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Under load for two different types of plastic pipe being The deflection temperature investigated Suppose we need to test whether the mean deflection temperature f...

Question

Under load for two different types of plastic pipe being The deflection temperature investigated Suppose we need to test whether the mean deflection temperature for type; 2 pipe exceeds that of type Two random samples of pipe specimens are tested, and the deflection temperatures observed are as follows (in %F)Type 1: 180, 185, 183,177Type 2: 200, 205,98, 202, 205(Use the style (0.00_ 0.00) in your solution) The two-sided 95% confidence interval on the difference in mean for the two typesApprox

under load for two different types of plastic pipe being The deflection temperature investigated Suppose we need to test whether the mean deflection temperature for type; 2 pipe exceeds that of type Two random samples of pipe specimens are tested, and the deflection temperatures observed are as follows (in %F) Type 1: 180, 185, 183, 177 Type 2: 200, 205, 98, 202, 205 (Use the style (0.00_ 0.00) in your solution) The two-sided 95% confidence interval on the difference in mean for the two types Approximate your answer decimals The test statistic Approximate your answer deciqals Reject the null hypothesis



Answers

The mean water temperature downstream from a power plant cooling tower discharge pipe should be no more than $100^{\circ} \mathrm{F}$. Past experience has indicated that the standard deviation of temperature is $2^{\circ} \mathrm{F}$. The water temperature is measured on nine randomly chosen days, and the average temperature is found to be $98^{\circ} \mathrm{F}$. (a) Is there evidence that the water temperature is acceptable at $\alpha=0.05 ?$ (b) What is the $P$ -value for this test? (c) What is the probability of accepting the null hypothesis at $\alpha=0.05$ if the water has a true mean temperature of $104^{\circ} \mathrm{F} ?$

All right. So we've got an all hypothesis of the ALPS, Our true mean population mean is going to be 100°F. Mhm. And the alternative that we're testing for is that it is going to be greater than 100°F. Yeah. And past experience tells us the standard deviation is 2°Fheit and that's gonna be it. So let's get into our test statistic that's gonna be our sample average minus our population average divided by population standard deviation divided by the square root of our sample size. So it's gonna be 98 -100 Divided by two and the Square Root of nine. And that is going to be a -3. So we'll go ahead and find out what our test statistic is. Critical value rather. And at the 5% significance level We're gonna find the score of 0.05 and it's 5% at 1.64. And so that's going to be a rejection region. And uh okay, it's gonna be some pretty strong evidence since we're looking at a right tail test And are critical value is 1.64. And we're landing at -3. So we definitely failed to reject the null hypothesis. And then for part B we can calculate that p value. So that's gonna be the p value is equal to 1- our feet of 1.64. That's gonna be one minus .014. So that's gonna be like 0.9 996 or 86 Which is uh greater than 5%. So we can nicely fail to reject the null hypothesis as well, and the probability of accepting the non hypothesis okay, At the 5% significance level, so it's going to be an application of our beta, so it's gonna be paid up is equal to our feet of our Z score at the significance level minus the delta square roots of and sample size divided by the population standard deviation. So this calculation is going to come out to be uh range like this 1.64 from before, and our delta is going to be the 4°, it is above 100 so four and r squared of nine as usual, and population deviation of two, And that's gonna be the feet of 1.6, 4 -6, which is equal to negative 4.36. And that is basically zero. We don't go out that far. So that's our probability of accepting the non hypothesis.

On all hypothesis is that the population standard deviations are the same. Our alternative is that the first one is greater than the second. We will be conducting this test at the 1% significance level. R F statistic is going to be s one squared divided by S two squared, This is going to be about 3.41 In Table eight. We can figure out our degrees of freedom. Our first degree of freedom is N -1, 31 -1 is 30 And our 2nd 1 is 16 -1, 15. This will be inputted as 30:15. We find out that F is at 10.1 is 3.21 321 is less than 341 For for reference, this is our f curve 321 and three or 3.41 is in this shaded region. Therefore, We reject the null hypothesis at the 1% significance level. The sigma population one is greater than the standard deviation of the population. To How I can calculate a 98% confidence interval. What we can do is find out the f of 0.01 which we know to be 3.21, And the f of 1 -0.01, which is simply going to be the reciprocal of 3.21, About 0.37. Mhm. The boundaries of the confidence interval are going to be given by the reciprocal of the square roots of half of the significant value here and Times S one divided by S two, this is going to be equal to 1.03, 1. Also the other one square root of f one minus alpha, half Times S one or S two is going to be equal to 3.037. So are 98% confidence interval is from 1.031 two, Mhm.

Okay, we're doing a test at the 5% significance level. Our and all hypothesis is that the two are going to be the same Are alternative hypothesis is that We have a 14.5 for s. one and a 30.4 for us to like this. Left hail test easy. Okay, so the next thing we do is calculate S one squared over S two squared. So that is going to be the following here. 14.5. Yeah. And 34. Yeah. Okay. No, this is going to be approximately equal to 0.228. Mhm. Our degrees of freedom or his fouls. We're not doing five and 15. It's going to be 10 and eight Since we have N -1 for both those much. And then we can go back here Yeah. To table eight And the f of 0.025 Which is half of five Is going to be 4.3. And this time around this is D F N and D. F. D. When we go of search for this in Table eight. Uh huh. Yeah. Now for 1 -25 which is 0.975 It is going to be won over 3.85 which is approximately 0.26. But you must be asking yourself wait, Why is it 3.85? Well, we have to switch these. That's just the way a left hail test rolls. Yeah. Okay. So now we can proceed with the rest of our problem. We found out this part. We found out that part. Yeah. Um That sets us up for the next part to conclude part A. We see that it is .228. Uh this is going to be less than 4.3. So just for illustration, I honestly I don't know what the f curve looks like here. So what we have is 4 3. That's our cut off value shaded region. To the left tail. Left tail test .228 gets us well within that rejection region. So we will reject Yeah, you know, hypothesis At the 10% significance level. Not for the next part. It's a little bit more fun. We get to construct a confidence interval. Um This confidence interval level is 95%. So we're just going to be doing this guy erase that every step. So what we have is 0.05 And 0.95. And we got a multi divide these by two were multiplied by one half. So FF .025 is going to be 4.3. Mhm. And f of 095 49 You know 975. So that yeah right there we go. Like this slight correction mm We get the f of nine or 0.975. This is going to be equal to one over 3.85, Which we discovered earlier to be easier to six. But and now we can calculate the other values. Okay for our first Valium it's going to be one over the square root of this guy over here. So one over 4.3 times 14.5 divided by 30.4 Is going to be 0.23 scared And the other one is going to be 0.935. So are 95% confidence acceptable Is going to be from .23 2.935.

Hello. Welcome to this lesson in this lesson. We have the body temperature that is believed to be 19.6 Very high. Uh Greece, very height. But somebody also has some kind of evidence that There shouldn't be that there should be 98.2. Okay so the first but we are writing the now higher protesters. And the alternate hypothesis for this. So for the and now we have The mean that is believed to be 98.6. So this is against the alternate hypothesis that is believed to be nine T 8.2. So just believe that the Mean is less than 19.6. If it is believed, if there is a sufficient evidence that or somebody claim somebody claims that the body temperature should be 98 point to any means that the claim is that The temperature is less than the established 98.6. Okay. Mhm. Let's go to the B. But where we will test this claim with an offer Of 0.05. So this is a one tailed test. So we have a Z 0.05 which is recalled to 1.64 because you're dealing with the half the negative side. The lower side that's a negative. So probably have it's And this is a rejection area that is negative 1.64. Mhm. Now we are testing this claim. So we calculate the Z. And see if that would fall within the reduction area. It will come about it. So this is the I mean that is claimed minus the hypothesis size me all over the standard invasion over the square. It's of the sample size. So here we have 98 points 285 nose obtained from the sample -3. The one that is established the believed one Oh look at the standard deviation of 0.62 five all of the square root of 50 two. Mhm. Now, at this point we have The whole of this as -3 63 44. So looking at this, If this is negative 1.64 then the calculated Z would be here, bob lee there. So that means a force we do deregulation area. If he had been greater than this, Probably if it had come to one or even 1.6 or something that is more than And they get to 1.6 which is more than a negative 1.64, you would have not rejected. Okay, so here we reject. Mhm. And now hypothesis. Yeah. So actually we go with the alternate protest that says that the mean is less than 19.6°. Very okay. Now, the third part we are constructing a confidence interval of 95% confidence interval to also see if that can be used to reject or not exact. Okay, so here we have an error only in an error. What you see called lazy school. I understand that division all of us current of So with this one we have 1.64 times the Standard distribution which is 0.6 25. Yeah. All over the square roots. Yeah. And so we have the arrow ass 0.14- one. And we are looking at a lower bound so mhm. Yeah. Mm. Yeah minus the mean is in this. So we have what? Mhm 98.6 -3 Arrow. Mm. Mhm. Okay. Yeah And that is 98.4 579. Okay. And this lower bound it's still greater than mhm. The lower bound is still greater than yeah the me which was found from the test. Okay so with this one to reject there now this is okay. Mhm. All right. Fast for time. This is the end of the lesson. Uh huh. Mm.


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