World Series Are the teams that play in the World Series evenl...

Question

World Series Are the teams that play in the World Series evenly matched? To win a World Series, a team must win four games. If the teams are evenly matched, we would expect the number of games played in the World Series to follow the distribution shown in the first two columns of the following table. The third column represents the actual number of games played in each World Series from 1930 to 2019. Do the data support the distribution that would exist if the teams are evenly matched and the outcome of each game is independent? Use the α = 0.05 level of significance.

Table showing number of games, expected probabilities, and observed frequencies for World Series outcomes from 1930 to 2019.

Accepted Answer

Welcome back, everyone. A common experiment in probability involves flipping a coin. If a coin is fair, we would expect a specific distribution of heads, age or tailst in a series of flips. Suppose we flip a coin 3 times. The expected probability distribution for the number of heads obtained if the coin is fair and each flip is independent, is shown in the 1st 2 columns of the following table. The third column represents the actual observed frequency of heads from 100 sets of 3 coin flips. Do these data support the distribution that would exist if the coin is fair and the outcome of each flip is independent? Use the 0.05 level of significance. A says yes and B says no. Now, in order for us to figure out if the data supports the distribution, we can use a chi square goodness of fit test to determine if the observed distribution of heads is consistent with the theoretically expected distribution for a fair coin. And if we're going to use this test first we need to state our hypothesis. So let's let the null hypothesis be that the observed distribution of the number of heads. OK. Of the number of heads. Follows The expected distribution expected binomial probability distribution. OK. For a fair coin. And no, if this is or not a hypothesis, OK, then our alternate hypothesis, HA is basically going to be the opposite of this. So we can just duplicate our statement here in the interest of time, but no, it would be that the observed distribution of the number of heads does not follow, does not follow. The expected binomial probability distribution for a fair coin. And now in this case. That means our decision rule then is going to be that if. Or calculated chi square value. Is greater than our critical chi squared value. Then we can reject the null hypothesis. If it is not, OK, then otherwise we would fail to reject the null hypothesis. So what this tells us so far is that we'll need a calculated chi square test statistic and to find a critical chi squared test statistic based on the parameters in our problem, OK? So let's start first with our calculated value. How do we calculate our chi square statistic? Well, recall that it can be found using the formula that says the calculated value. OK, is going to be equal to the sum of observed minus expected values squared. OK. All divided by the expected values. So what we'll need to do now, OK, is that we'll need to find the difference between the observed and expected values, square them and divide them by the expected values in our table. But the problem is here, if we go back to our table, we only have the probability and the observed frequencies that we can call O. So we'll need to calculate the expected frequencies, OK? Find their difference, square it, and then apply it in the chi square test statistic formula. So, since we have our probabilities here, let's create a column for our expected frequency, OK. And this expected frequency E is going to be equal to a sample size N. That is the number of trials, which is 100 multiplied by the expected probability P, OK? So if we were to apply this to our table here, OK. Then if we think about it, uh, for our first expected frequency for zero heads, it's going to be 100 and multiplied by the probability 0.125, which is 12.5. If we continue this for the remaining values, OK, then for one head, it's gonna be 0.375 multiplied by 100, which is 37.5. For two heads, it's gonna be the same, which is also 37.5. And then for three heads, it's going to be 0.125 multiplied by 100, which is 12.5. So now we have our values of O and E, where all our expected frequencies are greater than or equal to 5, meeting the conditions for the chi square test. Then we can create a column for O minus E. A column for O minus E squared. OK, so let me put that here. And then finally, we can create a column for O minus Esquared divided by our expected value E. And no, OK, let me just make this column here. All we need to do is to fill this out and oops, my apologies, that will help us to figure out our chi square statistic. Our chi square statistic for each value is basically our last column. So let's start with zero heads. Notice here in this case, since our observed frequencies 10, our expected value is 12.5, or minus E would be negative 2.5 squared, that would be 6.25, and when we divide that by the expected frequency of 12.5, then our chi square statistic would be 0.5. If we apply the same idea to the remaining number of heads, when it's one, the, sorry, O minus E would be 0.5 square that's 0.25, so the chi square statistic would be 0.0067, approximately. When it's two heads, O minus E is 4.5. Square that's 20.25, thus our statistic is 0.54. And finally, when it's three heads, that's gonna be negative 2.5 squared 6.25, and the statistic is going to be 0.5. So no, remember the chi square statistic is going to be the sum of all these values, OK? So if we go back to our formula. Then that is going to be equal to 0.5 plus 0.0067 plus 0.54 plus 0.5, which is gonna be approximately 1.5467 because remember 0.0067 was an approximate value. So this is our calculated chi square statistic. This is what we need to compare to our critical chi square value. Now for the critical chi squared value. OK, or it coulda great value. We can use, so let me write this instead using. A chi squared table. OK. Using a chi square table, OK. Then add the degrees of freedom, which would be equal to the number of categories, which is 40123, minus 14 minus 1, or 3, OK. And at a significance level alpha of 0.05, OK. Then, Or critical high squad value. Is going to be equal to or approximately equal to, rather, 7.815. So now that we have both of these values, we can compare. What do you notice? Well, here you probably notice that our calculated chi squad value of 1.5467. Is less than our critical chi squared value of 7.1.815 and based on our decision rule, that means we fail to reject the null hypothesis. In other words, at the 0.05 level of significance, there is insufficient evidence to conclude that the observed distribution of the number of heads is significantly different from the expected distribution for a fair coin. Thus, the observed data does support the hypothesis that the coin is fair and the flips are independent. A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Chi-Square Goodness-of-Fit Test

Probability Distribution

Level of Significance (α)

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