atRecall that to test a proportion, p, assuming all requirements are met, the z values will be calculated using the following formula,where r is the number of successes, n is the number of trials, p = - is the sample statistic, p is the population probability of%3Dsuccess, and q = 1 - p represents the population probability of failure.р - рz =pqIn order to use the normal distribution to estimate the p distribution, the number of trials n should be sufficiently large so thatboth np > 5 and ng > 5. We will first check that these requirements are met.We have a random sample of 20 binomial trials resulting in 9 successes and we wish to test the claim that the populationproportion of successes does not equal 0.50 using a significance level of 0.05. Therefore, we can define n, p, and q as follows.n =2020p = 0.500.50q = 1 - p = 0.500.50Step 2Now calculate np and ng for n = 20, p = 0.50, and q = 0.50.np =X (0.50)ng = 200We conclude that since ---Select---p distribution.the requirements have been met to use the normal distribution for theSubmitSkip (you cannot come back)

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Answered: Now calculate np and nq forn = 20, p =…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Please note the question is: What is the distribution of the t-test statistic if m, n -----> ∞ ?
The first thing we need to do is check to see if the distribution is approximately normal. 50(0.6) = 30 and ng = 50(0.4) = 20 np = пр Since np 10 and nq ≥ 10, we can conclude that is approximately a normal distribution with • mean Up • standard deviation op = = 0.6 and pq n = = 0.6(0.4) 50 = pq n 0.24 50 (which is very close to what we saw in our simulation, standard error =0.063). standard error found in the simulation (standard error = 0.030)? ор nearest thousandth). Now find the approximation for samples of size 250. Is the standard deviation close to the = 0.0693 (Round to the Is the standard deviation close to the standard deviation (standard error) found in the simulation?
stion 16: nts) A random sample of 200 psychology students was selected from a department's participant pool. A researchers determined each student's score on the NPI-16 and the number of his or her Facebook friends. The marginal distribution of NPI-16 scores was 4, with a standard deviation of 2. The marginal distribution "Facebook Friends" was 450, with a standard deviation of 200. The correlation between the two variables was .15. Assuming that NPI-16 is the predictor variable and Number of Facebook Friends is the outcome variable, compute the following: The lower limit of the 95% confidence interval around b. Note: Enter your answer to the second decimal place.
Consider the following set of three distributions, all of which are drawn to the same scale. Identify the two distributions that are normal. Of the two normal distributions, which one has the larger variation? Q 9 The two normal distributions are Ar 1. ^. where Ar 1) and (2) and (c) P) and (4) has the larger standard deviation.
ABC has compiled data on the Time spent by the Bachelor students to watch videos on TikTok. They confirm that the time spent of Bachelor Population (BP) of 1200 students follows a normal distribution with the standard deviation equal to15 minutes. XYZ choose from the population (BP) a random sample of 140 students, and he confirms that : - 60% from the sample (BS) have account to watch videos on TikTok - the average time spent to watch videos on TikTok equals 75 minutes Determine at 95% confidence level, the population (BP) proportion that have a TikTok account
The average time to run the 5K fun run is 21 minutes and the standard deviation is 2.6 minutes. 40 runners are randomly selected to run the 5K fun run. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of XX ~ N (21, 2.6) What is the distribution of ¯xx¯ ~ N (21, 0.4111) What is the distribution of ∑x∑x ~ N(840, 16.4438). If one randomly selected runner is timed, find the probability that this runner's time will be between 20.8833 and 21.4833 minutes. 0.0913 e. For the 40 runners, find the probability that their average time is between 20.8833 and 21.4833 minutes. 0.4913 f. Find the probability that the randomly selected 40 person team will have a total time less than 832. _________ For part e) and f), is the assumption of normal necessary? No The top 10% of all 40 person team relay races will compete in the championship round. These are the 10% lowest times. What is the longest total time that a relay team can have and…
An excess of high values for the cases distributed on a graph results in a positive skew in the data?
The average number of miles (in thousands) that a car's tire will function before needing replacement is 72 and the standard deviation is 18. Suppose that 50 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. 1. If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 73.9 and 76.9. 2. For the 50 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 73.9 and 76.9.

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Now calculate np and nq forn = 20, p = 0.50, and q = 0.50. np = x (0.50) %3D ng = 20 We conclude that since ---Select--- p distribution. the requirements have been met to use the normal distribution for the Submit Skip_ (you cannot come back)