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Test the claim that the mean GPA of Orange Coast students is significantly diflerent than the mean GPA of Coastline students at the 0.F significance level.The null ...

Question

Test the claim that the mean GPA of Orange Coast students is significantly diflerent than the mean GPA of Coastline students at the 0.F significance level.The null and alternative hypothesis would be:Ho: po > Uc Ho : Po <Pc Ho:Po = Pc Ho:Po > Pc Ho: po < Uc H : po < #c Hi:Po > PC Hi : Po 7 Pc Hj : Po < Pc Hi: #o > #cHo: po = Uc Hj : po # #cThe test is:left-tailed right-tailed two-tailedThe sample consisted of 45 Orange Coast students, with sample mean GPA of 2

Test the claim that the mean GPA of Orange Coast students is significantly diflerent than the mean GPA of Coastline students at the 0.F significance level. The null and alternative hypothesis would be: Ho: po > Uc Ho : Po <Pc Ho:Po = Pc Ho:Po > Pc Ho: po < Uc H : po < #c Hi:Po > PC Hi : Po 7 Pc Hj : Po < Pc Hi: #o > #c Ho: po = Uc Hj : po # #c The test is: left-tailed right-tailed two-tailed The sample consisted of 45 Orange Coast students, with sample mean GPA of 2.77 and standard deviation of 0.07, and 45 Coastline students, with sample mean GPA of 2.75 and standard deviation of 0.02 The test statistic is_ (t0 2 decimals) The p-value is: (to decimals) Based on this we: Fail t0 reject the null hypothesis Reject the null hypothesis



Answers

Testing Claims Using P-Values $(a)$ identify the claim and state $H_{0}$ and $H_{a}$, (b) use technology to find the P-value, (c) decide whether to reject or fail to reject the null hypothesis, and (d) interpret the decision in the context of the original claim. Assume the population is normally distributed. Class Size You receive a brochure from a large university. The brochure indicates that the mean class size for full-time faculty is fewer than 32 students. You want to test this claim. You randomly select 18 classes taught by full-time faculty and determine the class size of each. The results are shown in the table at the left. At $\alpha=0.05,$ can you support the university's claim?

20 every day. The non hypothesis is that new one in smaller than or equal to Muto and the alternative high Busses that me one is bigger than mute. So the critical values are the values in Table four, corresponding to mobility on 1.9. So that is even toe 1.3. So the rejection reasons contain all very is larger than 1.28 So this tested statistics X one bar 162 minus mu one minus muto over squared off symbol on the square over anyone plus Sigma squared over and to which approximately equal to four point always so if the value of testing is within the rejection rating than not, have also rejected. So as UH, two point or seven is bigger than 1.28 So you reject the not hypothesis, so there is sufficient evidence to support that.

18. So it's no if that new one is equal to mean to, and each one is that anyone is not equal commute. So the degree of freedom, the minimum off in one way, this one, which is 17 and 18 to minus one, which is 19th. So the video is 17, using all for April toe 0, 00.11 teen. So for table five, the critical, very equal toe postive or negative 1.74 So the rectory, then they contain all value smaller than negative 1.74 and all various larger than 1.74 So that is the statistic, if that makes one minus x two bar. So it's 56900 minus five 7800 minus me one minus me, too over square root off this one squared over in one, which is one toe. Once you lose your square over and once with the 18 lost, it's too square. So it's 8000 square over into approximately negative 4.267 So it's a very off statistics. Even in the final hypotheses rejected so and this value realize between the booze values. So we failed to reject the null hypothesis

Let us read this question. A random sample of 50 medical school applicants at the university has a mineral score of 31. So x bar is 31. My end is 50 on the multiple choice portions. Medical college admission test. Okay. A student says that the main draw school for schools applicants is more than 30. Is there enough evidence to support the students? Cling. Okay, So a student says that the mean raw score for its schools applicant is more than 30. He's saying that it is more than 30. So my new is more than 30. So let this be my alternatives. So what will be Mindulle? It is less than or equal to 30 right? This is my age. Not This is my ex alternative. Okay? Yeah. Population Standard division is 2.5. So let me simply do one thing. Let me directly like the formula. Now, this is going to be 31 minus 30 upon Standard Division is 2.5. This is 2.5 upon route over 50 upon router 50. Right. This is going to be route off 50 on If I use a calculator for this, this is going to be one divided by 2.5 multiplied by route off 50. It is 2.82 My statistic. This is my Z value for my Zeev value is coming out to be 2.8 to 8. Now, this is a right tail test because I'm checking for greater than 30. So this will be a right tail test. And my Alfa in this case is 0.1 And what is my critical value? 2.32 This is 2.32 This is my critical value on my critical well on myself. Statistic that I've calculated falls in the rejection reason off. Two point it Do it. So I will say that I will reject my age, not reject h not. Which means I would say that the claim that is being made by the student is actually correct. And this would be my answer

For this problem. We are given the following claim machines suspense amounts with standard deviation greater than 0.15 oz and we have the following information off to the side. So the first for this hypothesis tests the state or no hypothesis. So each not. We're talking about standard deviation. So sigma And it's gonna be our 1.5 oz are alternative since the claim is stating greater than it'll be sigma, Somebody greater than 0.15. So once we have our go on alternative hypothesis, let's find our testes cystic. Since we're talking about sanity, aviation, we're gonna be using the chi square distribution and the chi square testes cystic can be found by and before end minus one times as squared divided by sigma squared. Yeah, So this is going to be 27 -1. Yeah. Time 0.17 Squared Divide by 0.15 Squared. And the chi square tests should be approximately equal to 33 points 396. So I was the testes cystic, you can do the p value. There's two ways to do the people, you can either do since the sample size 27 we have a degrees of freedom. 26 But in the end -1. Since this degrees of freedom is relatively small, we can find it within the table or we can find the inequality arranged within the table. And since this is a right to test and the high score table, I'm using finds area right to the tip. Other critical values what we wanna do. We don't want to go down the degrees of freedom. Do we get to 26? We want to find the range where 33.396 is between the two numbers. So I found that is between 17 point 292. And it's gonna be less than 35 point 563. So now if we go all the way up we'll find that probably are the area to the right of the curve. So the 17.292 equivalent to 0.935.563 equivalent to 0.1. And now we have our P value. So RP values between 0.9 or 0.1 and 0.9. A more accurate where you can get your P value though is using software is like our you use the Arco p chi square which does it take the input of your chi square testes, cystic degrees of freedom and then lower dot L. Equals true or false. So the lord dot tails telling us whether it's a left tail test or right tail test. Since the right tail test, you want that to be false. If you plug in this information we should get to this p value is about 0.1 509. Which again agrees within the table method table inequality. So if you have the p value can make a conclusion by comparing your p value. What's your alpha. But you can also find your critical values. To find your critical value of critical values. Your again going to your chi square table is gonna be Since we're doing air to the right we just want our alpha. And so we are degrees of freedom. Remember that our alpha is 0.05. Our degrees of Freedom 26. So if we go to our high score table we can find were those two intersect and are critical values. 38 .885. All right. So, you can make conclusion with the critical values and your testes cystic. If you have a chi square distribution looking something like that, you're critical value. The states here Just 38.885. Okay, so you're critical value. We will reject any or don't have process. If it falls within this blue region, this is also called our rejection region. And if it falls anywhere within the non stated region, we will accept it. Which is also called our acceptance region. We see our test statistic as 33.396. So if you try to play set somewhere here that maybe belong around over there. Yeah. So that's archive square test. Yeah. And it falls within our acceptance region. So in this case for comparing test statistics and critical values we fail to reject If you want to compare with the P value and alpha. So a p value is greater than alpha. We failed to reject. H. Not. Yeah. and if P value is less than alpha, we reject H. Not again, we can look really quickly. We have AP value of 0.1509. And after a value of 0.05, our p value is greater than our alpha, so we failed to reject the null hypothesis. So in both cases we came to the same conclusion, which was we fail to reject H not.


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