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Please do Q1 before attempting this question. You can make useof the standard error from Q1, and the R command qt(prob,df) Thisgives the value such that the area to...

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Please do Q1 before attempting this question. You can make useof the standard error from Q1, and the R command qt(prob,df) Thisgives the value such that the area to the left of this value undera t-distribution curve with df degrees of freedom is equal to prob.Alternatively, you can obtain the multiplier needed using thestatistical tables. The margin of error for a 99%confidence interval for the slope of the regression line in thepopulation is?Q1 data below-Question 1Embryonic stem cells have the

Please do Q1 before attempting this question. You can make use of the standard error from Q1, and the R command qt(prob,df) This gives the value such that the area to the left of this value under a t-distribution curve with df degrees of freedom is equal to prob. Alternatively, you can obtain the multiplier needed using the statistical tables. The margin of error for a 99% confidence interval for the slope of the regression line in the population is? Q1 data below- Question 1 Embryonic stem cells have the capacity to produce neural progenitor cells, providing a potential means of repopulating cells lost due to spinal cord injury. However cell survival using existing methods is often low. Aiming to improve this, researchers investigated growing cells inside fibrin scaffolds and looked at the effect of two growth factors, neurotrophin-3 (NT3) and a platelet-derived growth factor (PDGF), on the outcomes. Mouse embryonic stem cells were cultured in 16 wells seeded with fibrin scaffolds to which various concentrations of the growth factors (in ng/mL) were added. With at least 5000 cells in each well, fluorescence-activated cell sorting was used after three days to count the number of living and dead cells in each well. The data from the 16 wells are presented below. The Alive variable gives the proportion of the cells alive after three days. PDGF,NT3,Cells,Alive 0,0,7520,0.479 0,0,8970,0.553 0,10,8310,0.632 0,10,7430,0.603 0,20,6930,0.500 0,20,8030,0.604 0,30,8520,0.812 0,30,7130,0.877 2,0,7720,0.655 2,0,7680,0.580 2,10,7500,0.681 2,10,8280,0.549 2,20,7740,0.746 2,20,7960,0.708 2,30,8620,0.817 2,30,7330,0.708 Please read the data into R as seen in previous lab sessions (see R-Summary.pdf for details) and call the resulting data frame welldata. We would now like to determine the P-value to test for an assocation between the proportion of cells alive and NT3 concentration. This is equivalent to testing whether or not the slope of a regression is zero (a two-sided hypothesis test). You would get the same p-value if you try to predict Alive from NT3 or NT3 from Alive. Here, the NT3 concentration can be measured immediately and in practice we would have to wait to measure the proportion of cells Alive. Hence it makes more sense to use Alive as the response (Y) variable and NT3 as the explanatory (X) variable. We will ignore the other variables in this analysis. You should work out the required test statistic using relevant formulae, your calculator and some R commands. You can then determine the corresponding p-value using R. Note that if you had to determine the p-value for a slope using tables, you would get a range, rather than a single answer. You can run a linear regression using NT3 as the X variable and Alive as the Y variable via the following R command. We will store the results in a structure called reg, which we then take a brief look at by just typing its name. Please use the commands as follows. Note that lm, which stands for linear model, is an R function which can be used to fit a linear regression to data. reg <- lm(Alive ~ NT3,welldata) reg The above only shows you the estimates of the intercept and slope, i.e. b0 and b1. Please make a note of these for Q2. However, there is a lot more information stored in reg. You can work out the standard deviation of the residuals (using the n-2 degrees of freedom) via the following R command, in which you need to enter n yourself: sqrt(sum(reg$residual^2)/(n-2)) You can work out the bottom line of the s.e.(b1) formula using the following R command: sqrt( sum( (welldata$NT3- mean(welldata$NT3))^2 ) ) You then have all the information needed to work out the standard error of the slope estimate (b1). Make a note of this for use with Q3. You can then work out the required p-value by making use of the following command, which gives the area to the left of a value t under a t-distribution curve with df degrees of freedom: pt(t,df)



Answers

There are several extensions of linear regression that apply to exponential growth and power law models. Problems 22, 23, 24, and 25 will outline some of these extensions. First of all, recall that a variable grows linearly over time if it adds a fixed increment during each equal period. Exponential growth occurs when a variable is multiplied by a fixed number during each period. This means that exponential growth increases by a fixed multiple or percentage of the previous amount. College algebra can be used to show that if a variable grows exponentially, then its logarithm grows linearly. The exponential growth model is $y=\alpha \beta^{x},$ where $\alpha$ and $\beta$ are fixed constants to be estimated from data. How do we know when we are dealing with exponential growth, and how can we estimate $\alpha$ and $\beta?$ Please read on. Populations of living things such as bacteria, locusts, fish, panda bears, and so on tend to grow (or decline) exponentially. However, these populations can be restricted by outside limitations such as food, space, pollution, disease, hunting, and so on. Suppose we have data pairs ( $x, y$ ) for which there is reason to believe the scatter plot is not linear but rather exponential, as described earlier. This means the increase in $y$ values begins rather slowly but then seems to explode. Note: For exponential growth models, we assume all $y>0$. Consider the following data, where $x=$ time in hours and $y=$ number of bacteria in a laboratory culture at the end of $\mathrm{x}$ hours. (a) Look at the Excel graph of the scatter diagram of the $(x, y)$ data pairs. Do you think a straight line will be a good fit to these data? Do the $y$ values seem almost to explode as time goes on? (b) Now consider a transformation $y^{\prime}=\log y .$ We are using common logarithms of base 10 (however, natural logarithms of base $e$ would work just as well). $$\begin{array}{l|lllll}\hline x & 1 & 2 & 3 & 4 & 5 \\\hline y^{\prime}=\log y & 0.477 & 1.079 & 1.342 & 1.748 & 2.161 \\\hline\end{array}$$ Look at the Excel graph of the scatter diagram of the $\left(x, y^{\prime}\right)$ data pairs and compare this diagram with the diagram in part (a). Which graph appears to better fit a straight line? (c) Use a calculator with regression keys to verify the linear regression equation for the $(x, y)$ data pairs, $\hat{y}=-50.3+32.3 x,$ with sample correlation coefficient $r=0.882$. (d) Use a calculator with regression keys to verify the linear regression equation for the $\left(x, y^{\prime}\right)$ data pairs, $y^{\prime}=0.150+0.404 x,$ with sample correlation coefficient $r=0.994 .$ The sample correlation coefficient $r=0.882$ for the $(x, y)$ pairs is not bad. But the sample correlation coefficient $r=0.994$ for the $\left(x, y^{\prime}\right)$ pairs is a lot better! (e) The exponential growth model is $y=\alpha \beta^{x} .$ Let us use the results of part (d) to estimate $\alpha$ and $\beta$ for this strain of laboratory bacteria. The equation $y^{\prime}=a+b x$ is the same as $\log y=a+b x \cdot$ If we raise both sides of this equation to the power 10 and use some college algebra, we get $y=10^{a}\left(10^{b}\right)^{x} \cdot$ Thus, $\alpha=10^{a}$ and $\beta=10^{b} .$ Use these results to approximate $\alpha$ and $\beta$ and write the exponential growth equation for our strain of bacteria. Note: The TI-84Plus/TI-83Plus/TI-nspire calculators fully support the exponential growth model. Place the original $x$ data in list $L 1$ and the corresponding $y$ data in list $L 2 .$ Then press STAT, followed by CALC, and scroll down to option 0: ExpReg. The output gives values for $\alpha, \beta$, and the sample correlation coefficient $r$.

Part one out of 990 students in the sample. The number that were never awarded a voucher. R 468. And you will find that number in our by something. All the cases, this is his name of the data set dollar sign. And select years equal equal zero. You are counting the total number of cases where select years is zero. How many students had a voucher available for four years. So similarly, you find the total number of cases where selected ears double equal four. And you will get 108 students. Then how many students actually attended a choice school for four years counting all the cases were variable choice years equal equal for you get 56 party. You you run a simple regression of choice ears on select years. This is what I get. I get there slope coefficient to be minus 1.837 And this one is significant. The minus sign is not what I expect. These regression is telling us that, mm, mm hmm. Uh huh. All right. This is actually part three. All right, sir. In Part three, you run a simple regression on uh of math percentile on choice years. And this is not what I expected. Because this result is telling me being able to attend choice years actually decreases math scores. And the variable um the estimate on choice ears does not change when I add the demographic variables black, hispanic and female. Okay, so I actually accidentally skipped part two. Yeah. Let's move down to the end. In part to you. You are examining whether select ears is a good instrument for choice ears and the answer is yes. This is a regression result. This equation is telling us being selected for a voucher increases the chance of the students attending the school of choice. And the relationship is strong. Almost 1 to 1 and also highly significant. Select years can be a good instrument, assuming that it does not correlate with the error terms in the choice years. Um Now in the mass score equation let's continue with Part four. Why might choice years be endogenous in the given equation, I could think of two reasons. First one admitted bearable bias. It is possible that we have not accounted for some factors that affect both mass scores and the number of years students attending the school of choices, for example, intellectual ability and the second reason is reverse causality. The problem does not give more context on this, but it is possible that mass scores effect the ability to attend the schools of Choice, Part 56 N seven and eight. Um So for the last four parts, I put the results together and they are here in one table. This is part five, Part six and part seven. Yeah. In part five you estimate the equation in part four by instrumental variables. And Selected Years is an instrument for Choice Years does using I. V. Produced a positive effect of attending a charter school? The answer is no because the sign of the main estimate is negative. What do you make of the coefficients of the other explanatory variables? We have black, hispanic and female. As you can see the first few have negative estimates meaning students that are back and of hispanic origin tend to have significantly lower math scores. The standard error is way smaller than the estimate for variable female female dummy. This estimate is telling us being female does not affect mass scores controlling for race origin and whether attending choice schools. I wouldn't write that down about six. We at M n c E 90 the math score in 1992 the equation and we will estimate the equation by L. S. N. I. V. Now we have a positive as expected estimate on choice ears. The I. V. Estimate has a greater magnitude comparing to the old LS estimate and is actually statistically significant. I. Ve estimate is practically large each year in a school uh in a child's school is worth almost two on the math percentile score. Okay again, I will write that down about seven. So the results which has got is not entirely convincing because we have a smaller sample. So you can see from your statistical package when they run the regression in part six, they would say that 662 observations deleted due to missing. And we know that it is the missing values of the new variable M N C E 90 on a part part eight. We replaced choice years with four dummies choice here one, 23 and four. And we generate its instruments. They're instruments select years 1 to 4. We estimate the equation by ivy. This is what I get. Yeah. So we don't have choice years anymore. We don't have a constant anymore. Choice here's one is not significant. Do not significant. Three not significant. Only choice. Air Force is significant and that means being in only being in the choice school for the whole four years. The maximum amount of time matters for the math scores. So that's it and I think it's an interesting result. And you should read the paper if you have a chance to.

The following is solution video to number 26 which is two sample T tests comparing meantime loss in the workforce with stressors and intimidators. And you just got to be really careful here. The only kind of weird thing is They kind of switch up the order on you so X one represents the intimidators. In the next two represents the stressors. And we want to see if the stressors, so I'm actually gonna write that down. So this is intimidators and this is stressors. Okay, So it says is there enough evidence to suggest that the stressor is greater than the intimidator? So keeping that in mind, we're going to write less than some you won Is less than you two, Even though it says greater than but the orders just switched, if that makes sense? So the greater than in this case means less than for our alternative, since the order is opposite. Okay, so the first part of this is just verifying that these means. And standard deviations are four and 2.38 for intimidators, and then 5.5 and 2.784 stressors. So I went ahead and it says to use a calculator, so I'm using a T 84 if you go to stat and then edit here are the list, so L one is the intimidator and then L two is the stressor. So if you go back to stat over to couch and then it's one of our stats, we can change that to L one and this will be the four and the 238 that was over here. So four and 238 And then go back to stat couch. One bar stats, change that to L two calculate, and that's where we get to 55 and the 27855 to 78 So we verified that that is the way to do it. So then the second part of it is to kind of sort of not formally conduct a hypothesis test but essentially come up with the conclusion here and it is gonna be a two sample T test. We're gonna use the calculator since we already have the data input anyway. So it's two sample T test for the alpha is point oh five. And the alternative we already talked about, the alternative is less than not greater than but less than so. Um if you go back to stat and then we're gonna air over two tests and it's the fourth one down is the T. Test. Two sample T test because we don't know the standard deviation for the population, We don't know sigma, we only know as um so sincerity of the data, I'm just going to keep it as data instead of doing the summary stats. L one is the first list. L two is the second list and you can keep those frequencies as one and then we're gonna change this alternative to less than you two. And then pulled is usually no unless they tell you otherwise and then we're gonna go and calculate and you can have that test statistic there if you want but you really don't need to because the p value is all you need. So it's 0.14. So let's write that down. So the p values 0.14 and what we do is we explicitly compare the P value with the alpha. In this case 0.14 is greater than point oh five. So that means whenever the p values greater than alpha, we fail to reject H not so we failed to reject h. Not whenever the p values greater than alpha at the p values less than alpha, then we will reject the null hypothesis. So keep in mind that the null hypothesis is generally always saying that these two means are equal. We can we're failing to reject that, so we're accepting that to be the truth. So let's go ahead and complete this hypothesis test. And we're gonna say there is not sufficient evidence to suggest that the mean time lost due to stressors is greater than the meantime lost due to intimidators. That's spelled right? I have no idea. Looks like it's not so intimidators. Okay, that's why math teachers aren't spelling enthusiast. So intimidators um now you could also say there is not sufficient evidence to suggest that the meantime lost due to intimidators is less than the meantime loss due to stressors. If you want to keep to that less than thing. I just wrote it like this because that's the way the book had it. So there is not enough evidence to say that these two means are different or that one means more than the other. So we're accepting the truth that these two means are probably about the same.

In this part, you are going to show some statistics of the variable net F. A thistle variable indicates the Net total financial assets in thousands of dollars. This variable has a mean of 19 up seven. The standard deviation is 63 Boyne Nice six. The minimum level is minus 502.3 and the maximum level is 1000 536 18 Remember, these numbers are in 1000. In the second part, you are going to test whether the average net total asset differs by 401 k eligibility studies and you can use a two sided alternative. The null hypothesis of this question is the average of net S A for their group that is eligible for for one cave equal that f A of the group that is not Elizabeth. For 41 K, we would do a T test and the T test. We will get a value of 13 11 The P value off the test is very small is way smaller than 0.1 So we are able to reject the non hypotheses that two groups have equal net total financial asset. The difference between the two groups in that total financial asset is well intact. The value of Met Toto A set the average one for the group that is eligible for 401 K, which is 30 points 54 These numbers are generated by their statistical software. Then we subtract from it their average net F A of the group that is not eligible for for a one K, which is 11 going 68 What we get is roughly 18 0.86 So the group that is eligible for for one K has a larger average net total financial asset, and it is larger by the average of the groups ineligible for 41 K by 18.8 $6000. We will estimate a mentally near regression model for net total financial assets. That includes income age eligibility for 401 K, and we also have age and income included in the regression as quadratic form. Okay, we will look at the coefficient of the variable E for a one K. As an indicator of the estimated dollar effect of the eligibility status, the coefficient of E 41 k is nine point 705 It means that if the family is illegible 4401 k, their net total financial asset when increased by $9705 we will add to the model estimated in Part three, the interaction terms of a legibility status and age minus 41. And another term is the interaction of eligibility status and age minus 41 square. This is the regression result. When we look at the interaction terms, we see that only this term it's significant. The term with Asian minus 41 the term with Asia minus 41 square, is not significant. In future models, we can safely drop the last term. Women compare the estimated effect of the allege ability status between model in Part three and model in Part four. The left panel show the regression results from the model. In part, we and the right panel show the results from the model. In Part four, you can tell the difference of the coefficient of E 401 K between two models. In monetary, it is 9.75 A model for it is 9.960 You should note that the meaning of the coefficient of E 401 k, the first between two models in model three. It is the effect for all ages. A model for it is the effect when age equals 40. 1 way will include family size dummies In the regression equation, you can see that in our equation we now have F s two s, three s four and fs five. These are their families side dummies F s two takes a value of one if the family has to people access to equal zero otherwise, after three takes a value of one. If the family has three people and takes a value of zero otherwise ever as far is for family of four. And Verse five is for family with more than five people. Mhm. So you don't cfs one in the equation. That is because Fs one or a family of one is the base group We have to base group So all the coefficient of the dummies should be interpreted as the difference in net total asset compared to the base group. So among the four dummies, you can see that Fs two is not significant. F s 32 Fs five are significant and the level of significance of these dummies increases with their size of the families. So have a sweet It's significant at the 5% level. But s four and s wide are significant at the 1% level. Even the P value you can say that compared to the base group, a family of a single person having three or more people in the family associate with a greater average value of net total financial assets. We can also conduct a F test to see the joint significance of their family size dummies. The null hypothesis is the coefficient of the family size dummy equals each other and equals zero. We will get an F statistic with two degrees of freedom four and NYT housing 265. The statistic is 5.44 and the P value it's very, very small is way smaller than 0.1 So we are able to reject the null hypothesis. In other words, the family size dummies are jointly significant. At the 1% level, we will do a child test forward coefficient equality across five family sized categories For the test statistic. We we need some of square residual from the unrestricted models and re restricted model. The restricted model comes from Part six and we use the phone sample to estimated The sum of square residual for the respective model is 30 million yeah, 213,000 and 115 for the and restricted models. They are five of them. They are estimated separately for each category of family size. The sum of square residual for and restricted models is the sum of the sum of square residual of all five models. Adding them up. You wouldn't get a total of 29 million, 980,000 and 959 now. Given that we have there some of square residual for the restricted model and unrestricted models, we can calculate the F statistic or the child test statistic the child has is based on the F distribution. That's why we have an F statistic. This is their formula for the F statistic. We will take the sum of square residual of the restricted model minus the sum of square of residual of the unrestricted model divided by their sum of square of residual of the unrestricted model altogether multiply with the ratio of the second degree of freedom divided by the first degree of freedom. These degrees of freedom are given from their problem. We have D F one. The first degree of freedom is the number of constraints were going to test, which is 20 and the second degrees of freedom is given as NYT housing and 245. The F statistic would be three point 57 and we can back up the P value using your statistical software. It is a very small number 0.0 03 So we are able to reject the null hypotheses at the 1% level, which means the coefficients are not equal across five family size categories.


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