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Decide whether the experiment is binomial experiment. If it is, identify success, specify the values of n, P, and q and list the possible values of the random varia...

Question

Decide whether the experiment is binomial experiment. If it is, identify success, specify the values of n, P, and q and list the possible values of the random variable If it is not binomial experiment) explain why. EXPERIMENT: A survey asks 1400 chief financial oflicers, "Has the economy forced you to postpone redluce the amount of vacation YOu plan to take this year?" Thirty-one percent of those surveyed sny they postponing or reducing the amount of vacation: Twenty oflicers particip

Decide whether the experiment is binomial experiment. If it is, identify success, specify the values of n, P, and q and list the possible values of the random variable If it is not binomial experiment) explain why. EXPERIMENT: A survey asks 1400 chief financial oflicers, "Has the economy forced you to postpone redluce the amount of vacation YOu plan to take this year?" Thirty-one percent of those surveyed sny they postponing or reducing the amount of vacation: Twenty oflicers participating. the survey Are randomly welected , The random variable represents the number of oflicers who are postponing or redlucing the nmount of vucation: (Source: Robert Half Manugement Rexources)



Answers

determine whether the experiment is a binomial experiment. If it is, identify a success, specify the values of $n, p,$ and $q,$ and list the possible values of the random variable $x .$ If it is not a binomial experiment, explain why. Generation A survey found that $68 \%$ of adults ages 18 to 25 think that their generation is unique and distinct. Twelve adults ages 18 to 25 are randomly selected. The random variable represents the number of adults ages 18 to 25 who think that their generation is unique and distinct.

In this question were posed an experiment and asked if it's a binomial experiment or not, the experiment is we want to select eight households and determined the number of of those households that have a dedicated gaming console. Now, is this a binomial experiment? That's our first question is this'll binomial what we know about a binomial experiment that is composed of n trials? And all of those trials have our independent with the same probability p of success, So we can see right away from the question that there are n trials of something right. We're looking at eight houses. So eight different households now Are those households independent? And do they have the same probability of success? Well, if one household has a dedicated gaming console, do we know that another randomly select household will also have a game in council? Know the probability of one house having a gaming council is not affected by the by the another house having a game in council. So these are in fact, independent. And finally, we need to know that the probability of success is equal for each house. We're told that 49% of households have dedicated gaming consoles, and that is true over all the households. So our probability success P is 0.49 So, yes, this is in fact, a binomial experiment, and we can say what our parameters are. We already noted that we have eight households or eight trials, so n is equal to eight. And we said that the probability of success P is equal to 0.49 And if we want to find Q, which is just the probability of failure, remember, Q is just always going to be equal to one minus p or, in this case, 0.51 So these are your final answers. This is indeed a binomial experiment.

This question proposes an experiment and asks us, Is it a binomial experiment? The experiment it proposes is like a lottery in a lottery. We choose six numbers at random between one and 40 so we have six numbers between one and 40 that we select one and 40. Now. If these numbers are match up with the lottery numbers, then that constitutes a success. So our question is, how many numbers match up with the lottery numbers now? For this to be a binomial experiment, we need a couple things. We need an individual trials, and each of those trials needs to have an independent and identical probability of success. So P is independent and identical. That means if I succeed on my first trial or if I fail on my first trial, it does not affect my probability of success on my second trial. So let's start looking at this. But as it looking at this as if it were binomial experiment and say what would constitute a success, a successful trial? Well, a successful trial would be if I selected a lottery numbers, say, the first number I selected, Um, I selected a three and the actual lottery number was indeed Ah three. So say I selected the three and the actual lottery number was a three. If the's air equal, then that constitutes a success. But say I instead selected a 17 and the actual number waas 32. That would not constitute a success because they were not equal. So success would be choosing the correct lottery number. And since there are 40 to choose from and only one correct, the probability of success here is 1/40. Now we have to make sure that there are and trials here and that those trials are independent. Well, we know the end trials is true because we have Thio select six numbers, right? So for each of the six numbers, we need 12345 and six. We will have the same probability of success because each number is could be the same or different. We don't know. So let's say I chose 35 nine, 40 12 and two For each of those, I have a one in 40 chance of being correct. Now, one more thing we have to look at is, Are these trials independent? Well, what happens does the probability of being correct on number to change if I was correct on number one. Well, whether or not I knew number one or not, I will always still have a one in 40 chance of selecting number two correctly. So that means that one is completely independent of to, No matter what I what happens on number one, I have a one in 40 chance of being correct on number two. So these trials are indeed independent. So we have a six trials, all of which are independent and have an identical probability of success. This is indeed a binomial experiment. So let's start naming our parameters. If we have six trials, we know that end is equal to six, and we said that are probably that probability of success is equal to 1/40. Soapy is just 1/40. Now we have another parameter que, which is just the probability of failure, which is always gonna be one minus p. And in this case it will be 39/40. So these are the parameters of our binomial experiment that goes along with winning or losing the lottery

We want to conduct a chi square test variants and construct a confidence interval as follows. We have a population acts with variance, sigma squared equals 47.1. A random sample we obtained from, the population has n equals 15. For sample size and a square example variants equals 83.2. And why we want to test the claim sigma squared is greater than 47.1 at 5% significance. We proceed through steps a through it, it's all first and a week later, Alpha, which is our confidence or significance level in our hypotheses are known variants in our claim. Thus we have alpha equals 0.5 H nine. Sigma squared equals 47.1. H. A single square is greater than 47.1 and b we calculate our chi square value our degree of freedom and state the assumptions we're making a better distribution. This gives chi squared equals m minus one squared over sigma squared equals 24.7304 The degree freedom is m minus one equals 14. And we're assuming X is normally distributed. Next. We estimate the P value based on archive square value, our degree freedom from a chi square table. This gives P approximately 140.25 or less than 0.25 Thus we can conclude that we reject H not because P is less than equal to alpha, which means that we have evidence to support the alternative provinces H N. Next and two. We constructed 95% confidence interval for the variance sigma square. This interval has the formula on the left. We already know and square where we need X U squared. Nextel square X you an Excel squared or the chi squared values that split archive square distribution, distribution of degree freedom 14 and the two tails with value 20.25 for 95% confidence. Thus excuse square in Excel Square to find on the right taken from a chi square table, And we can conclude that 44.6 is less than chi squared is less than 206.9.

In this question. We want to conduct a chi square test of variance and construct confidence intervals as follows. We have a population acts with variance. Sigma squared equals 1 36.2 and around sample. Besides an equal eight and samples identity, aviation or experience 1 15.1 in one. We test the claims sigma squared is less than 1 36.2 at 1% significance. So we have to proceed through a three to solve for this test first. And a we state the alpha level of hypotheses offers the confidence of significant so all vehicles 30.1 H alpha sigma squared equals 1 36.2 H. A single square is less than 1 36 point to our claim. Next. And be we calculate the chi square value degree of freedom and state the assumptions we're making for our distribution this is chi squared equals m minus one squared over signal squared or 5.92 The degree freedom is n minus one equals seven. And we assume x is normally distributed. Next we estimate the p value and see using our chi squared value for the appropriate degree freedom seven from a chi square table. We obtained P between 70.1 point nine. Thus we conclude in D any that we fail to reject asian off because P is greater than alpha, which ultimately means we lack evidence to support the alternative hypothesis AJ Now we can instruct our 90% confidence interval for Sigma Square. So the confidence interval formula is on the left. We have NNS square but we still need excuse square, nextel square, excuse square next elsewhere to the chi square value for our degree of freedom that split The two tails of the chi squared distribution and the values of .05 or areas of .05. This gives a few squared equals 14.7 X squared equals 2.17 from a table. Thus, we conclude that our interval is 57.3 is less than signal square is less than 371.3.


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