Discuss the assumptions and check the conditions to create a confidence interval for the true mean weight of all the oranges harvested this season at Sun Day Orchards.
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- The Weschler Intelligence Scale for Children (WISC) is an intelligence test designed for children between the ages of 6 and 16. The test is standardized so that the mean score for all children is 100 and the standard deviation is 15. Suppose that the administrators of a very large and competitive school district wish to estimate the mean WISC score for all students enrolled in their programs for gifted and talented children. They obtained a random sample of 40 students currently enrolled in at least one program for gifted and talented children. The test scores for this sample are as follows: 117,142,112,99,107,109,109,121,121,125,116,123,110,105,110,111,95,128,103,106, 155,98,100,123,107,137,115,127,102,106,121,113,109,128,103,105,130,134,112,102 Click to download the data in your preferred format. CrunchIt! CSV Excel JMP Mac Text Minitab PC Text R SPSS TI Calc Use this data to calculate the mean WISC score, x¯, for these 40 students. Next, compute the…In a large city, researchers selected a random sample of 120 individuals who have multiple televisions in their house connected to a streaming device. Each individual was asked how many hours of streaming content they watch in a typical week. The sample had a mean of 12.5 hours. For which of the following populations is 12.5 hours a reasonable estimate of the mean hours of weekly streaming content watched? (A) All individuals who regularly watch streaming content (B) All individuals from this city who regularly watch streaming content (C) All individuals from this city with multiple streaming devices in their house (D) All individuals from this city with multiple streaming devices in their house who regularly watch streaming content (E) All 120 individuals in this city with multiple streaming devices in their home who regularly watch streaming contentOld Faithful is a geyser located in Yellowstone National Park in Wyoming, USA. Millions of travelers come from afar to witness Old Faithful's eruptions each year. Travelers to Yellowstone are told that Old Faithful is expected to erupt, on average, every 7272 minutes, but park scientists suspect that this average is incorrect. To investigate, the scientists determine the waiting time until the next eruption for a random sample of 272272 Old Faithful eruptions. The waiting times have a mean of 70.897 minutes and a standard deviation of 13.595 minutes. You will use these data to carry out a significance test at the ?α = 0.05 level.
- The breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1775 pounds and a standard deviation of 70 pounds. The Español company believes that, due to an improvement in the manufacturing process, the mean breaking strength, µ, of the cables is now greater than 1775 pounds. To see if this is the case, 16 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1806 pounds. Assume that the population is normally distributed. Can we support, at the 0.10 level of significance, the claim that the population mean breaking strength of the newly- manufactured cables is greater than 1775 pounds? Assume that the population standard deviation has not changed. Perform a one-tailed test. Then complete the parts below. 00 Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H,…The lifetimes of a certain type of car battery are normally distributed with a mean 5.9 years and standard deviation 0.4 year. The batteries are guaranteed to last at least 5 years. What proportion of batteries fail to meet the guarantee? In other words, what proportion of batteries last less than 5 years?Judy is a psychology student who participates in three competitions: Swimming, cycling, and running in one event. In the last competition, Judy completed the swimming race in 13 minutes and 20 seconds (800 seconds), cycling in 33 minutes and 52 seconds (2032 seconds), and the running race in 17 minutes and 3 seconds (1023 seconds. Completion times for participants in Swimming are normally distributed with a mean of / = 725 seconds and a standard deviation o - 50 seconds. Completion times for participants in Cycling* are normally distributed with a mean of = 2182 seconds and a standard deviation o= 75 seconds. And completion times for participants in Running are normally distributed with a mean of u = 1083 seconds and a standard deviation o = 60 seconds. In which competition race (swimming, cycling, or running) was Judy's performance better, relative to others on average who took part in that competition? Justify your answer, quoting. relevant statistics as part of your explanation
- A sample of 100 postal employees found that the mean time these employees had worked for the postal service was eight years. Assume we know the standard deviation of the population of times postal service employees have spent with the postal service is five years. Find the 98% confidence interval for the mean time the population of postal service employees have spent with the postal service.A fast-food franchise, looking to reduce wait times for customers, launched a pilot program where customers could use a smartphone app to place orders. The mean wait times (in seconds) for in-store and drive-through customers at 10 stores participating in the pilot program and at 10 stores not in the program are recorded. Assume that the population standard deviation of mean wait times is 30 seconds for both groups of stores and that the mean wait times are normally distributed for both groups of stores. Let the mean wait times of the stores not in the program be the first sample, and let the mean wait times of the stores in the program be the second sample. The franchise conducts a two-mean hypothesis test at the 0.05 level of significance, to test if there is evidence that the smartphone app reduces wait times. (a) H0:μ1=μ2; Ha:μ1>μ2, which is a right-tailed test. Mean Wait Time: Not In Program Mean Wait Time: In Program 275 176 284 218 307 261 290 252 351 238…An industrial oil heater is tested by its ability to heat up a large room on a cold day and in particular, the rate at which it reaches a desired average temperature. The size of the room and the room temperature prior to switching on the heater were identified as two important factors of the rate at which the average room temperature increases. The experiment was replicated 3 times for each combination. The time (in minutes) till an average temperature increase of 10°C was measured is tabulated below: Factor B Large Classroom (200m2) Large Auditorium (400m2) (1) Treatment Combination (2) Factor A 7°C (1) 8,9,8 22,18,21 12°C (2) 9,10,12 19,29,25 Let factor A be the initial temperature and factor B be size of the room.
- The breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1825 pounds and a standard deviation of 70 pounds. The company believes that, due to an improvement in the manufacturing process, the mean breaking strength, μ, of the cables is now greater than 1825 pounds. To see if this is the case, 17 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1844 pounds. Assume that the population is normally distributed. Can we support, at the 0.05 level of significance, the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1825 pounds? Assume that the population standard deviation has not changed. Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H0…An excess of high values for the cases distributed on a graph results in a positive skew in the data?