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- Researchers are trying to determine if there is a difference in the average lifespan of two species of fruit fly: Ceratitis capitata and Drosophila melanogaster. They measure the lifespans of two random samples of each species. The SRS of Ceratitis capitata was of size 105 and had an average lifespan of 51 days with a standard deviation of 7 days. The SRS of Drosophila melanogaster was of size 95 and had an average lifespan of 48 days with a standard deviation of 6 days. let μ1 denote the mean lifespan of Ceratitis capitata and μ2 denote the mean lifespan of Drosophila melanogaster. Compute the standard error of the statistic xˉ1−xˉ2x. Round your answer to four decimal places.Researchers are trying to determine if there is a difference in the average lifespan of two species of fruit fly: Ceratitis capitata and Drosophila melanogaster. They measure the lifespans of two random samples of each species. The SRS of Ceratitis capitata was of size 105 and had an average lifespan of 51 days with a standard deviation of 7 days. The SRS of Drosophila melanogaster was of size 95 and had an average lifespan of 48 days with a standard deviation of 6 days. Let μ1denote the mean lifespan of Ceratitis capitata and μ2 denote the mean lifespan of Drosophila melanogaster. The researchers will conduct a test of significance at level α=0.01 to test if these two species have different average lifespans. The PP-value for this test of significance isResearchers are trying to determine if there is a difference in the average lifespan of two species of fruit fly: Ceratitis capitata and Drosophila melanogaster. They measure the lifespans of two random samples of each species. The SRS of Ceratitis capitata was of size 105 and had an average lifespan of 51 days with a standard deviation of 7 days. The SRS of Drosophila melanogaster was of size 95 and had an average lifespan of 48 days with a standard deviation of 6 days. Let μ1 denote the mean lifespan of Ceratitis capitata and μ2 denote the mean lifespan of Drosophila melanogaster. Provide an estimate for the difference in the mean lifetimes of these two species.
- Researchers are trying to determine if there is a difference in the average lifespan of two species of fruit fly: Ceratitis capitata and Drosophila melanogaster. They measure the lifespans of two random samples of each species. The SRS of Ceratitis capitata was of size 105 and had an average lifespan of 51 days with a standard deviation of 7 days. The SRS of Drosophila melanogaster was of size 95 and had an average lifespan of 48 days with a standard deviation of 6 days.A comparison is made between two bus lines to determine if arrival times of their regular buses from Denver to Durango are off schedule by the same amount of time. For 51 randomly selected runs, bus line A was observed to be off schedule an average time of 53 minutes, with standard deviation 15 minutes. For 60 randomly selected runs, bus line B was observed to be off schedule an average of 60 minutes, with standard deviation 11 minutes. Do the data indicate a significant difference in average off-schedule times? Use a 5% level of significance. What are we testing in this problem? difference of proportionsdifference of means single meanpaired differencesingle proportion (a) What is the level of significance?State the null and alternate hypotheses. H0: ?1 = ?2; H1: ?1 < ?2H0: ?1 = ?2; H1: ?1 ≠ ?2 H0: ?1 > ?2; H1: ?1 = ?2H0: ?1 = ?2; H1: ?1 > ?2 (b) What sampling distribution will you use? What assumptions are you making? The Student's t. We assume that both population…A comparison is made between two bus lines to determine if arrival times of their regular buses from Denver to Durango are off schedule by the same amount of time. For 51 randomly selected runs, bus line A was observed to be off schedule an average time of 53 minutes, with standard deviation 19 minutes. For 61 randomly selected runs, bus line B was observed to be off schedule an average of 61 minutes, with standard deviation 17 minutes. Do the data indicate a significant difference in average off-schedule times? Use a 5% level of significance. 1) What is the value of the sample test statistic? (Test the difference ?1 − ?2. Round your answer to three decimal places.) Estimate the P-value. 2) P-value > 0.500 0.250 < P-value < 0.500 0.100 < P-value < 0.250 0.050 < P-value < 0.100 0.010 < P-value < 0.050 P-value < 0.010 3) Sketch the sampling distribution and show the area corresponding to the P-value.
- A comparison is made between two bus lines to determine if arrival times of their regular buses from Denver to Durango are off schedule by the same amount of time. For 51 randomly selected runs, bus line A was observed to be off schedule an average time of 53 minutes, with standard deviation 17 minutes. For 60 randomly selected runs, bus line B was observed to be off schedule an average of 62 minutes, with standard deviation 11 minutes. Do the data indicate a significant difference in average off-schedule times? Use a 5% level of significance.What are we testing in this problem? single mean or paired difference or difference of means or difference of proportions or a ingle proportion What is the level of significance?State the null and alternate hypotheses. H0: ?1 ≠ ?2; H1: ?1 = ?2 H0: ?1 = ?2; H1: ?1 ≠ ?2 H0: ?1 ≤ ?2; H1: ?1 > ?2 H0: ?1 ≥ ?2; H1: ?1 < ?2 What sampling distribution will you use? What assumptions are you making? a.The standard normal. We assume that both…A former patient, who had both wisdom teeth extracted by two different dentists, claims that the average operating time (in minutes) performed by Dr. Reyes on all his patients is greater than the average operating time (in minutes) performed by Dr. Tupaz on all her patients. A sample of 102 patients attended to by Dr. Reyes had a mean operating time of 95.32 minutes with a population standard deviation of 29.27 minutes. On the other hand, 145 patients who went to the clinic of Dr. Tupaz gave a mean operating time of 84.11 minutes with a population standard deviation of 28.06 minutes. Assuming that both distributions are normal, is there sufficient evidence to support the former patient’s claim at alpha = 0.10? Use the p-value method.An automotive corporation claimed that their sedan brand of car can travel an average mean of 35 km in a liter of gasoline and if the standard deviation of 2 kilometers. The consumers made a test using 64 cars of the same brand and it had resulted to 32 kilometers for a liter of gasoline. Which claim would be accepted at 5% level of significance at 1 tailed test?
- The management at The Bank of the United States claims that the mean waiting time for all customers at its branches is less than that at the Public Bank, which is its main competitor. A business consulting firm took a sample of 100 customers from the The Bank of the United States and found that they waited an average of 4.5 minutes before being served. Another sample of 200 customers taken from the Public Bank showed that these customers waited an average of 4.75 minutes before being served. Assume that the standard deviations for the two populations are 1.2 and 1.5 minutes, respectively. Suppose you must test the hypothesis at 5% significance level whether the claim of the management of the The Bank of the United States is true. 2.DETERMINE I) the alternative hypothesis to be tested: II) the value of test statistic III) the p-value for this problem and the conclusionThe credit manager of a large department store claims that the mean balance for the store’s charge account customers is R3,280. An independent auditor selects a random sample of 18 accounts and finds a mean balance of R4,090.64 and a standard deviation of R1,470. If the manager’s claim is not supported by these data, the auditor intends to examine all charge account balances. If the population of account balances is assumed to be approximately normally distributed, what action should the auditor take? (Remember to use the 4-step process.)he manager of a large resort hotel, has been receiving complaints from some guests that they are not being provided with prompt service upon approaching the front desk. In particular, he is concerned that the desk staff might be providing female guests with less prompt service than their male counterparts. For a sample of 35 male guests, he found thatit takes an average of 15.2 seconds, with a standard deviations 5.9 seconds for them to be greeted after their arrival at the front desk. For a sample of 37 female guests, the mean and standard deviation are 17.4 seconds and 6.4 seconds, respectively. Assuming the two population are independent and the population variances are equal, hypotheses examine whether the population mean time for serving female guests might actually be greater than that of serving male-guests. Suppose that the pooled variance is 37.9729. The test statistic is