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To determine if this distribution is Poisson, we use the chi-square test for goodness and ? = 0.01.

Step 1: To calculate chi-square, we need to know the number of calls per 5-minute interval. This is determined from the data by dividing the number of calls by the number of minutes in a 5-minute interval. In this case there were 9 calls in 5 minutes. This gives us a chi-square value of 0.09.

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Springfield Emergency Medical Service maintains records of emergency calls. A study of 150 five-minute time slots gave the following distribution of the number of calls. For example, no calls were made during 18 of the five minute intervals. Use the chi-square test for goodness and ?=0.01 to determine if this distribution is a Poisson distribution. Find the observed chi-square value. Round answer to 2 decimal places.Number of calls (per 5 minute interval)Frequency0181282473214165116 or more9

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Hello, in this video we are going to focus on the records of emergency care and waste in the study. A time interval of 150 gives us the distribution of a number of coins, so we need to check if the distribution is poisonous, and we also need to find the value of c squared. So here you could write the null hypothesis that the distribution follows the poison and it's going to be here and we can here that it's going to follow and so this is going to be the poison. Now, here we will also talk about the alternative hypothesis. So the alternative hypothesis is like this, and it's the exact opposite of the null hypothesis, so here the distribution doesn't follow what's stated in the null hypothesis. So we can write here, no poison follows. So this is our null hypothesis and we can say an alternative hypothesis. Now let's see what the lambda value is here, so the lambda value is nothing but the sum here will be f times x divided by will be the sum of f. . Where f gives us the value of the frequency and x the number of calls, here precisely this value will be such that we know that the frequency times the value of x is 0 18 Pit. So it's going to be, and we know that if the value of x is 1, the frequency is 28. And likewise, it's going to continue to the last term, where the last term is a number of calls, which is 6 or more, and basically it's going to be here 6 times. We will have 9 where the frequency is 9 points. So in the denominator we are now going to get the value of the sum of f equal to 1 and 50 so the lambda value here will be equal to 2.4 and now we need to first compute the hypothesis probabilities related to each class using Poison Distribution. , where the parameter value is equal to 2.4, so here we assume that it follows Poison. What happens then, and in this case we can write. The probability of x is equal to taking a smaller x. So here's e to the negative lambda and lambda to the x divided by x, factorial. So, based on that exact information, let's calculate the values of p 1 p 2 and here we can write p. 1 here is no more than equal. Basically, we have to start with p 0, so it will be here, p, 0 and it will be here. The probability of x is 0, and here we have a power of minus 2.4, and here is 2.4, and here is x. So the value of x is 0 and the denominator is 0 factorial and this is going to equal 0.090 7 here. And now we can say here that the value of p 1 is equal to nothing more than the probability that x is equal to 1, and here it is equal to e raised to the negative power 2.4. And here it is 2.4 raised to 1 in the denominator. We're going to have 1 factorial, so this will give us the value which is basically 0.2177 and we can similarly calculate all the other values. Now just write the values, so it will stay here. P, 2 and p. The value 2 is nothing more than the probability that x is equal to 2, and if we insert that exact information into the equation we wrote, it will give us 0 points. 261. 3 and similarly we can talk about the p-value of 3 here, so the p-value of 3 is nothing more than equal to the probability that x is equal to 3, so here it will be equal to 0.2090, and so here we can write the p 4 value which is going to be equal to the probability that x is equal to 4 and which here is basically 0.1254 and now we can write the p 5 point value here. So the value of p 5 is nothing more than the probability that x is equal to 5, and therefore here it is equal to e, again raised to the power of minus 2.4 points. Let's write the formula down once so we don't forget, so here it goes, 5 divided by 5 factorial and so it equals 0.060 2. And now we can say here that p has a value of 6. Just focus on that p-value of 6, which here is basically the probability that x is greater than or equal to 6, correct. It could be written as 1 minus the probability that x is less than or equal to 5, so it will be equal to 1 here. Minus the sum of x equals 0, and here it's 5. Now it's going to increase e by at least 2.4 here, and then it's 2.4 to the x here y in the denominator. We have x, factorial, and it can be written like this. This expression will be right here, equal to 0.0 3 5 7, and that's the p-value of 6. Now let's try to find the value of the expected frequencies of each class, so the expected frequencies of each class the formula, which can be written as ei, and that equals n, which is the sample size multiplied by p. So here we can see that the value of n is not more than equal to 150 because that's our sample size and there are so many frequencies that it just denotes the sample size. So there shouldn't be any confusion about its value now. Just like we did for the p 1 p 0 values, we need to write to 0 and then we need to write to 1, and it goes to e 6. So here it's going to be 150 times the p 0 value, which is 0.0907, and there will be about 13,605 coming in, and we can similarly write the values directly. So here is 32.6 and 5 and then here we get the value of e 2 as equal. Basically here it's going to be 39,195 and now it's going to be a 3 here and this is equal to 30, 1.35 and now here we have a value of 4 and this is going to be equal to 18 point 81 and now we can get the value of a Type 5 and a 6, then the value of i will be here, equal to 9.03, and the value of 6 will be here, equal to 5.355, and so on based on this exact information. We can calculate the chi square sub value here. Sum, what is the subtotal of chi-square? Here it is equal to or minus a divided by e, where o is the observed frequency, which we used to call f. This will be exactly that expression, and we know that the chi-square test statistic is nothing more than the sum of these values, which is the subsum of chi-square. So here we can write that cy squared is equal to the sum of or minus a divided by a to the power of 2, and it's going to be the same here. We can write these values. Basically, the first term is 18 and here it's negative 13,605 and it's raised to the power of 2 plus here. It will be here, 28 minus 32,655, and it will be raised to the power of 2 here and it will continue to be written. So finally we can write that we get the expression, so we forgot to write the value of e. Basically it will be 13,605 and this will be 30 2,655 and now we can write the last term here. So the last term will be here: 9 minus. It's going to be 5.355 here, and it's going to be raised to the power of 2 here, divided by 5.355, and then the value of this chi-square here is going to equal 10.3 85, and then we have the c-squared value, which used in this very problem it was vaulted so we can put it in a box to make it stand out and now we can talk about degrees of freedom so degrees of freedom has the formula which can be written here as degrees of freedom equals K minus 1 and then it becomes here: 7 minus 1, which equals 6, so now we use the chi-square table and the degrees of freedom value is 6. The critical value obtained at the level of significance of is here assuming the significance level is 0.01. In fact, given the problem, there's no need to assume that. In fact, now we can say that the value of the fundamental critical value here is going to be equal to 16.812, and this is our critical value. It can be replaced here as a critic, and then it will pay off here. So from this information, we can say that this is the calculated value that we just calculated. It's 10 points. 35385 is less than the critical value, which is 16 points. 812 the decision is that we do not reject the null hypothesis at this same level of significance. So let's write the decision here and that decision will be that we don't reject the null hypothesis. This is h, zero given the alpha value of 0.01, from which it can be concluded that the distribution of a number of calls follows the distribution of poison. So here we can finally write the answer. From there the distribution follows and so it will stay here and we can write again the chi-square value that was requested, then chi-square value. We hit 10,385 here and this and everything will be the answer to this very problem and we can put this in a box to highlight it and it will look like this. So that's it for this Villeo. Thank you so much.