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Chapter 09, Section 9.Z Problcn 0202randam SMp135 observations Droduced samolc Mean population @ sboubor normal;Find the critical Jnm obsurvu valvesfor the followin...

Question

Chapter 09, Section 9.Z Problcn 0202randam SMp135 observations Droduced samolc Mean population @ sboubor normal;Find the critical Jnm obsurvu valvesfor the following test hypothesis using0.02. The population standard deviation KNoxtnJnd theHo:28 versus Hi: / # 28_ Round Your ansvens to twro Jecimal DiacesZeritkal IcrZenteZobscrvca

Chapter 09, Section 9.Z Problcn 0202 randam SMp 135 observations Droduced samolc Mean population @ sboubor normal; Find the critical Jnm obsurvu valves for the following test hypothesis using 0.02. The population standard deviation KNoxtn Jnd the Ho: 28 versus Hi: / # 28_ Round Your ansvens to twro Jecimal Diaces Zeritkal Icr Zente Zobscrvca



Answers

Use Table 5 in Appendix $B$ to find the critical value(s) for the alternative hypothesis, level of significance $\alpha,$ and sample sizes $n_{1}$ and $n_{2}$. Assume that the samples are random and independent, the populations are normally distributed, and that the population variances are (a) equal and (b) not equal. $$H_{a}: \mu_{1}<\mu_{2}, \alpha=0.10, n_{1}=30, n_{2}=32$$

All right, and it's moved to question 1 84 Um, so in this question, our sample size can is equal to 80. The now hypothesis ISS, the population mean is equal Teoh 200. And if we use me to represent the population mean it ISS equipment and to say our no hypothesis is is mu equaled Teoh 200. All right, part be the alternative hypothesis ISS population mean is different from 200. Or we can just employees say our mute is now equal to 200. All right, um parte see, what's the value for Alfa? You hear our alphas? He co two Thera playing there a one because we're interested in on Thea Aeroplane there a one level of significance. All right, parte de, what's the is that value for Alfa divided by two? So here it is, the critical value off Seth for the to tout situation as we desired. So we can go Teoh table for four b and find the corresponding Call him for the, um, Alfa Value. Which is there a point there? What? And we will have the critical value equaled Teoh 2.58 Okay, in part B What is the value from you May sound the null hypothesis is so, as we said in part a, our mu is 200. All right, on, then. What is a value for ex parte here? Export is how a simple mean And according to the question, our simple mean is equal to 205. All right, and, uh uh, this standard deviation, which is sick a lot. Yes, he called to 15. It is given in a question and, uh, cart age ciccolo off export. Or so it is a standard arrow off the mean and it is equal to speak a lot of item by square root off school which is 15 divided by a squared off 80 and IHS, he could chew one point 68 All right. And we need to calculate the past a statistic that star. So according to the formula that star is, you go to your ex four minus the hypothesized value for you over sick. A lot divided by square enough. And and here the denominator is also he quote Teoh seeking law export, which is the center arrow for the mean all right way won't just plug in the numbers, Um, using the original formula. So the simple me is to go five minus 200. Go far 15 divided by a squared. Rode off Katie. And what we have, IHS two point 98 So this is ours. Test a statistic. Oh, here, we need Teoh. Make the decision. Can We've already have the value for the Zet off Gulf over to it as you go to 2.58 And we also have our test, a statistic that star is equal to 2.98 So to make the comparison between though and we found power is that star is greater than the that alfa over to you. Which means Oh, where is that star falls into the critical region. So we're going Teoh reject. Well, no hypothesis. All right. And if we draw a curve, Um imagine the critical value. Sad elf over two is here. And then as he go to 2.58 here we have the negative one negative 2.58 And the area for this How in this towel they ended up together is equal Teoh health, which is their applying? There are one, but our test of statistics that star and somewhere here and as 2.98 So obviously, um, can't falls into the critical ranging. So we're going Teoh reject the null hypothesis.

In this problem, we are asked to determine the critical values for several hypothesis tests about the difference between two means with unknown population variances. So for parts A three D were given the alternative hypothesis. We're given the sample sizes for two samples that are used to test the hypothesis and were given the significance level Alfa. And another thing to note is we're told that the population variances are unknown, which means the difference of our sample. Averages are sample means are distributed according to the T distribution, and so the critical value is based on the T distribution. So for part A, we have a no hypothesis new one minus mewtwo not equal to zero. We have and one is equal to 26 and two is equal to 16 and Alfa is 0.5 So we can look are critical value up in the tea table based on this information. But first, let's find the degrees of freedom. Since we're calculating manually, we're just going to take the smaller sample size and subtract one. So that is 15 and we're also going to note that the alternative hypothesis is, ah, not equal to hypothesis which means that this is a two tailed test, which means that an area of 0.5 it's distributed into both the upper and lower tail of the distribution. So looking this up, we have 15 degrees of freedom, an area of 0.5 and two tails. So are critical. Value is 2.131 and I've made a positive remind us 2.131 because for a two tailed test, we're looking for critical values. So let's say this is positive 2.131 and this is minus 2.131 in the two tailed test. We're going to reject the null hypothesis if our test statistic falls within the critical region in either tail. And so we're looking for test scores that are greater than 2.131 or less than minus 2.131 And for part B. So I'm not gonna write out all the information for Part B. It's just it's an alternative hypothesis of new one. Minus mewtwo is less than zero, so this is a lower tail test. The smaller sample sizes and two is equal to 27. So the degrees of freedom is 26 and the significance level is 0.1 So it's a lower tail test, which means that we're looking for an area of 0.1 in one tail, and the degrees of freedom is 26 and so are critical. Value is minus 2.479 and that's minus because this is a lower tail test. And that's based on the sign in the alternative hypothesis that we're looking for something less than zero. I thought the alternative hypothesis is that new one minus mewtwo is less than zero, so we're interested in the lower tail of the distribution. And for part C, the alternative hypothesis is new. One minus you, too, is greater than zero. The smaller sample size is and one which is eight, which means that the degrees of freedom is seven. Alfa is 0.10 so this is an upper tail test. New one minus mewtwo is greater than zero, so it's seven 0.10 in one tail, and that gives 1.415 And that's positive, since it's an upper tail test and then for party. The alternative hypothesis is mu one minus. Mewtwo is not equal to 10 and we have a smaller sample. Is is en one, which is 14. So that implies the degrees of freedom is 13 and the significance level is 0.5 So again, the alternative hypothesis is new. One minus mewtwo is not equal to 10 so you can have a night prosthesis on the difference of means being not equal to zero are greater than zero are less than zero or any other number it doesn't actually change with. The critical value is it would change with the test statistic comes out as. But it doesn't change the problem in terms of the critical value once it's been standardised to the T distribution. So this is just, ah two sided test again. So going to the tea table, we have 0.5 in two tails then with 13 0.5 and two tails, gives us 2.160 plus or minus 2.16 year old

Four. The alternative, my boss is with me one, because I mean to all for equal to open or one and one is equal to 12 on into is equal to 50. So determined the degree of freedom in the variance. Oh, it's a very incident equals toe equal toe in one waas in two minus tools is 25. If the degree of freedom is the minimum off in one in two. Yes, which is equal to the minimum off that with minus one. So 11 F 15 management from 14 to 11. So the critical value for the responding degrees of freedom on also equal to open 11 tail from table five appendix B. The test is statistics A is equal to 2.718 critical value and for me in a critical value, 2.6 to 4

Problem number five. So I didn't mind the degrees of freedom. The population variances are equal degree of freedom Equal toe in one plus in two minus two is equal to seven plus 11. Minus two is 16 and is the variances are not equal. So the minimum off in one minus 1 56 and then the two minus one, which is 10 toe, actually six. So the critical value in the row off degree of freedom with the current off off ableto open toe 51 teen Using table five, they created a very little to 94.746 and the critical value is equal to negative 1.94 for three.


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