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Suppose yOU use computer technology aimulac Icsu under Ite nu hypolhesis Afler 3000 simtations you fnd that 50 of the 3000 simulatior produced 36 ol Itte 3000 slmul...

Question

Suppose yOU use computer technology aimulac Icsu under Ite nu hypolhesis Afler 3000 simtations you fnd that 50 of the 3000 simulatior produced 36 ol Itte 3000 slmulalions produccd a aillerence less tan Ihe opposite Tlnt dnlcrenze more than OUr observed ditference Wknnul POidr observed ditierence Luknriz PolsrUserCrts 0f the simulatlonJodie the p-valucBased on tne p-value now much evidence there Ihat the null moddgood iit tor Ihe obscrved results?Little Some Slrong Very Shong Extremely Slrong

Suppose yOU use computer technology aimulac Icsu under Ite nu hypolhesis Afler 3000 simtations you fnd that 50 of the 3000 simulatior produced 36 ol Itte 3000 slmulalions produccd a aillerence less tan Ihe opposite Tlnt dnlcrenze more than OUr observed ditference Wknnul POidr observed ditierence Luknriz Polsr Use rCrts 0f the simulatlon Jodie the p-valuc Based on tne p-value now much evidence there Ihat the null modd good iit tor Ihe obscrved results? Little Some Slrong Very Shong Extremely Slrong



Answers

Use the data in LOANAPP for this exercise.
(i) How many observations have $o b r a t>40,$ that is, other debt obligations more than 40$\%$ of total income?
(ii) Reestimate the model in part (iii) of Computer Exercise $\mathrm{C8}$ , excluding observations with obrat $>40 .$ What happens to the estimate and $t$ statistic on white?
(iii) Does it appear that the estimate of $\beta_{\text {white}}$ is overly sensitive to the sample used?

We are thinking about hypothesis tests and we want to figure out which of the following P values defined stronger evidence to reject the null hypothesis, H not P equals 0.2 or P equals 0.3 To start off with, let's remember that in a hypothesis test you're going to observe some data with the score Z not on a normal distribution. And the P value is the probability you can observe something as or more extreme than Xena. So if the probability you're going to observe the area under the normal curve outside of your critical values or not. So in this particular sense we have that a smaller P corresponds to having more extreme of an observation. And since the smaller opinion is more extreme that means it's more inconsistent with your null hypothesis. The smaller the P value, the more likely you're going to fall underneath your critical significance values Alpha. So given this information that means P equals 0.2 is a stronger evidence Because 0.02 is smaller than 0.03. And so will be more inconsistent with any h. Not

In this experiment, we have, um, some samples of human skin cells and were either inducing auto factory or we are inhibiting auto factory of the cells and comparing the level of melanin in each sample to control. Can we explain these results? Okay, so we have, um if we induce auto factory for skin has less melanin so inducing auto factory, so 40 reduces melanin, concentration and inhibiting. What if Peggy increases it? As you can see, the inhibited one has a high percentage melon. The induced one has a lower percentage of melon compared to control. And we also have to check if the results are significance. So for the inducing or to Fiji, we can see that we have a P value of less than that point, not one. And for, um, inhibiting the automated, we have a P value of less than not point, not five. So in both cases, we're below the standard value of 9.95 But we used to statistical significance, so we can say that these results are statistically significant. And as for what conclusions we can draw from MS based purely on this, or we can say is that also, Fagih can affect the amount of melanin in skin cells

This question. We need to find the P value. Oh, for ah, just in proportions. So what you have to do is it was many tab. Okay, Andi goto the one sample proportion and choose the others. Ah, option here. Some earth data and fill as described in the question. So it is 73 off sexists and the number of friends. 100. And you have to specify, um, the proportion. Andi, don't forget to change their tentative here. Then it look OK, Then you're gonna find the p value for this. This is a quarto born.

Okay, so we are conducting a one tale test at the 1% significance level Arnold hypothesis is that they're the same. And the alternative is that the first is less than the second the computer have statistic is going to be s one squared divided by S two squared. So this turns out to be approximately 0.17 Our degrees of freedom are one less than each of the sample size is making 18 the numerator and 50 the denominator, We can go to the back of the book in table eight to figure out what are critical value is. Since this is a left tail test, our f statistic is going to be f of one minus point to one, which is going to be equal to F of 099 which has a statistic of 2.78 Of course we take the reciprocal of that. This turns out to be approximately 0.36 So for illustration purposes, we have our F curve, whatever it looks like, and 0.36 is going to be somewhere right here, and 0.17 is going to be clearly on this side in the rejection region. Therefore, we reject the null hypothesis at the 1% significance level.


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