 B. H0:p233.00 C. H0:p=33.00 HA: p $33.00 HA: p 33.00 HA: p = 33.00 HA: p 233.00 Calculate the test statistic. El = D (Round to two decimal places as needed.) The director of a state agency believes that the average starting salary for clerical employees in the state is less than $27,000 per year. To test her hypothesis, she has collected a simple random sample of 100 starting clerical salaries from across the state and found that the sample mean is $26,750. a. State the appropriate null and alternative hypotheses. b. Assuming the population standard deviation is known to be $3,000 and the signicance level for the test is to be 0.025, what is the critical value (stated in dollars)? 0. Referring to your answer in part b, what conclusion should be reached with respect to the null hypothesis? d. Referring to your answer in part c, which of the two statistical errors might have been made in this case? Explain. E) a. State the appropriate null and alternative hypotheses. Choose the correct answer below. 0 A. Ho: p=26,750 0 B. H0:0220,750 o c. H02u527,000 H A: p at 26,750 HA: 0 27,000 0 D. H0: 0227.000 0 E. Ho:u=27,000 O F. H0:u526,750 HA: 0 25,750 b. What conclusion should be reached based on your answer to part b? 0 Do not reject the null hypothesis. 0 Reject the null hypothesis. c. Which of the two statistical errors might have been made in this case? Explain. O A. A Type II error might have been made because the null hypothesis was not rejected. O B. A Type I error might have been made because the null hypothesis was rejected. O C. A Type I error might have been made because the null hypothesis was not rejected. O D. A Type ll error might have been made because the null hypothesis was rejected.The quality-control manager at a compact fluorescent light bulb (CFL) factory needs to determine whether the mean life of a large shipment of CFLs is equal to 7,544 hours. The population standard deviation is 900 hours. A random sample of 81 light bulbs indicates a sample mean life of 7,294 hours. a. At the 0.05 level of significance, is there evidence that the mean life is different from 7,544 hours? b. Compute the p-value and interpret its meaning. c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs. d. Compare the results of (a) and (c). What conclusions do you reach? . . . a. Let u be the population mean. Determine the null hypothesis, Ho, and the alternative hypothesis, H,. HO: H= What is the test statistic? Test Statisitc = (Round to two decimal places as needed.) What is/are the critical value(s)? (Round to two decimal places as needed. Use a comma to separate answers as needed.)What is the final conclusion? O A. Fail to reject Ho. There is sufficient evidence to prove that the mean life is different from 7,544 hours. O B. Reject Ho. There is not sufficient evidence to prove that the mean life is different from 7,544 hours. O C. Reject Ho. There is sufficient evidence to prove that the mean life is different from 7,544 hours. O D. Fail to reject Ho. There is not sufficient evidence to prove that the mean life is different from 7,544 hours. b. What is the p-value? (Round to three decimal places as needed.) Interpret the meaning of the p-value. Choose the correct answer below. O A. Fail to reject Ho. There is sufficient evidence to prove that the mean life is different from 7,544 hours. O B. Reject Ho. There is not sufficient evidence to prove that the mean life is different from 7,544 hours. O C. Fail to reject Ho. There is not sufficient evidence to prove that the mean life is different from 7,544 hours. O D. Reject Ho. There is sufficient evidence to prove that the mean life is different from 7,544 hours.c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs. Sus (Round to one decimal place as needed.) d. Compare the results of (a) and (c). What conclusions do you reach? O A. The results of (a) and (c) are the same: there is sufficient evidence to prove that the mean life is different from 7,544 hours. O B. The results of (a) and (c) are not the same: there is sufficient evidence to prove that the mean life is different from 7,544 hours. O C. The results of (a) and (c) are not the same: there is not sufficient evidence to prove that the mean life is different from 7,544 hours. O D. The results of (a) and (c) are the same: there is not sufficient evidence to prove that the mean life is different from 7,544 hours