a simple random sample of size n = 49 is obtained from a population ha is skewed right wih = 83 and = 7
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- The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,092 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 189 of their customers and calculates that these customers used an average of 10,628 kWh of electricity last year. Assuming that the population standard deviation is 2970 kWh, is there sufficient evidence to support the power company's claim at the 0.02 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision. Answer O Tables E Keypad Keyboard Shortcuts We reject the null hypothesis and conclude that there is sufficient evidChce at a 0.02 level of significance to support the power company's claim that the mean amount of electricity for their residents is more than the national average. We fail to reject the null hypothesis and…Shown below is a randomization dotplot for sample differences in student pulse rates during a quiz versus listening to a lecture. Use the randomization dotplot to answer the questions that follow: The randomization dotplot is based on 5000 simulated samples under the null hypothesis assumption that there is no difference in mean pulse rates during a quiz versus during a lecture. The alternative hypothesis is that the difference is greater than zero (or that the mean pulse rate is higher during a quiz). If we assume the null hypothesis is true, of the choices below, which of these sample differences would be the least likely to occur (or, to say it another way, which would give the strongest evidence in favor of the alternative hypothesis?…Magnetic surveying is one technique used by archaeologists to determine anomalies arising from variations in magnetic susceptibility. Unusual changes in magnetic susceptibility might (or might not) indicate an important archaeological discovery. Let x be a random variable that represents a magnetic susceptibility (MS) reading for a randomly chosen site at an archaeological research location. A random sample of 120 sites gave the readings shown in the table below. Magnetic Susceptibility Readings,centimeter-gram-second ✕ 10−6 (cmg ✕ 10−6) Comment MagneticSusceptibility Number ofReadings EstimatedProbability "cool" 0 ≤ x < 10 30 30/120 = 0.25 "neutral" 10 ≤ x < 20 54 54/120 = 0.45 "warm" 20 ≤ x < 30 24 24/120 = 0.20 "very interesting" 30 ≤ x < 40 6 6/120 = 0.05 "hot spot" 40 ≤ x 6 6/120 = 0.05 Suppose you are working in a "warm" region in which all MS readings are 20 or higher. In this same region, what is the probability that you will find a…
- One cable company claims that it has excellent customer service. In fact, the company advertises that a technician will arrive within 55 minutes after a service call is placed. One frustrated customer believes this is not accurate, claiming that it takes over 55 minutes for the cable technician to arrive. The customer asks a simple random sample of 4 other cable customers how long it has taken for the cable technician to arrive when they have called for one. The sample mean for this group is 65.7 minutes with a standard deviation of 5.9 minutes. Assume that the population distribution is approximately normal. Test the customer's claim at the 0.10 level of significance. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.One cable company claims that it has excellent customer service. In fact, the company advertises that a technician will arrive within 35 minutes after a service call is placed. One frustrated customer believes this is not accurate, claiming that it takes over 35 minutes for the cable technician to arrive. The customer asks a simple random sample of 4 other cable customers how long it has taken for the cable technician to arrive when they have called for one. The sample mean for this group is 40.2 minutes with a standard deviation of 5.5 minutes. Assume that the population distribution is approximately normal. Test the customer's claim at the 0.005 level of significance. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho Ha = 35 35When engaging in weight control (fitness/fat burning) types of exercise, a person is expected to attain about 60% of their maximum heart rate. For 20-year-olds, this rate is approximately 120 bpm. A simple random sample of 100 20-
- 2. A partially completed ANOVA table for a completely randomized design is shown here: Source SS df | MS F Treatments 18.4 6 Error Total 45.2 41 a) Complete the ANOVA table. b) How many treatments are involved in the experiment? c) Do the data provide sufficient evidence to indicate a difference among the population means? Test, using a = 0.10. d) Find the approximate p-value for the test in part c).One cable company claims that it has excellent customer service. In fact, the company advertises that a technician will arrive within 60 minutes after a service call is placed. One frustrated customer believes this is not accurate, claiming that it takes over 60 minutes for the cable technician to arrive. The customer asks a simple random sample of 25 other cable customers how long it has taken for the cable technician to arrive when they have called for one. The sample mean for this group is 63.4 minutes with a standard deviation of 15.1 minutes. Assume that the population distribution is approximately normal. Test the customer's claim at the 0.025 level of significance. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places. Answer 囲 Tables E Keypad Keyboard ShortcutsThe U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,951 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIIA's reported average. To test their claim, the company chooses a random sample of 184 of their customers and calculates that these customers used an average of 11,162 kWh of electricity last year. Assuming that the population standard deviation is 1263 kWh, is there sufficient evidence to support the power company's claim at the 0.05 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision. Answer E Tables 国 Keypad Keyboard Shortcuts We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.05 level of significance to support the power company's claim that the mean amount of electricity for their residents is more than the national average. We fail to reject the null…
- Magnetic surveying is one technique used by archaeologists to determine anomalies arising from variations in magnetic susceptibility. Unusual changes in magnetic susceptibility might (or might not) indicate an important archaeological discovery. Let x be a random variable that represents a magnetic susceptibility (MS) reading for a randomly chosen site at an archaeological research location. A random sample of 120 sites gave the readings shown in the table below. Magnetic Susceptibility Readings,centimeter-gram-second ✕ 10−6 (cmg ✕ 10−6) Comment MagneticSusceptibility Number ofReadings EstimatedProbability "cool" 0 ≤ x < 10 36 36/120 = 0.30 "neutral" 10 ≤ x < 20 48 48/120 = 0.40 "warm" 20 ≤ x < 30 18 18/120 = 0.15 "very interesting" 30 ≤ x < 40 12 12/120 = 0.10 "hot spot" 40 ≤ x 6 6/120 = 0.05 Suppose you are working in a 'warm' region in which all MS readings are 20 or higher. In this same region, what is the probability that you will find…Annie is concerned over a report that "a woman over age 4040 has a better chance of being killed by a terrorist than of getting married." A study found that the likelihood of marriage for a never-previously-wed, 4040-year-old university-educated American woman was 2.9%2.9%. To demonstrate that this percentage is too small, Annie uses her resources at the Baltimore Sun to conduct a simple random sample of 516516 never-previously-wed, university-educated, American women who were single at the beginning of their 4040s and who are now 4545. Of these women, 2121 report now being married. Does this evidence support Annie’s claim, at the 0.100.10 level of significance, that the chances of getting married for this group is greater than 2.9%2.9%? Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.A simple random sample of size n = 200 drivers with a valid driver's license is asked if they drive an American-made automobile. Of the 200 drivers surveyed, 123 responded that they drive an American-made automobile. Determine if a majority of those with a valid driver's license drive an American-made automobile at the x = 0.05 level of significance. A. Hypothesis test on a population proportion OB. Hypothesis test on a population mean OC. Hypothesis test on a population standard deviation Determine the null and alternative hypotheses. Ho ▼ ▼ H₁ (Type integers or decimals. Do not round.) Clear all Check answer Help me solve this # 5:14 PM 6/18/2022 Type here to search 3 4 View an example R 5 G O B Get more help. E H N 8 1 M K P 96°F Mostly sunny
