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Suppose that the New England Colonials baseball team is equally likely to win any particular game as not tO win it. Suppose also that we choose random sample of 20 ...

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Suppose that the New England Colonials baseball team is equally likely to win any particular game as not tO win it. Suppose also that we choose random sample of 20 Colonials gamesAnswer the following:(If necessary, consult a list_of formulas )(a) Estimate the number of games in the sample that the Colonials win bY giving the mean of the relevant distribution (that is the expectation of the relevant random vanable). Do not round your response_(b) Quantify the uncertainty of your estimate by givin

Suppose that the New England Colonials baseball team is equally likely to win any particular game as not tO win it. Suppose also that we choose random sample of 20 Colonials games Answer the following: (If necessary, consult a list_of formulas ) (a) Estimate the number of games in the sample that the Colonials win bY giving the mean of the relevant distribution (that is the expectation of the relevant random vanable). Do not round your response_ (b) Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decima places_



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The World Series in baseball is won by the first team to win four games (ignoring the 1903 and 1919-1921 World Series, when it was a best of nine). As found on the Major League Baseball Web site in World Series Overview, historically, the lengths of the World Series are as given in the following table. $$\begin{array}{|c|c|c|} \hline \begin{array}{c} \text { Number } \\ \text { of games } \end{array} & \text { Frequency } & \begin{array}{c} \text { Relative } \\ \text { frequency } \end{array} \\ \hline 4 & 20 & 0.20 \\ 5 & 23 & 0.23 \\ 6 & 22 & 0.22 \\ 7 & 35 & 0.35 \\ \hline \end{array}$$ Let $X$ denote the number of games that it takes to complete a World Series, and let $Y$ denote the number of games that it took to complete a randomly selected World Series from among those considered in the table. a. Determine the mean and standard deviation of the random variable $Y$. Interpret your results. b. Provide an estimate for the mean and standard deviation of the random variable $X .$ Explain your reasoning.

Consider a binomial experiment with our equals 3139 successes and and equals 5792 trials we want to construct a 99% confidence interval for the population proportion P. We proceeded the following three steps below to complete first step a We check the requirements to use a normal distribution to approximate the confidence interval. P So the testes is 60 hat is all over and equals 600.5419 mp. Had equals 31 39 to have equal 2063 are both greater than five to the requirements have in fact met next. We construct the interval, we have margin of error equals critical Z score rupee hat you had over end Since we are 99% confidence the Z score is 2.58 which means our margin of error is 0.169 which means that our confidence interval is P hat minus e. 4.2 point 5 to 51 is less than P is less than 510.5588 Finally, we interpret this to mean that we are 99% confident. The probability of success in a random trial is between P hat plus or minus.

Okay, So for this question, we're told that we have normally distributed random variable with me and zero standard deviation 13.861 And we won't know the probability that X is greater than negative three. So this is equal to one, minus the probability that X is less than negative three. And so, if we standardize are very well here, this is the probability that a standard normalcy is less than or equal to make a three minus 13 point 861 She is negative. Three minus zero, divided by 13.861 s O negative three divided by 13.61 gives us point negative point to 164 So we're gonna solve this by looking up our value on RZ table. We'll need the negative side of it because we've got a negative number here. It's let's side of the way a little bit and we're looking for negative 2.2. If we round Negative too. 0.2, this here. So one man s point for 1 to 9, and this would give us 0.5 eight 71 as the final answer

We want to conduct a hypothesis test dealing with some data collected from Major League Baseball World Siri's um instead of the chart going in a more horizontal direction, I'm gonna make it go in a more vertical direction. So sometimes world Siri's lasts only four games. Other times they last five games. Sometimes they last six games, and sometimes they last seven games. So the study was to take 105 Major League Baseball world Siri's and break them down by Was it over in 24 games, five games, six games or seven games, and the observed data for the contests were that 21 of those 105 games ended after game 4 23 ended after game five 23 ended after Game six and 38 ended after Game seven. And that is what we would refer Thio as are observed data. Now we have some expected proportions. We expect two out of every 16 to go on Lee four games. We expect four out of 16 to go five games. We expect five out of 16 to go six games and we expect five out of 16 games to go seven games and again. Our goal here is to run a hypothesis test. Well, in order to run a hypothesis test, we are going to have to construct, um, are null hypothesis. And our alternative hypothesis so are no hypothesis is going to be that the actual number of games fits the distribution by the expected proportions. Yes. And the alternative to that, or the alternative hypothesis would be the actual number of games does not fit the distribution mhm by the expected proportions. And in order to run this, we're going to be running a goodness of fit test. And to do a goodness of fit test, we're going to have to find a chi square test statistic. And the Chi Square test statistic is found using the formula. The sum of observed minus expected quantity squared, divided by expected. So we're going to have to go back up to our chart, and we're going to have to calculate the expected number of games, not just the proportion. So we want expected games. Well, if we are expecting two out of 16, then how Maney would that have been out of 105? So if I do my cross products, I'd get 16 X equals 210 so the expected number would have been 13.1 to 5. Then I'm gonna do the same thing for the 4/16. I would expect four out of every 16 games or world Siri's to go to five games. So how much is that out of the 105 that I included in this study? And if I cross multiply that I would get 16. X equals 420 and the result would be 26 point 25 And then I'll do the same thing for the five out of 16. So if we had five out of 16, that would be what out of 105. And again, if I cross multiply 16 X equals 525 and I get an expected value to be 32.8125 And that's the same for both lasting six games and or seven games because they had the same expected proportions. So we're ready to find that Chi Square statistic, and we're going to have to create a new column in our chart, and the new column is going to be called observed minus expected quantity squared, divided by expected. So in order to calculate those, the easiest way to do it is to put the information into our graphing calculator. So I'm gonna bring in our graphing calculator, and I'm going to clear anything that I might have in my list. So I know I've been using list one list to enlist three quite regularly, so I'm gonna make sure they're clear before we start. So it hits, stat, mhm and edit. And I thought I cleared out my lists, but let's try it one more time. There we go. So we're going to put the observed values into list one? Yeah. And we're going to put the expected values into list too. Yeah. Mm hmm. And we'll put the, um, stuff there. Let's try one more time. 32.8125 and then the final one. Okay, so we have all our data, so we want the calculator to give us information into list three. So we're going to sit on top of list three, and we're going to tell it to take all of the values, the observed values that we've placed enlist one, subtract all the expected values that we've placed Enlist to square that difference and then divide it by all the expected values enlist to, and we're going to get a variety of decimal answers, and I'm going to write them down in the chart, and I'm going to go out to three decimal places with each. But I'm going to keep the full decimal in the calculator. So we will have for the 21 games, will have four point 7 to 5 for the 23 games will have 402 for the 20 or for six games or the Siri's lasting six games, we'd get 2.934 and for Siri's lasting seven games, we would get 70.8 20 Now, remember what the Chi Square test statistic said to Dio. We need to add up these values, and the easiest way to add those values up is to bring our calculator back in, quit out of the list mode and ask second stat math. Let's sum up all those values that were enlist three. And in doing so, we get a Chi Square test statistic as a dust small of 8.8819 so our Chi Square test statistic is 8.8819 Now. The next part of the hypothesis test is to calculate a P value and to calculate RPI value. What we're really asking is, what's the probability that Chi Square would be greater than that test statistic? And I always like to draw a picture to kind of model that. So we know we are using a chi square distribution and chi Square distributions are generally skewed to the right, and they are based on the degrees of freedom and our degrees of freedom are found by doing K minus one. And K represents the number of categories that we have divided our data into. And if we go back to our chart, we see that we have broken our data into four different categories. How Maney of the Siri's lasted four games, how Maney lasted 56 and seven. So our K value is going to be four, resulting in our degrees of freedom being three. And that degrees of freedom also tells us what the average of the distribution is. So our average of this chi square distribution is going to be three and will always find our average slightly to the right of the peak. So down here on our horizontal axis are Chi square axis. We would find three in that location. Now to find that p value we're looking for. What's the likelihood of the probability that Chi Square is greater than this value? Well, that value would be further to the right from three, and we'd have eight 0.8819 and we're looking for the likelihood or the probability we're running greater than that. So this would be where the RPI value would be, and it's It's that area in that right tail. And the best way to find that area is to use the chi square cumulative density function. And when you use that function, it asks you for the lower boundary, the upper boundary and then the degrees of freedom. So in our instance, our lower boundary is the 8.8819 The upper boundary is found way out in that tail, and that tail, um, keeps going. So we're going to use a very large number. We're gonna say 10 to the 99th power and then our degrees of freedom we calculated to be three. So I'm going to bring in my graphing calculator again, and I'm going to show you where you can find that Chi Square distribution. So you're going to hit your second button and your Vares button and it's number eight in my menu. And again, it asks for that lower value. That lower value was 8.8819 We said our upper value is going to be 10 to the 99th Power, and our degrees of freedom was three. So our P value, where the probability that the chi square would be greater than 8.8819 ends up being 0.309 We're almost done with our hypothesis test. Now that we have a P value, it's time to make our decision. And we want to run this hypothesis test at a significance level of 0.5 And we describe our significance level utilizing the variable Alfa, and we will make the decision to reject the null hypothesis. If that's the statement, if your Alfa is greater than your P value, So in our instance are Alfa was 0.5 and it is indeed greater than RPI value because our P value was 0.309 So our decision will be that we're going to reject the null hypothesis. So let's go back up to that null hypothesis and we're rejecting it. So basically, we're throwing that statement away. And by throwing that statement away, then we're throwing our support toward the alternative hypothesis, so we can then make the conclusion that there is sufficient evidence right to conclude that the actual number of games played in a major league baseball world. Siri's does not fit the stated expected proportions. So we're saying that those


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