True or False 1. The t test is sometimes invalid in small samples because the t statistic does nothave a t distribution when there exists heteroskedasticity in error terms. However, wecan still use the t-test even if the error terms are heteroskedastic because we can eliminateheteroskedasticity from error term by GLS (generalized least square) estimator.2. Suppose a regression of Y on X is run and the t-statistic for testing H,: B, = 0 hasa p-value of 0.0295 associated with it. Using a 5% level of significance, the nullhypothesis should be rejected.3. If the correlation between X and Y is 0.9, then the R value for a simple linearregression ofY on X will be 90%.4. Even though least squares estimators and maximum likelihood estimators involvedifferent estimation procedures, in the simple linear regression estimation, they providethe exact same estimators for Bo. Bi and o5. In sufficiently large samples, regression coefficients will always be significantlydifferent from zero6. OLS procedure will generate unbiased but inefficient estimator in the presence ofserial correlation.

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A snack food manufacturer estimates that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.23. A dietician is asked to test this claim and finds that a random sample of 24 servings has a variance of 1.27. At α=0.10, is there enough evidence to reject the manufacturer's claim? Assume the population is normally distributed. (b) Find the critical value(s). I do not know how to calculate the critical value with my calculator. I have a TI83+.
We plan to run a two-tailed t-test. Our sample size is 8, and we've decided upon an alpha level of 0.01. From the t-table, what is the t critical value(s)?
Texas Instruments produces computer chips in production runs of 1 million at a time. It has found that the fraction of defective modules can be very different in different production runs. These differences are caused by small variations in the set-up of each production run. Managers have observed that defective rates are roughly triangular, with a lower bound of 0%, and an upper bound of 50%. Defects more likely to be near 10% than any other single value in their range. Now suppose that we have taken a sample of 10 modules, and 2 of them are defective. i. What is the conditional probability of a defective rate less than 25% in this production run?ii. For what number M would you say that the defective rate is equally likely to be above or below M?
An urban community wants to show that the incidence of breast cancer is higher in their locality than in a neighboring rural area. (PCB levels were found to be higher in the soil of the urban community). If you find that in the urban community 20 out of 200 adult women have breast cancer and that in the rural community 10 out of 150 adult women have it, could you conclude, at a significance level of 0.05, that breast cancer is more prevalent in the urban community?1. The parameter of interest is:2. The hypotheses for this test are:3. The calculated test statistic is:4. The critical region is:5. Draw the critical region (make decision):6. It can be concluded that:
Let X1, ..., Xn be a sample from an exponential population with parameter λ.(a) Find the maximum likelihood estimator for λ. (b) Is the estimator unbiased?(c) Is the estimator consistent?
Under alpha 0.05 should you reject or retain the null hypothesis
Suppose X, Y, Z are iid observations from a Poisson distribution with parameter λ, which is unknown. Consider the 3 estimators T1 = X + Y − Z, T2 = 2X + Y + Z 4 , T3 = 3X + Y + Z 5 . (a) Which among the above estimators are unbiased? (b) Among the class of unbiased estimators, which has the minimum variance?
The records of a casualty insurance company show that, in the past, its clients have had a mean of 1.7 auto accidents per day with a variance of 0.0025 . The actuaries of the company claim that the variance of the number of accidents per day is no longer equal to 0.0025. Suppose that we want to carry out a hypothesis test to see if there is support for the actuaries' claim. State the null hypothesis H0 and the alternative hypothesis H1 that we would use for this test. H0 : ? H1 : ?
A random sample of size n = 4 taken from a normal population with σ2 = 9 is used to test H0 : µ = µ0against H1 : µ = µ1 where µ1 > µ0. e null hypothesis will be rejected if X > µ ¯0 + 2.5. Find the levelof significance of the critical region
John rejected a null hypothesis in a right-tailed test for a mean at alpha = 0.025 because the critical T value was 2.000 and his calculated T value was 2.345. We can be sure that: a.) John did not commit type 1 error B.) John committed neither type 1 nor type 11 error c.) none of the presented answers is correct because none can definitely be concluded d.) John did not commit type 11 error
When one is testing H0: µ= µ0 on the basis of data from a sample of size n from a normally distributed population with a known variance of σ2 , the test statistic is ____________________________________________. When one is testing H0: µ= µ0 on the basis of data from a sample of size n from a normally distributed population with a unknown variance, the test statistic is ___________________. The null hypothesis contains a statement of ______________________. The statement μ≥0 0 is an inappropriate statement for the ___________ hypothesis. The rejection region consists of those values of the ___________ that will cause rejection of the null hypothesis. The null hypothesis and the alternative hypothesis are ___________ of each other. Given H0: µ= µ0, then Ha: ___________. Given H0: µ ≤ µ0, then Ha: ___________. Given H0: µ ≥ µ0, then Ha: ___________. A statement of what you wish to conclude goes in the ___________. A market analyst believes that more than 30% of the adults in a…
As what we learned, heteroskedasticity arises because of non-constant variance of the error terms. We said proportional heteroskedasticity exists when the error variance takes the following structure: var(et)=σt^2=σ^2xt But as we know, that is only one of many forms of heteroskedasticity. To get rid of that specific form of heteroskedasticity using Generalized Least Squares, we employed a specific correction – we divided by the square root of our independent variable x. And the reason why that specific correction worked, and yielded a variance of our GLS estimates that was sigma-squared, was because of the following math: var(et*)=var(et/(square root of xt))=1/xt * var(et)= 1/xt σ^2 xt= σ^2 Where var(et)=σ^2 according to our LS assumptions. In other words, dividing everything by the square root of x made this correction work to give us sigma squared at the end of the expression. But if we have a different form of heteroskedasticity (i.e. a difference variance structure), we have to do…

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