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QI: The genril partnt of a limited-partnenship fin has told # polential Investor thut the mean monthly rent for three-bedroom apartnxents in the city is 587 $. The ...

Question

QI: The genril partnt of a limited-partnenship fin has told # polential Investor thut the mean monthly rent for three-bedroom apartnxents in the city is 587 $. The investor rndomly sclarts 32 tne-becroom #partments in the cily and detennin > their monthly Fents: Tl following uable disphys the monthly rent for the 32 apannents obuined Do Ihe data Sugg"st thut the gcneTal partner claim inconert? Perfon the approprale hypothesis test A1 th: 0.05 level ol significant. 289 560 726 65 586 657

QI: The genril partnt of a limited-partnenship fin has told # polential Investor thut the mean monthly rent for three-bedroom apartnxents in the city is 587 $. The investor rndomly sclarts 32 tne-becroom #partments in the cily and detennin > their monthly Fents: Tl following uable disphys the monthly rent for the 32 apannents obuined Do Ihe data Sugg"st thut the gcneTal partner claim inconert? Perfon the approprale hypothesis test A1 th: 0.05 level ol significant. 289 560 726 65 586 657 565 676 656 577 663 729 745 597 659 626 450 669 675 603 45 661 610 64 595 598 42 02) A semiconductor manufacturer produces controllers used in Jutomobile engine applications. The customer requires that the process fallout or fraction defective at critical manufacturing step not exceed 0.05 and that the manulacturer demonstrate process eapability this level of quality using 0-0.05 The semiconductor manufacturer takes rndom sample of 200 devices and finds that four of them are defective. Can the manufacturer demonstrate process capability for the customer? Q3) An Izod impact test was performed on 30 specimens of PVC pipe. The sample mean is |.25 and the sample standard deviation is 0.25. Find 9990 lower and upper conlidence bounds on Izod impact strength



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Testing the Difference Between Two Means (a) identify the claim and state $H_{0}$ and $H_{a}$, (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic t, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed. If convenient, use technology. Tensile Strength An cngineer wants to compare the tensile strengths of steel bars that are produced using a conventional method and an experimental method. (The tensile strength of a metal is a measure of its ability to resist tearing when pulled lengthwise. To do so, the engineer randomly selects steel bars that are manufactured using each method and records the tensile strengths (in newtons per square millimeter) listed below. Experimental Method: \begin{tabular}{|cccccccc} \hline 395 & 389 & 421 & 394 & 407 & 411 & 389 & 402 \end{tabular} 422 $\begin{array}{rrrrrrrr}416 & 402 & 408 & 400 & 386 & 411 & 405 & 389\end{array}$ Conventional Method: \begin{tabular}{rrrrrrr} 362 & 352 & 380 & 382 & 413 & 384 & 400 \\ \hline \end{tabular} $379 \quad 384 \quad 388$ $378 \quad 419$ $372 \quad 383$ At $\alpha=0.10,$ can the engineer support the claim that the experimental method produces steel with a greater mean tensile strength? Assume the population variances are not equal.

The following is a nova test based on the mean salaries for different metropolitan areas. So the alternative or the null hypothesis is that all the means are the same. So there are six metropolitan areas, I think it goes Chicago, Dallas Miami, Denver san Diego and Seattle. Uh So the null hypothesis is that all the means are the same. And then the alternative is that at least one of them is different. The second step is to find the critical value and you can do that using either software or a table, But they're essentially three things you need. The first thing is your alpha value, your significance level and that's usually given to you the problem and that's .05. Then you need the degrees of freedom for the numerator and the degrees of freedom for the denominator. And the way you find that Is the degrees of freedom for the numerator is the number of categories -1. So there were six cities that we looked at our metropolitan areas, so 6 -1° of freedom would be five for the numerator. And then for the denominators, the total number of data values minus the number of categories. So there were 36 data values minus the six metropolitan areas. So 30 is your degrees of freedom for the denominator. So that should be enough to use a table. But I use a calculator and I wrote a program in here called inverse. F. I'm not going to show you how to how to write the program. You can youtube it if you wish. Um But this is what I do. So um I put in my area which is my alpha value, my degrees of freedom is five and then my degrees of freedom for the denominator is 30 and That gives me my critical value. About 2.534 2534 is my critical value. I call f. star. So 2.534. Okay so anything greater than 2.534. We reject the annual hypothesis that all the means are the same And anything less than 2.534. We failed to reject meaning the h not is true. Okay so the second step is to find the F statistic and there's a formula but it's a bit of a mess. I always use software you know technology is a great thing. So if you go to stat and you can type in your data values. So these are the mean salaries um So again L1 I think was Chicago and then this is the mean salary for Dallas Miami Denver San Diego and Seattle. So there are six categories. And if you go to stat tests and then we're gonna go to the Unova test and then you just type in your columns separated by commas remember there were six columns, six data columns that we used and we need to make sure that all of them are in there and last one and then also you know make sure you separate those by commons, otherwise it's going to read it wrong. So then um that gives us everything we need. So the F. Is the F statistic, that's the third step. So we're looking at this it's about 2.281 as our F. Value. So two point 281 is our f statistic Which is actually barely in the non rejection region 2.281. So that means we fail to reject. Okay and also we can verify that with this p value here. So the p values 0.7 which is a pretty small p value, but it's still in this case greater than the alpha value. So the alpha value remembers point oh five, so it's barely greater than the alpha value. And whenever it's greater than the alpha value, uh we failed to reject, I should probably put H not there, so we failed to reject H not whenever the P values greater than the alpha. Okay. So then the last step is to summarize everything with actual words. So what does this all mean? It just means that there is not sufficient evidence, there is not sufficient. I guess you could say statistical evidence to suggest that the mean salaries from the different metropolitan areas are different. Okay. And that's the five step process for an Innova one way and over test

19, which in eight it's notice that me one is equal to Muto and each one is that me. One is not equal to Muto, so don't remind. The degree of freedom is equal to n one plus and two minus two is 10 plus 13. Minus two is 21 So the critical value corresponding. Tau alpha 4.1 with degree freedom 21 1 to date. So you think Temple five Critical Various possible negative 2.831 The second reason they contain all advantages. Smaller than negative 2.831 and larger than 2.83 More, uh, standard deviation is the square root off n minus one. So mine bye 22.3 or 12 square plus 12, which is into minus one times 14.5122 square over 10 plus 13 minus two to stem plus 13 minus two, which is 18.262 So the distance, which is X one minus six to so it 368.313 89.5385 Over square Rode off your father 18 0.262 is one over and one plus one over and to which approx negative 2.765 So if the value off the is in the rejection regions in the non deposit is reelected, so as the negative is bigger than negative 2.831 and smaller than 2.831 So we failed to reject, okay, they're not hypothesis, so there is no sufficient evidence to support, okay?

Okay. So what we're gonna be looking at is the comparison of I. Q. S. Of kids that went through some sort of lung cancer therapy and those that didn't. So what we're gonna be doing is taking a right tail test since our first group is bigger than the other one, the average is bigger for the other one. So we're gonna split this down right here and this is gonna be at the 1% significance level. So if our tests the state falls in this region we will reject the hypothesis. Mhm. So our test statistic is gonna be given by T. Equals this equation right here and we're going to substitute our values right there. So T. Is going to be equal to 84.4 minus 78.2. And then we're gonna take the sp which is the pool standard deviations sample size for the first one in 74 minus 1, 73. With a standard deviation of 12.6 squared. And for the other one it's 72 minus one. So 71 times the standard deviation of 15 squared. Yeah. And then we're gonna divide by the two adjusted sample sizes, Okay? Which is gonna be our degrees of freedom. Mhm. At 1% significant level. So we'll get a critical value later. But first our full statistic is 13.84 So go and put that in there. Take the square root of the solid reciprocal sample sizes. Mhm. And what we get for our t statistic is 2.71 And when we look in the back of the book we look between 102 100 degrees of freedom since 144 is about between them and we find out that it's 2.35 so obviously 2.71 is greater than 2.35 meaning that we do land in that critical region zone. So at least 1% significance level. We reject the null hypothesis. There is the this means that the accused of the kids that went through that therapy are lower.


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