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7.211Compute tne test statistic_ (Round to two decimal 5 aces (eeded;,/Using evel of significance 0f0 05, wnat isthe critical value? (Round two decima 302sQeeded;Fi...

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For Exercises I through 25, perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Sick Days A manager states that in his factory, the average number of days per year missed by the employees due to illness is less than the national average of $10 .$ The following data show the number of days missed by 40 randomly selected employees last year. Is there sufficient evidence to believe the manager's statement at $\alpha=0.05 ? \sigma=3.63 .$ Use the $P$ -value method. $$ \begin{array}{cccccccc}{0} & {6} & {12} & {3} & {3} & {5} & {4} & {1} \\ {3} & {9} & {6} & {0} & {7} & {6} & {3} & {4} \\ {7} & {4} & {7} & {1} & {0} & {8} & {12} & {3} \\ {2} & {5} & {10} & {5} & {15} & {3} & {2} & {5} \\ {3} & {11} & {8} & {2} & {2} & {4} & {1} & {9}\end{array} $$

In a problem. Eight. We're going to be comparing the number of games one by each league. We have two legs here the American League and the National League for Baseball games for the years 1970 to 1993 at the 0.5 level of significance, We're going to see whether there is sufficient evidence to conclude that with a difference in the number of kings. So the first step is to state the non Harper. This is then get the critical values wrong. The data then one called the values that to make a decision, whether to reject or real, to reject the night policies. So in our case, the null hypothesis change not is that there is no difference in the number wins and the alternative put. This is a case that there is a difference in the number of wins and both cases comparing the theme to leaks. But next we just did the critical Bali Andi, In our case, the critical value at 0.5 level of significance is plus or minus 1.96 Next, we're going to give the ranking full each a win, and he had the rankings for the American League and for the National League. So after getting the ranking, we need to find out the sample size of a small A sample. And in this case, the smaller sample is for the national legs. Onda. We have 11 wins and 11 samples and then for the American League, we have 12. So we're going to get Are we going to sound the rungs for the National League? So it's going to be 2.5 sound all the way out to 22.5 and the sum is 125. Next, we get Newmar from substituting the values off anyone and entered into the formula. And that yields 132. You are is obtained by substituting evolution to this formula, and we get 16.24 the value of Zell calculated values that is going to be obtained from the substitution of the values that we have just obtained. That is 120 five minus 132 divided by 16.24 And when you want that, don't get negative 0.43 08 now we can compare that one, but you off that to the critical value offset and the negative. Negative. 0.43 08 He is great. German negative. 1.96 If you can sketch the distribution, you find that the critical value is negative. One 1.96 And this is the critical region. While I watch critical, our calculated value is zero negative 0.4. So negative 0.4308 does not fall in the critical region and hens we make the decision to feel to reject. Then my life with this is And in this case since we've failed to reject the knowledge by this is we can't conclude that these not enough evidence to reject the claim that there is no difference in the number winds in the tunings.

Little this is from 13.31. We're going to be using the chi squared goodness of fit test. We'll be talking about dice one day in particular. The no hypothesis is that to die is not voted and the alternative is that die he's loaded are alpha value is 0.05 and now we need to figure out the probabilities. So the probability of landing a die being ruled in landing on a one is the same, is the probability that he lands on to That. It lands on three for five and six When the Dye is not loaded which is 16 since we have a six sided die, The values are 1, 2, 3, 4, five and six and uh 150 trials were done and out of those this is observed And it was 150 trials 23 Times of London on one 26 on two 23 on three 21 on 431 on 5 26 on six. No, the expected counts is equal to the sample size times of probability, Which in this case it's 150 trials multiplied by the 16 for which is probably That die lands on one and simplifying we'll get 25, you would have to do the same thing for everything else. So This would be 25 as well and so on. So that's right that down. Okay now we need to find what chi squared is. So chi squared will be good to some value. We need to figure out where the degrees of freedom are and we need to find the p value which is denoted as probability that okay squared is greater than a certain value. It's equal to something. Okay. In order to get the chi squared value though, we need to figure out what will be actually using these numbers right here. So we need to input the observed values and the expected values in a website called stacked ology dot org. And after that, after we get the chi square value and we have the degrees of freedom or use the other website called Salt by single judge. Learning to get our p value. Okay, let's get to it. As you can see on this first part we have All the observed counts are inputted in the expected values and our chi squared test statistic is 0.88. As shown, We'll get your .8 or degrees of freedom is just the number of categories which are six And subtracted by one should get is five. And we need to find the probability that chi squared is greater than The Chi Square Statistic, which is your .8. We use the website called salt and as you can see we have in put it that we have a chi square distribution Degrees of freedom are five And the probably the X. is greater than or equal to 0.88 Is going to be 09 seven, 0.9. Now we need to interpret what is happening because pierre value Is equal to 097 which is growing there equals 0.05. We do not reject. Was it no, meaning the data do not provide sufficient evidence to conclude that die is loaded.

Critics were trying to test whether or not the median of paid attendance at a certain athletic event is 3000 or not. So, um, we going we're going to state our hypothesis and step one. So one step one, we have our hypothesis are hor uh, the meeting that we believe is true is the median attendances 3000. So it's this one here and are all deny. Hypothesis is the median attendance is not gonna equal to not be able to 3000. So that's this one here. All right, in step two. That too, then goes on to talk about the critical value of the test. So in step two, we have the critical value, or CV. Just remember, anytime I use C v, I'm talking about critical value where end is going to be 20. The Alfa level the significance level. You're looking at this point, all five. It's a two tailed tests, and again, the reason why this is going to be a two tailed test is because you're using this not equal to signs you're looking at both sides of the 3000. Um, according to Table J. When you look it up, it's equal to five. So the critical value is going to be five. All right, Step three, we're going to try to find out what the test value is. So what you do is you take each data point and I'm not gonna go through all but I have them listed the same way they're listed in the question. So the first data point of 6000 210 And since that's greater than your 3000 it gets a plus, I But that's this one here. So you get a plus sign. The next one is 31 50. Since that's greater than 3000 it gets a plus time. The next one is 2700 since it's greater than. But since it's less than 3000 that one gets a minus sign so you can continue with that. So the next two are both greater than so pluses. In the next row, we started with 35 40. It's a plus, a plus, and then you get your three minuses for the next three pieces. We started with three minuses in that third row. Then we have 20 3700 which is a plus 6030 years of plus and then finally let last rope. It takes on these values and then what you're supposed to do. And this is what makes the scientists so easy is or simple is you just simply count up your pluses. And you have 10. You can up your negatives, you have 10. And so you're gonna take the less of the lesser of these values. They're both the same. So you're gonna take the value itself, and we get the true value for the test value of the statistic to be taxed. All right, so now we move on to step four, and we have to make a decision. So since the test value is equal to 10 which is greater than the critical value, which is equal to five, we do not reject HR. So there's our decision. Um, and then the next piece instead. Five. We want to make our

For this question, we have eight randomly selected identical twins for whom there post rates were measured. And we're interested to know at this significance level of 0.1 if their post rates differ and were also asked to construct a 99% confidence interval for the difference. So we're given Alfa is equal to 0.1 so we can start off by declaring our hypotheses. No hypothesis is that there is no difference. So the mean of the differences where the sample is zero, and the alternative hypothesis is that there is some difference so that the mean of the differences is not equal to zero, since we're using the P value method for this question, which is requested in the question, we start with our test value, and that's given according to this formula. And so we have end the sample size from the question that's eight, and so we need to find the average of the differences from the sample and the sample standard deviation for the differences. So I've done this in Excel, So I copied and pasted the information from the question into these two columns and then in the third column. I calculated the differences between the first and second column and here I've averaged the differences using excels, average function and in the cell, I calculated the sample standard deviation for the differences using excels, standard deviation function for samples. So we have 1.25 divided by 3.62 over the square root of eight, and that comes out to zero point 978 So now if we want to estimate the P value from a table, we can estimate the degrees of freedom as being and minus one just seven. So we're going to use a significance level of 0.1 It's a two tailed test and the degrees of freedom is seven and the test value is 0.978 which means that the area and two tails is going to be something greater than 0.2. So now we simply compare our P value to Alfa and we could see that it is bigger than Alfa, which is 0.1 and therefore we failed to reject the no hypothesis and we can make a conclusion by saying that there is insufficient evidence to support the case that there is a difference in the post rates of identical twins. And now the next step in the question is to create a 99% confidence interval for the difference. So that is given by this equation. It's the average difference, plus or minus T sub alfa over to times the different sample difference standard deviation over the square root of the sample size. So we already have this value, which is right here, and we have sample standard deviation and we also have the sample size. So we just have to find t sub ALF over to. And we find this using our degrees of freedom, which we found to be seven and are a significance level for a 99% confidence interval. Alfa is 0.1 So moving back to the tea table we have seven degrees of freedom and an area of 0.1 in two tails, which means that T sub ALF over two is about 3.5. So now we substitute in these values we get 1.25 plus or minus 3.5 times 3.62 divided by the square root of eight and this gives us a range for the difference of minus 3.22 to positive 5.72 So we can also stated like this minus 3.22 is less than the mean difference, which would be lesson 5.72 And we can make a summary statement by saying that we're 99% confident that the true difference in pulses between identical twins is between minus 3.22 and 5.72 and because this range includes zero, it also agrees with their decision to not reject the no hypothesis.


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