Draw a random sample without replacement of size 7 from a popülation of size 836.
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Q: The U.S. Energy Information Administration claimed that U.S. residential customers used an average…
A: Given,sample size(n)=153sample mean(x¯)=10,586standard deviation(σ)=2509α=0.05
Q: The U.S. Energy Information Administration claimed that U.S. residential customers used an average…
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Q: The U.S. Energy Information Administration claimed that U.S. residential customers used an average…
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Q: The U.S. Energy Information Administration claimed that U.S. residential customers used an average…
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Q: The U.S. Energy Information Administration claimed that U.S. residential customers used an average…
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Q: The U.S. Energy Information Administration claimed that U.S. residential customers used an average…
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- The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,608 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 187 of their customers and calculates that these customers used an average of 10,737kWh of electricity last year. Assuming that the population standard deviation is 1220kWh, is there sufficient evidence to support the power company's claim at the 0.05 level of significance? Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,476 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 153 of their customers and calculates that these customers used an average of 10,767 kWh of electricity last year. Assuming that the population standard deviation is 2478 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.The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,069 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 171 of their customers and calculates that these customers used an average of 10,461 kWh of electricity last year. Assuming that the population standard deviation is 2973 kWh, is there sufficient evidence to support the power company's claim at the 0.01 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision.
- The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,941 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 115 of their customers and calculates that these customers used an average of 11,425 kWh of electricity last year. Assuming that the population standard deviation is 3217 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.The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,069 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 171 of their customers and calculates that these customers used an average of 10,461 kWh of electricity last year. Assuming that the population standard deviation is 2973 kWh, is there sufficient evidence to support the power company's claim at the 0.01 level of significance? Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.2. Reporting on cheats What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: "You witness two students cheating on a quiz. Do you go to the professor?" Only 11% answered "Yes." The dotplot shows the proportion who would go to the professor in each of 1000 random samples of size 172 from a population where 11% would go to the professor. 17 0.1 0.15 0.2 0.25 Simulated sample proportion who would say "Yes" 0 0.05 Distribution of simulated proportion # samples 1000 mean 0.11 SD 0.024 (a) Use the results of the simulation to approximate the margin of error for Gallup's estimate of the proportion of U.S. adults who were satisfied with the way things were going in the United States at the time of the poll. (b) Interpret the margin of error.
- A lawn mower company will produce 1,500 lawn mowers by 2020. In an effort to determine how much maintenance-free consumers will buy when they buy lawn mowers, the company decided to conduct a study over the past 1 year of these mowers. First, a random sample was taken by contacting 200 grass mowers. The company owner provides an 800 number and is asked to contact the company when the first repair is needed for the lawn mower. Owners who no longer use lawn mowers to cut their grass are either disqualified or not sampled. After 1 year, it turned out that 187 owners had reported for the first repair. However, the other 74 disqualified themselves. The average number per year until the first improvement was 5.3 times for samples deemed feasible. It is believed that the standard deviation is 1.28 per year. If the company wants to advertise the lawnmower and with a confidence level of 95%, what is the estimated annual mean value of repair-free lawn mower for this lawn mower? What is the…A biologist wants to determine if different temperatures (15oC, 25oC, or 35oC) and amounts of sunlight (partial or full) will affect the growth of plants. He will test each combination of temperature and sunlight by randomly assigning 15 plants to each of the combinations. What type of sampling is described in this study? one sample paired data two samples more than two samplesAssume a population with 5 elements: X1-0, X2-1, X3-2. X4-3,X5-4. Calculate mean and variance of X for random samples of size 3 taken with replacement Provide a numerical answer only.
- A biologist argues that a certain species of fish found in Lake Tanzania in Africa should be classified in a different genus than the currently accepted genus for the species. A simple random sample of standard lengths of fully grown fish from the species in Lake Tanzania is given below: {116.2, 100.3, 118, 152.8, 141.3, 132.8, 95.6, 111.5, 95.2, 95.7} The population of alls such lengths is assumed to be close to normally distributed. Calculate the following interval estimates using this sample (note that the only one of the methods below can be used for a single sample in practice). Round answers to one decimal place accuracy. A 99% confidence interval for the mean of all fully grown fish of this type: type your answer... A 99% prediction interval for the length of one fully grown fish of this type is: type your answer... type your answer... type your answer...An adventure company runs two obstacle courses, Fundash and Coolsprint, with similar designs. Since Fundash was built on rougher terrain, the designer of the courses suspects that the mean completion time of Fundash is greater than the mean completion time of Coolsprint. To test this, she selects 205 Fundash runners and 260 Coolsprint runners. (Consider these as independent random samples of the Fundash and Coolspring runners.) The 205 Fundash runners complete the course with a mean time of 77.6 minutes and a standard deviation of 8.2 minutes. The 260 individuals complete Coolsprint with a mean time of 76.0 minutes and a standard deviation of 7.7 minutes. Assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean completion time, u,, of Fundash is greater than…An adventure company runs two obstacle courses, Fundash and Coolsprint, with similar designs. Since Fundash was built on rougher terrain, the designer of the courses suspects that the mean completion time of Fundash is greater than the mean completion time of Coolsprint. To test this, she selects 230 Fundash runners and 210 Coolsprint runners. (Consider these as independent random samples of the Fundash and Coolspring runners.) The 230 Fundash runners complete the course with a mean time of 77.9 minutes and a standard deviation of 6.6 minutes. The 210 individuals complete Coolsprint with a mean time of 76.6 minutes and a standard deviation of 7.3 minutes. Assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean completion time, U1, of Fundash is greater than…