Based on a USA Today poll, 10% of the population believes that college is no longer a good investment Assuming 10% rate, 12 people are randomly selected. Verify that this is a binomial experiment and identify the symbols: x, n, and p
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Based on a USA Today poll, 10% of the population believes that college is no longer a good investment
Assuming 10% rate, 12 people are randomly selected. Verify that this is a binomial experiment
and identify the symbols: x, n, and p
investment
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- A local school board wants to estimate the difference in the proportion of households with school-aged children that would support starting the school year a week earlier, and the proportion of households without school-aged children that would support starting the school year a week earlier. They survey a random sample of 40 households with school-aged children about whether they would support starting the school year a week earlier, and 38 households respond yes. They survey a random sample of 45 households that do not have school-aged children, and 25 respond yes. The school board plans to construct a 90% confidence interval for the difference in proportions of households who would support starting the school year a week earlier. Are the conditions for inference met? A) Yes, the conditions for inference are met. B) No, the 10% condition is not met. C) No, the randomness condition is not met. D) No, the Large Counts Condition is not met.What is the number of successes in a binomial experiment called?Solve the last question please
- You are hiking through the woods, and come across some tracks; based on your expe- rience, you estimate there is a 40% chance the tracks are from a dog, a 30% chance the tracks are from a wolf, a 20% chance the tracks are from a coyote, and a 10% chance the tracks are from some other animal. Your friend, also a nature lover, notices some white hairs in the tracks. They know that 20% of dogs have white hair, 50% of wolves have white hair, 10% of coyotes have white hair, and 30% of other animals in the area have white hair. What are the odds of these tracks coming from a dog, wolf, coyote, or other animal?According to a job website, each job opening on average attracted 250 résumés in 2016. The job market improved in 2017 compared to 2016, which means that more people will likely be switching jobs but also fewer unemployed workers remain in the job market. To find out which trend is stronger, a random sample of 20 employers in a region was taken. Each employer reported how many résumés they received in 2017 for each job opening. Their answers are shown in the accompanying table. Using a = 0.05, complete parts a through d. Click the icon to view the data on résumés received. a. State the null and alternative hypotheses. Determine the null hypothesis, Ho, and the alternative hypothesis, H₁. Ho: H₁: (Type integers or decimals. Do not round.)Scenario 1: A surgeon routinely performs 4 surgeries. The chance that he performs surgery “A” is 30%. A random sample of 15 patients is selected. Let the variable X be defined as the number of patients that have surgery "A". Scenario 2: Thirteen (13) cards are randomly selected from a standard deck of cards without replacement. Let the variable X be defined as the number of spades selected. Directions: Pick a scenario describe a binomial experiment. Recall the random variable X is for the number of successes in a binomial experiment. • Be certain to state the criteria for a binomial distribution and how it does or does not meet each of the five conditions. Once your answer is posted, you will be able to see your classmates' responses.
- Xu and Garcia (2008)conducted a research study demonstrating that 8-month-old infants appear to recognize which samples are likely to be obtained from a population and which are not. In the study, the infants watched as a sample of n = 5 ping-pong balls was selected from a large box. In one condition, the sample consisted of 1 red ball and 4 white balls. After the sample was selected, the front panel of the box was removed to reveal the contents. In the expected condition, the box contained primarily white balls like the sample, and the infants looked at it for an average of M = 7.5 seconds. In the unexpected condition, the box had primarily red balls, unlike the sample, and the infants looked at it for M = 9.9. The researchers interpreted the results as demonstrating that the infants found the unexpected result surprising and, therefore, more interesting than the expected result. Assuming that the standard error for both means is σM = 1 second, draw a bar graph showing the two sample…Itranscript One male and one female dam rat pup were randomly selected from 8 litters to perform the swim maze. Each pup was placed in the water at one end of the maze and allowed to swim until it escaped at the opposite end. If the pup failed to escape after a certain period of time, it was placed at the beginning of the maze and given another chance. The experiment was repeated until each pup accomplished three successful escapes. The table to the right reports the number of swims required by each pup. Is there sufficient evidence of a difference between the mean number of swims required by male and female pups? Use a=0.01. Comment on the assumptions required for the test to be valid. B. t> Litter 1 2 3 OC. t 10 12 Female 11 [6526675 10SWIMMER’S FLEXIBILITY STUDY Swimming requires complete body movement as defined in sports science. A swimmer’s flexibility helps in his/her movement in water, thus making him/her a faster swimmer. In a study conducted to determine if there is an improvement in swim speed after doing flexibility exercise, 15 swimmers of the same characteristics were randomly selected. They were asked to do a 25-m freestyle and their swim speed (in seconds) were recorded. After this, a flexibility exercise was done. All the swimmers were then asked to do another 25-m freestyle and their swim speed (in seconds) were also recorded. The data collected are given below and can also be found in the worksheet “swim_speed”of excel file “exer3_data.xlsx”. R outputs can be found in the file “exer3_outputs.pdf”.(Use before – after in your computations Swimmer 1 2 3 4 5 6 7 8 Before 16.87 19.42 20.04 22.82 24.24 17.76 23.91 17.17 After 15.82 18.47 20.43 21.76 23.88 18.12 20.96 16.03 Swimmer 9 10 11 12…
- A dowser has correctly located water for a well 1 out of 2 times in Jones County. In Jones County, someone who is just guessing has a 40% chance of locating water for a well. Does this sample provide sufficient evidence that the dowser can locate water and is not just guessing?A local school board wants to estimate the difference in the proportion of households with school-aged children that would support starting the school year a week earlier, and the proportion of households without school-aged children that would support starting the school year a week earlier. They survey a random sample of 40 households with school-aged children about whether they would support starting the school year a week earlier, and 30 households respond yes. They survey a random sample of 45 households that do not have school-aged children, and 25 respond yes. Assuming the conditions for inference have been met, what is the 90% confidence interval for the difference in proportions of households that would support starting the school year a week earlier? Find the z-table here. 0.75(1-0.75), 0.56(1-0.56) O (0.75 - 0.56) ±1.65 45 40 0.75(1-0.75) 0.56(1-0.56) O (0.75 - 0.56) ±1.96, 85 85 O (0.75 - 0.56) ±1.65, 0.75(1-0.75) 0.56(1-0.56) + 40 45 O (0.75 - 0.56) ±1.96 0.75(1-0.75)…A local school board believes there is a difference in the proportion of households with school-aged children that would support starting the school year a week earlier, and the proportion of households without school- aged children that would support starting the school year a week earlier. They survey a random sample of 40 households with school-aged children about whether they would support starting the school year a week earlier, and 38 households respond yes. They survey a random sample of 45 households that do not have school-aged children, and 25 respond yes. Let p3= the true proportion of households with school-aged children that would support starting the school year a week early and pw= the true proportion of households without school-aged children that would support starting the school year a week earlier. The P-value for this significance test is 0.000034. Which of the following is the correct conclusion for this test of the hypotheses Ho: P;- Pw=0 and H, P,- Pw0 at the a =…