Description
y6z4
____
y5z simplify
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Explanation & Answer
y6z4
Means
= y*6*z*4
Now separate constant terms and variables, we get
= 6*4*y*z
= 24yz
Hence y6z4 = 24yz
y5z
Means
= y*5*z
Now separate constant terms and variable, we get
= 5*y*z
= 5yz
Hence y5z = 5yz
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PSY325 Ashford Standard Normal Distribution and Primer in Probability Quiz
Based on past experience, the probability that the sun will rise tomorrow is ________. If instead of z = (x – M)/s the f ...
PSY325 Ashford Standard Normal Distribution and Primer in Probability Quiz
Based on past experience, the probability that the sun will rise tomorrow is ________. If instead of z = (x – M)/s the formula is x = z(s) + M, what ...
Mode and Median probability, statistics homework help
1. Read the webpage, https://en.wikipedia.org/wiki/Binomial_distribution , down to the section, Mode and Me ...
Mode and Median probability, statistics homework help
1. Read the webpage, https://en.wikipedia.org/wiki/Binomial_distribution , down to the section, Mode and Median. Read thoroughly the example about the biased coin that comes up heads with probability of 0.3. This variable can be described as, X~Bi(n=6, p=0.3). Use this example to answer the following questions.a. What is the expected value of this variable? b. What is the expected number of heads if this coin is tossed 6 times? Same answer as in part 2.a. but in different words.c. What is the probability that the number of heads out of 6 tosses of this unfair coin is less than 2?d. What is the probability that the number of heads out of 6 tosses of this unfair coin is more than 2?e. What is the probability that the number of heads out of 6 tosses of this unfair coin is at least 2?f. What is the probability that the number of heads out of 6 tosses of this unfair coin is at most 2?2. Assume that 25% of all accounting audits identify an accounting error and that 8 audits were completed. The variable of interest is the number of audits that identify an accounting error out of the 8 audits. The distribution of this variable is described as X~Bi(n=8, p=0.25). Use this information to answer the questions below. Use this online calculator, if you want: http://stattrek.com/online-calculator/binomial.aspxa. What is the expected number of audits that will identify accounting errors?b. What is the probability of 2 or more accounting errors being identified in the 8 audits?c. What is the probability of 3 or more accounting errors being identified in the 8 audits?d. What is the probability of 4 or more accounting errors being identified in the 8 audits?e. What is the probability of 5 or more accounting errors being identified in the 8 audits?f. What is the probability of 6 or more accounting errors being identified in the 8 audits?g. What is the probability of 7 or more accounting errors being identified in the 8 audits?h. What is the probability of 8 or more accounting errors being identified in the 8 audits?
Answer 16 Question Multiple Choice Statistic Assignment
Here the following are the questions, please when answering give brief sentence/reason to why you chose your answer. Show ...
Answer 16 Question Multiple Choice Statistic Assignment
Here the following are the questions, please when answering give brief sentence/reason to why you chose your answer. Show work if there is work that can be shown so I can have a better understanding.- Ignore the boxes due to odd formatting.Question 1Select one answer.10 pointsA school district claims that the normal attendance rate for their schools is 95%. An educational advocate believes that the true figure is lower. She chooses a school day in October and chooses 120 random students from the district. On that day, 12.5% of the students missed school.Can she conduct a hypothesis test to determine whether the proportion of students who attend school is lower than 0.95? Yes, because the sample is random, so it represents the students in the district. Yes, because (120)(0.125) and (120)(0.875) are both at least 10. This means the normal model is a good fit for the sampling distribution. No, because even though (120)(0.95) is at least 10, (120)(0.05) is less than 10. This means the normal model is NOT a good fit for the sampling distribution.Question 2Select one answer.10 pointsStudent debt default: In a CNN article, 7% of college graduates say they don’t plan on ever being able to entirely repay their student loans. Financial aid office staff at a private university conducted a study to determine whether the proportion at their university is higher. They surveyed 500 randomly selected recent college graduates and determined that 12% of them don’t plan on ever being able to entirely repay their student loans.Which of the following are the appropriate null and alternative hypotheses for this research question? H0: p = 0.07Ha: p ≠ 0.07 H0: p = 0.07Ha: p > 0.07 H0: p = 0.12Ha: p ≠ 0.12 H0: p = 0.12Ha: p > 0.12Question 3Select one answer.10 pointsIn survey conducted by Quinnipiac University from October 25-31, 2011, 47% of a sample of 2,294 registered voters approved of the job Barack Obama was doing as president.What is the 99% confidence interval for the proportion of all registered voters who approved of the job Barack Obama was doing as president? (0.460, 0.480) (0.453, 0.487) (0.450, 0.490) (0.443, 0.497)Question 4Select one answer.10 pointsAn interactive poll on the front page of the CNN website in October 2011 asked if readers would consider voting for Herman Cain, a Republican presidential candidate. A statistics student used the information from the poll to calculate the 95% confidence interval. He got (0.53, 0.59). He also conducted a hypothesis test. He found very strong evidence that more than half of voters would consider voting for Herman Cain. To what population do these conclusions apply? They apply to all likely voters. They apply to the population of those who visit the CNN website. The results do not apply to any population because this was a voluntary response sample. They apply to the population of those who visited the CNN website while the poll was on the front page.Question 5Select one answer.10 pointsSample size: A researcher is trying to decide how many people to survey. Which of the following sample sizes will result in a confidence interval with the largest width? 300 700 1000 The width depends on the sample proportion.Question 6Select one answer.10 pointsA tire manufacturer has a 60,000 mile warranty for tread life. The company wants to make sure the average tire lasts longer than 60,000 miles. The manufacturer tests 250 tires and finds the mean life for these tires to be 64,500 miles.What is the alternative hypothesis being tested in this example? The proportion of tires that are worn out after 60,000 miles is greater than 1/2. The mean tire life is equal to 64,500 miles. The mean tire life is less than 60,000 miles. The mean tire life is greater than 60,000 miles.Question 7Select one answer.10 pointsDelta Flights: According to the Bureau of Transportation Statistics, 77.4% of Delta Airline’s flights arrived on time in 2010. The company is trying to improve on-time arrivals. They test the hypotheses H0: p = 0.774 versus Ha: p > 0.774.They calculate a P‑value of 0.03. Using a significance level of 0.05, which of the following is the best explanation for how to use the P‑value to reach a conclusion in this case? Since the P‑value is less than the significance level, we reject the null hypothesis. Since the P‑value is less than the significance level, we fail to reject the null hypothesis. Since the P‑value is less than the significance level, we accept the null hypothesis.Question 8Select one answer.10 pointsGun rights vs. gun control: In a December 2014 report, “For the first time in more than two decades of Pew Research Center surveys, there is more support for gun rights than gun control.” According to a Pew Research survey, 52% of Americans say that protecting gun rights is more important than controlling gun ownership. Gun rights advocates in a conservative city believe that the percentage is higher among city residents.They survey 200 city residents and find that 130 say that protecting gun rights is more important than controlling gun ownership. What is the test statistic?http://www.people-press.org/2014/12/10/growing-pub... Z = -3.68 Z = 3.68 Z = -3.85 Z = 3.85Question 9Select one answer.10 pointsSexual assault in college: In a Washington Post/Kaiser Family Foundation poll conducted from January through March 2015, 46% of adults (ages 17‑26) who attended college during the past 4 years say it’s unclear whether sexual activity when both people have not given explicit agreement is sexual assault. The survey methodology section states that the margin of error is ±3.5% at the 95% confidence level.What does this margin of error tell you about the results of this poll? We are 95% confident that the sample proportion is off by 3.5%. We are confident that 95% of the responses are within 3.5% of the population proportion. We are 95% confident that the population proportion is within 3.5% of the sample proportion of 46%. We are confident that population proportion is within 3.5% of the sample proportion of 46%.Question 10Select one answer.10 pointsIn a study of the nicotine patch, 21% of those who used the patch for 2 months reported no smoking incidents in the following year. The 95% confidence interval is (17.4%, 24.8%).Which of the following is an appropriate interpretation of the 95% confidence interval? There is a 95% probability that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%. We can be 95% confident that the proportion of the sample who would report no smoking incidents in the following year is between 17.4% and 24.8%. 95% of samples will have between 17.4% and 24.8% who would report no smoking incidents in the following year. We can be 95% confident that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.Question 11Select one answer.10 pointsA 2014 study by the reputable Gallup organization estimates that 34% of U.S. adults are worried about money. We want to know if the proportion is greater this year. We select a random sample of 100 U.S. adults this year and find that 40% are worried about money.After carrying out the hypothesis test for p = 0.34 compared to p > 0.34, we obtain a P‑value of 0.10. Which of the following interpretations of the P‑value is correct? There is a 10% chance that 34% of U.S. adults are worried about money this year. There is a 10% chance that 40% of U.S. adults are worried about money this year. There is a 10% chance that 40% or more of U.S. adults are worried about money this year if 34% were worried about money in 2014. There is a 10% chance that a sample of 100 U.S. adults will have 40% or more worried about money if 34% of the population is worried about money this year.Question 12Select one answer.10 pointsIn a study of a new treatment for cold sores, researchers randomly assigned 2209 patients to use of the new topical medication or a placebo in a double-blind experiment. Cold sores healed more quickly with the new medication and the improvement was statistically significant at the 0.01 level. Results were published the Journal of the American Medical Association in 1997. (http://jama.jamanetwork.com/article.aspx?articleid=415844)Which of the following is an appropriate conclusion? With a large sample, statistically significant results suggest a large improvement over the placebo. With a large sample, statistically significant results may come from a small improvement over the placebo. Regardless of the sample size, a statistically significant result means there is a meaningful difference in cold sore relief.Question 13Select one answer.10 pointsOne population proportion test: Which of the following situations involves testing a claim about a single population proportion? The Centers for Disease Control estimates that 22.8% of Americans (ages 18 to 24) get 6 or less hours of sleep per night. A researcher believes that the figure for college students is higher than this. The mean SAT math score for Florida is 514. An educational researcher is concerned that this average may be lower in rural counties. A Statistics student wants to determine whether there is a difference in the average number of credit hours male and female students are taking. A growing practice among some parents is called “redshirting,” Redshirting means holding a child back a year from starting kindergarten even though he or she is eligible by age. Many states use August 31 as the cutoff for the 5th birthday in order for a child to start kindergarten. A researcher is curious if the proportion of boys with August birthdays who are redshirted is different than the proportion of girls with August birthdays who are redshirted.Question 14Select one answer.10 pointsBirthdays of hockey players: In Malcolm Gladwell’s book Outliers, he shares the work of Canadian psychologist Roger Barnsley, who noticed that a disproportionately high percentage of elite ice-hockey players have birthdays between January and March. A group of statistics students would like to test if this is true for the Los Angeles Kings 2010–2015 rosters (22 out of 57 ). After debating whether this set of hockey players can be viewed as a random sample of hockey players, they decide to run a hypothesis test anyway to practice finding the P‑value. They test the hypotheses H0: p = 0.25 versus Ha: p > 0.25. The P‑value is small enough to reject the null hypothesis.Which of the following is an appropriate conclusion (if we assume the sample is random)? The data provides strong evidence to conclude that the proportion of LA Kings hockey players who have birthdays between January and March is greater than 0.25. The data does not provide strong evidence to conclude that the proportion of LA Kings hockey players who have birthdays between January and March is greater than 0.25. The data provides strong evidence to conclude that the proportion of LA Kings hockey players who have birthdays between January and March is equal to 0.25. The probability that the proportion of LA Kings hockey players who have birthdays between January and March is greater than 0.25. is equal to the level of significance, 0.05.Question 15Select one answer.10 pointsPolice body cameras: A survey of New York State residents asked about police officers having to wear video cameras on duty. The question stated, “Do you agree or disagree that police officers should carry video cameras for the purposes of filming their activities while on duty?” Most (88%) respondents agreed with this statement.Do Californians share the same opinion? A California-based civil rights group conducted a similar survey by randomly selecting 500 California residents, and 425 agreed that police officers should carry video cameras for the purposes of filming their activities while on duty.Conduct a hypothesis test to determine if the proportion of California residents who agree is different from New York residents. Use a 5% significance level to make your decision. Use the applet to determine the P‑value.http://www.futurity.org/police-poll-new-york-94275...Use an applet if necessary.Which of the following is an appropriate conclusion based on the results? The survey provides strong evidence that the proportion of California residents who agree that police officers should carry video cameras for the purposes of filming their activities while on duty is significantly different from the proportion of New York residents. The survey suggests that the proportion of California residents who agree that police officers should carry video cameras for the purposes of filming their activities while on duty is 85%. Of the California residents surveyed, 85% agree that police officers should carry video cameras for the purposes of filming their activities while on duty, but this is not strong enough evidence to conclude that the proportion of California residents who agree is significantly different from the proportion of New York residents. This survey suggests that 88% of California residents surveyed agree that police officers should carry video cameras for the purposes of filming their activities while on duty.Question 16Select one answer.10 pointsGender and College Students: According to the U.S. Department of Education, approximately 57% of students attending colleges in the U.S. are female. A statistics student is curious whether this is true at her college. She tests the hypotheses H0: p = 0.57 versus Ha: p ≠ 0.57.She plans to use a significance level of 0.05. She calculates her test statistic to be 1.42. Using the applet, which is the P‑value? Use an applet. P‑value = 0.078 P‑value = 0.156 P‑value = 0.922 P‑value = 0.05
TAMU Connecting the Chain Rule to Integration by Substitution Lab Report
I need someone to complete this final and send it back to me no later than 11/19/20 by noon. If you could do this for me i ...
TAMU Connecting the Chain Rule to Integration by Substitution Lab Report
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Spc In Template
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PSY325 Ashford Standard Normal Distribution and Primer in Probability Quiz
Based on past experience, the probability that the sun will rise tomorrow is ________. If instead of z = (x – M)/s the f ...
PSY325 Ashford Standard Normal Distribution and Primer in Probability Quiz
Based on past experience, the probability that the sun will rise tomorrow is ________. If instead of z = (x – M)/s the formula is x = z(s) + M, what ...
Mode and Median probability, statistics homework help
1. Read the webpage, https://en.wikipedia.org/wiki/Binomial_distribution , down to the section, Mode and Me ...
Mode and Median probability, statistics homework help
1. Read the webpage, https://en.wikipedia.org/wiki/Binomial_distribution , down to the section, Mode and Median. Read thoroughly the example about the biased coin that comes up heads with probability of 0.3. This variable can be described as, X~Bi(n=6, p=0.3). Use this example to answer the following questions.a. What is the expected value of this variable? b. What is the expected number of heads if this coin is tossed 6 times? Same answer as in part 2.a. but in different words.c. What is the probability that the number of heads out of 6 tosses of this unfair coin is less than 2?d. What is the probability that the number of heads out of 6 tosses of this unfair coin is more than 2?e. What is the probability that the number of heads out of 6 tosses of this unfair coin is at least 2?f. What is the probability that the number of heads out of 6 tosses of this unfair coin is at most 2?2. Assume that 25% of all accounting audits identify an accounting error and that 8 audits were completed. The variable of interest is the number of audits that identify an accounting error out of the 8 audits. The distribution of this variable is described as X~Bi(n=8, p=0.25). Use this information to answer the questions below. Use this online calculator, if you want: http://stattrek.com/online-calculator/binomial.aspxa. What is the expected number of audits that will identify accounting errors?b. What is the probability of 2 or more accounting errors being identified in the 8 audits?c. What is the probability of 3 or more accounting errors being identified in the 8 audits?d. What is the probability of 4 or more accounting errors being identified in the 8 audits?e. What is the probability of 5 or more accounting errors being identified in the 8 audits?f. What is the probability of 6 or more accounting errors being identified in the 8 audits?g. What is the probability of 7 or more accounting errors being identified in the 8 audits?h. What is the probability of 8 or more accounting errors being identified in the 8 audits?
Answer 16 Question Multiple Choice Statistic Assignment
Here the following are the questions, please when answering give brief sentence/reason to why you chose your answer. Show ...
Answer 16 Question Multiple Choice Statistic Assignment
Here the following are the questions, please when answering give brief sentence/reason to why you chose your answer. Show work if there is work that can be shown so I can have a better understanding.- Ignore the boxes due to odd formatting.Question 1Select one answer.10 pointsA school district claims that the normal attendance rate for their schools is 95%. An educational advocate believes that the true figure is lower. She chooses a school day in October and chooses 120 random students from the district. On that day, 12.5% of the students missed school.Can she conduct a hypothesis test to determine whether the proportion of students who attend school is lower than 0.95? Yes, because the sample is random, so it represents the students in the district. Yes, because (120)(0.125) and (120)(0.875) are both at least 10. This means the normal model is a good fit for the sampling distribution. No, because even though (120)(0.95) is at least 10, (120)(0.05) is less than 10. This means the normal model is NOT a good fit for the sampling distribution.Question 2Select one answer.10 pointsStudent debt default: In a CNN article, 7% of college graduates say they don’t plan on ever being able to entirely repay their student loans. Financial aid office staff at a private university conducted a study to determine whether the proportion at their university is higher. They surveyed 500 randomly selected recent college graduates and determined that 12% of them don’t plan on ever being able to entirely repay their student loans.Which of the following are the appropriate null and alternative hypotheses for this research question? H0: p = 0.07Ha: p ≠ 0.07 H0: p = 0.07Ha: p > 0.07 H0: p = 0.12Ha: p ≠ 0.12 H0: p = 0.12Ha: p > 0.12Question 3Select one answer.10 pointsIn survey conducted by Quinnipiac University from October 25-31, 2011, 47% of a sample of 2,294 registered voters approved of the job Barack Obama was doing as president.What is the 99% confidence interval for the proportion of all registered voters who approved of the job Barack Obama was doing as president? (0.460, 0.480) (0.453, 0.487) (0.450, 0.490) (0.443, 0.497)Question 4Select one answer.10 pointsAn interactive poll on the front page of the CNN website in October 2011 asked if readers would consider voting for Herman Cain, a Republican presidential candidate. A statistics student used the information from the poll to calculate the 95% confidence interval. He got (0.53, 0.59). He also conducted a hypothesis test. He found very strong evidence that more than half of voters would consider voting for Herman Cain. To what population do these conclusions apply? They apply to all likely voters. They apply to the population of those who visit the CNN website. The results do not apply to any population because this was a voluntary response sample. They apply to the population of those who visited the CNN website while the poll was on the front page.Question 5Select one answer.10 pointsSample size: A researcher is trying to decide how many people to survey. Which of the following sample sizes will result in a confidence interval with the largest width? 300 700 1000 The width depends on the sample proportion.Question 6Select one answer.10 pointsA tire manufacturer has a 60,000 mile warranty for tread life. The company wants to make sure the average tire lasts longer than 60,000 miles. The manufacturer tests 250 tires and finds the mean life for these tires to be 64,500 miles.What is the alternative hypothesis being tested in this example? The proportion of tires that are worn out after 60,000 miles is greater than 1/2. The mean tire life is equal to 64,500 miles. The mean tire life is less than 60,000 miles. The mean tire life is greater than 60,000 miles.Question 7Select one answer.10 pointsDelta Flights: According to the Bureau of Transportation Statistics, 77.4% of Delta Airline’s flights arrived on time in 2010. The company is trying to improve on-time arrivals. They test the hypotheses H0: p = 0.774 versus Ha: p > 0.774.They calculate a P‑value of 0.03. Using a significance level of 0.05, which of the following is the best explanation for how to use the P‑value to reach a conclusion in this case? Since the P‑value is less than the significance level, we reject the null hypothesis. Since the P‑value is less than the significance level, we fail to reject the null hypothesis. Since the P‑value is less than the significance level, we accept the null hypothesis.Question 8Select one answer.10 pointsGun rights vs. gun control: In a December 2014 report, “For the first time in more than two decades of Pew Research Center surveys, there is more support for gun rights than gun control.” According to a Pew Research survey, 52% of Americans say that protecting gun rights is more important than controlling gun ownership. Gun rights advocates in a conservative city believe that the percentage is higher among city residents.They survey 200 city residents and find that 130 say that protecting gun rights is more important than controlling gun ownership. What is the test statistic?http://www.people-press.org/2014/12/10/growing-pub... Z = -3.68 Z = 3.68 Z = -3.85 Z = 3.85Question 9Select one answer.10 pointsSexual assault in college: In a Washington Post/Kaiser Family Foundation poll conducted from January through March 2015, 46% of adults (ages 17‑26) who attended college during the past 4 years say it’s unclear whether sexual activity when both people have not given explicit agreement is sexual assault. The survey methodology section states that the margin of error is ±3.5% at the 95% confidence level.What does this margin of error tell you about the results of this poll? We are 95% confident that the sample proportion is off by 3.5%. We are confident that 95% of the responses are within 3.5% of the population proportion. We are 95% confident that the population proportion is within 3.5% of the sample proportion of 46%. We are confident that population proportion is within 3.5% of the sample proportion of 46%.Question 10Select one answer.10 pointsIn a study of the nicotine patch, 21% of those who used the patch for 2 months reported no smoking incidents in the following year. The 95% confidence interval is (17.4%, 24.8%).Which of the following is an appropriate interpretation of the 95% confidence interval? There is a 95% probability that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%. We can be 95% confident that the proportion of the sample who would report no smoking incidents in the following year is between 17.4% and 24.8%. 95% of samples will have between 17.4% and 24.8% who would report no smoking incidents in the following year. We can be 95% confident that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.Question 11Select one answer.10 pointsA 2014 study by the reputable Gallup organization estimates that 34% of U.S. adults are worried about money. We want to know if the proportion is greater this year. We select a random sample of 100 U.S. adults this year and find that 40% are worried about money.After carrying out the hypothesis test for p = 0.34 compared to p > 0.34, we obtain a P‑value of 0.10. Which of the following interpretations of the P‑value is correct? There is a 10% chance that 34% of U.S. adults are worried about money this year. There is a 10% chance that 40% of U.S. adults are worried about money this year. There is a 10% chance that 40% or more of U.S. adults are worried about money this year if 34% were worried about money in 2014. There is a 10% chance that a sample of 100 U.S. adults will have 40% or more worried about money if 34% of the population is worried about money this year.Question 12Select one answer.10 pointsIn a study of a new treatment for cold sores, researchers randomly assigned 2209 patients to use of the new topical medication or a placebo in a double-blind experiment. Cold sores healed more quickly with the new medication and the improvement was statistically significant at the 0.01 level. Results were published the Journal of the American Medical Association in 1997. (http://jama.jamanetwork.com/article.aspx?articleid=415844)Which of the following is an appropriate conclusion? With a large sample, statistically significant results suggest a large improvement over the placebo. With a large sample, statistically significant results may come from a small improvement over the placebo. Regardless of the sample size, a statistically significant result means there is a meaningful difference in cold sore relief.Question 13Select one answer.10 pointsOne population proportion test: Which of the following situations involves testing a claim about a single population proportion? The Centers for Disease Control estimates that 22.8% of Americans (ages 18 to 24) get 6 or less hours of sleep per night. A researcher believes that the figure for college students is higher than this. The mean SAT math score for Florida is 514. An educational researcher is concerned that this average may be lower in rural counties. A Statistics student wants to determine whether there is a difference in the average number of credit hours male and female students are taking. A growing practice among some parents is called “redshirting,” Redshirting means holding a child back a year from starting kindergarten even though he or she is eligible by age. Many states use August 31 as the cutoff for the 5th birthday in order for a child to start kindergarten. A researcher is curious if the proportion of boys with August birthdays who are redshirted is different than the proportion of girls with August birthdays who are redshirted.Question 14Select one answer.10 pointsBirthdays of hockey players: In Malcolm Gladwell’s book Outliers, he shares the work of Canadian psychologist Roger Barnsley, who noticed that a disproportionately high percentage of elite ice-hockey players have birthdays between January and March. A group of statistics students would like to test if this is true for the Los Angeles Kings 2010–2015 rosters (22 out of 57 ). After debating whether this set of hockey players can be viewed as a random sample of hockey players, they decide to run a hypothesis test anyway to practice finding the P‑value. They test the hypotheses H0: p = 0.25 versus Ha: p > 0.25. The P‑value is small enough to reject the null hypothesis.Which of the following is an appropriate conclusion (if we assume the sample is random)? The data provides strong evidence to conclude that the proportion of LA Kings hockey players who have birthdays between January and March is greater than 0.25. The data does not provide strong evidence to conclude that the proportion of LA Kings hockey players who have birthdays between January and March is greater than 0.25. The data provides strong evidence to conclude that the proportion of LA Kings hockey players who have birthdays between January and March is equal to 0.25. The probability that the proportion of LA Kings hockey players who have birthdays between January and March is greater than 0.25. is equal to the level of significance, 0.05.Question 15Select one answer.10 pointsPolice body cameras: A survey of New York State residents asked about police officers having to wear video cameras on duty. The question stated, “Do you agree or disagree that police officers should carry video cameras for the purposes of filming their activities while on duty?” Most (88%) respondents agreed with this statement.Do Californians share the same opinion? A California-based civil rights group conducted a similar survey by randomly selecting 500 California residents, and 425 agreed that police officers should carry video cameras for the purposes of filming their activities while on duty.Conduct a hypothesis test to determine if the proportion of California residents who agree is different from New York residents. Use a 5% significance level to make your decision. Use the applet to determine the P‑value.http://www.futurity.org/police-poll-new-york-94275...Use an applet if necessary.Which of the following is an appropriate conclusion based on the results? The survey provides strong evidence that the proportion of California residents who agree that police officers should carry video cameras for the purposes of filming their activities while on duty is significantly different from the proportion of New York residents. The survey suggests that the proportion of California residents who agree that police officers should carry video cameras for the purposes of filming their activities while on duty is 85%. Of the California residents surveyed, 85% agree that police officers should carry video cameras for the purposes of filming their activities while on duty, but this is not strong enough evidence to conclude that the proportion of California residents who agree is significantly different from the proportion of New York residents. This survey suggests that 88% of California residents surveyed agree that police officers should carry video cameras for the purposes of filming their activities while on duty.Question 16Select one answer.10 pointsGender and College Students: According to the U.S. Department of Education, approximately 57% of students attending colleges in the U.S. are female. A statistics student is curious whether this is true at her college. She tests the hypotheses H0: p = 0.57 versus Ha: p ≠ 0.57.She plans to use a significance level of 0.05. She calculates her test statistic to be 1.42. Using the applet, which is the P‑value? Use an applet. P‑value = 0.078 P‑value = 0.156 P‑value = 0.922 P‑value = 0.05
TAMU Connecting the Chain Rule to Integration by Substitution Lab Report
I need someone to complete this final and send it back to me no later than 11/19/20 by noon. If you could do this for me i ...
TAMU Connecting the Chain Rule to Integration by Substitution Lab Report
I need someone to complete this final and send it back to me no later than 11/19/20 by noon. If you could do this for me it would be greatly appreciated!
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