Consider the potential outcomes for each individual i: JY₁i Yoi Y₁ = D₁ = 1 Di = 0 1. Write the formula for the treatment effect for each individual i. 2. Show that: E [Y₁ | D₁ = 1] E [Y₁ | D₁ = 0] = ATE + Bias
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![Consider the potential outcomes for each individual i:
[Y₁i D₁ = 1
Yoi
Di = 0
Y₁ =
1. Write the formula for the treatment effect for each individual i.
2. Show that:
E [Yi | D₁ = 1] E [Yi | D₁ = 0] = ATE + Bias](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F402693e2-a79e-44ab-81c0-693b0f662e5b%2F960815fa-443d-4648-9569-b9973237d281%2Fg3z84tn_processed.jpeg&w=3840&q=75)
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- Repeat Example 5 when microphone A receives the sound 4 seconds before microphone B.For b there are two cases and for c I have to plug the initial data into the odeThe Beer-Lambert law relates the absorbance A of a solution to the concentration C of a species in solution by A = MLC, where L is the path length and M is the molar absorption coefficient. Assume that C = 1.25 ± 0.03 mol/cm3, L = 1.2 ± 0.1 cm, and A = 1.30 ± 0.05.a) Estimate M and find the uncertainty in the estimate.b) Which would provide a greater reduction in the uncertainty in M: reducing the uncertainty in C to 0.01 mol/cm³, reducing the uncertainty in L to 0.05 cm, or reducing the uncertainty in A to 0.01?
- The article in the ASCE Journal of Energy Engineering (1999, Vol. 125, pp.59-75) describes a study of the thermal inertia properties of autoclaved aerated concrete used as a building material. Five samples of the material were tested in a structure, and the average interior temperatures (°C) reported were as follows: 23.01, 22.22, 22.04, 22.62, and 22.59. Test that the average interior temperature is equal to 22.5°C using alpha (a) = 0.05. This problem is a test on what population parameter? What is the null and alternative hypothesis? What are the Significance level and type of test? What standardized test statistic will be used? What is the standard test statistic? What is the Statistical Decision? What is the statistical decision in the statement form?Is there a way to solve this w/o using L'Hospital's rule?M1= 37 SS1=360 n1=11 M2= 43 SS2=576 n2=10 Assume two failed test and alpha= .05 What is the critical value?
- Solve An article in the ASCE Journal of Energy Engineering (1999, Vol. 125, pp.59-75) describes a study of the thermal inertia properties of autoclaved aerated concrete used as a building material. Five samples of the material were tested in a structure, and the average interior temperatures (°C) reported were as follows: 23.01, 22.22, 22.04, 22.62, and 22.59. Test that the average interior temperature is equal to 22.5°C using alpha (a) = 0.05. 1.)This problem is a test on what population parameter? a.Variance/ Standard Deviation b.Mean c.Population Proportion d.None of the above 2.)What is the null and alternative 3 points hypothesis? a.Ho / (theta = 22.5) , Ha: (0 # 22.5) b.Ho / (theta > 22.5) , Ha: (0 # 22.5) c.Ho / (theta < 22.5) , Ha: (theta >= 22.5) d.None of the above 3.)What are the Significance level 3 points and type of test? alpha = 0.05 two-tailed alpha = 0.95 two-tailed alpha = 0.95 one-tailed None of the above 4.)What standardized test statistic will…Let Z~N(0, 1) and X~N(mu, delta2). a. Calculate E[Z4].b. Calculate E[X4].A certain brand of upright freezer is available in three different rated capacities: 16ft3, 18 ft3, and 20 ft3. Let X = the rated capacity of a freezer of this brand sold at acertain store. Suppose that X has pmfx 16 18 20p(x) .2 .5 3a. Compute E(X), E(X2), and V(X).b. If the price of a freezer having capacity X is 70X – 650, what is the expectedprice paid by the next customer to buy a freezer?c. What is the variance of the price paid by the next customer?d. Suppose that although the rated capacity of a freezer is X, the actual capacityis h(X) = X - .008X2. What is the expected actual capacity of the freezer purchasedby the next custom
- The particular solution of PVI is:In testing the hypotheses H0: p = 0.5 vs Ha: p <= 0.5, suppose you get a p-value 0.022. Thenyou realize that the one sided alternative is too restrictive and re-do the test with a two sidedalternative Ha : p =/= 0.5.Based on the new p-value, what will be your conclusion ? A. Reject H0 at alpha = 0.01 but not at alpha = 0.05; 0.1.B. Reject H0 at alpha = 0.05 and alpha = 0.1 but not at alpha = 0.01.C. Fail to reject H0 at all the above alpha values.D. Reject H0 at alpha = 0.05 and alpha = 0.01 but not at alpha = 0.1.E. Reject H0 at all the above alpha values.When testing: H0: (alpha)1^2=(alpha)2^2. HA: (alpha)1^2>(alpha)2^2. At (alpha)=0.01, n1=5, n2=6. s1^2=15750 and s2^2=10920, what critical value do we use?