1. Given that yı = 3.2, y2 = 1.3, y3 = 4.1 is a random sample from yeu/o fr (y;0)= 604 y 20, calculate the maximum likelihood estimate for 0.
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Q: For 50 randomly selected speed dates, attractiveness ratings by males of their female date partners…
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Q: For 50 randomly selected speed dates, attractiveness ratings by males of their female date partners…
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Q: For 50 randomly selected speed dates, attractiveness ratings by males of their female date partners…
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Q: Heights (cm) and weights (kg) are measured for 100 randomly selected adult males, and range from…
A: Note that P-value = 0.013 < 0.05 = alpha. Thus there is a linear relation between x and y.
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- A random sample is obtained from a population with a mean of µ 20. After a treatment is administered to the individuals in the sample, the sample mean is M = 21.3 with a variance of s2 = 9.Consider the following population model that satisfies the CNLRM assumptions: Y = βX; + u in particular assume that: ui ~ Ν (0,1) What is the mean and variance of B₁? Ε (βι) = 0,6%, = nΣα 22 x Στ Ο Ο Ο ΣΧ Ε (βι) = 0,6% = nΣ # Ε(β) = 1,6 = Ε(βι) = 1,6% = nΣ # (31) {" 2 Σ βι η ΣΧ Cannot be determined with the information provided.For 50 randomly selected speed dates, attractiveness ratings by males of their female date partners (x) are recorded along with the attractiveness ratings by females of their male date partners (y); the ratings range from 1 to 10. The 50 paired ratings yield x= 6.4, y = 6.0, r= - 0.267, P-value = 0.061, and y= 7.90- 0.296x. Find the best predicted value of y (attractiveness rating by female of male) for a date in which the attractiveness rating by the male of the female is x= 4. Use a 0.01 significance level. The best predicted value of y when x= 4 is. (Round to one decimal place as needed.)
- Let m denote margin of error, n sample size and σ standard deviation. m= zα/2(σn) Solve the above equation for n.A snack food manufacturer estimates that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.34. A dietician is asked to test this claim and finds that a random sample of 16 servings has a variance of 1.22. At α=0.05, is there enough evidence to reject the manufacturer's claim? Assume the population is normally distributed. Complete parts (a) through (e) below.Consider the following population model that satisfies the CNLRM assumptions: Y; = β2X; + ui in particular assume that: u ~ N (0,1) Win What is the mean and variance of 3₁ ? 2Σ o? σ' Ο Ε(βι)=0,6 = {" Ο Ε(βι)=0,6 – Σ Ε(βι)=1,6 = EX (βι) = 1,6% = ΣΧ Στ x σ βι n Ο Ο E ΣΧ ηΣ Cannot be determined with the information provided.
- With an American penny, the likelihood of getting H when it is spun on edge is 0.3. If X is the random variable where X(H ) = 1, X(T ) = −1, find the expected value E(X), the variance, Var(X), and express X in its standard form.A random sample of 10 observations from population A has sample mean of 152.3 and a sample standard deviation of 1.83. Another random sample of 8 observations from population B has a sample standard deviation of 1.94. Assuming equal variances in those two populations, a 99% confidence interval for μA − μB is (-0.19, 4.99), where μA is the mean in population A and μB is the mean in population B. (a) What is the sample mean of the observations from population B? (b) If we test H0 : μA ≤ μB against Ha : μA > μB, using α = 0.02, what is your conclusion?An engineer wants to know if producing metal bars using a new experimental treatment rather than the conventional treatment makes a difference in the tensile strength of the bars (the ability to resist tearing when pulled lengthwise). At α=0.10, answer parts (a) through (e). Assume the population variances are equal and the samples are random. If convenient, use technology to solve the problem. Treatment Tensile strengths (newtons per square millimeter) Experimental 449 354 450 360 433 388 400 Conventional 370 376 374 424 378 450 438 404 352 376 (a) Identify the claim and state H0 and Ha. The claim is "The new treatment ▼ makes a difference does not make a difference in the tensile strength of the bars." What are H0 and Ha? The null hypothesis, H0, is ▼ mu 1 equals mu 2μ1=μ2 mu 1 less than or equals mu 2μ1≤μ2 mu 1 greater than or equals mu 2μ1≥μ2 . The alternative hypothesis, Ha,…
- A random sample of size 36 from a population with known variance, o = 9, yields a sample mean of x = 17. Find ß, for testing the hypothesis H,: u = 15 versus H1 : =16. Assume a 0.05. %3D %3DHeights (cm) and weights (kg) are measured for 100 randomly selected adult males, and range from heights of 138 to 190 cm and weights of 40 to 150 kg. Let the predictor variable x be the first variable given. The 100 paired measurements yield x = 167.88 cm, y = 81.51 kg, r=0.195, P-value = 0.052, and y= - 105 + 1.05x. Find the best predicted value of y (weight) given an adult male who is 184 cm tall. Use a 0.01 significance level. The best predicted value of y for an adult male who is 184 cm tall is kg. (Round to two decimal places as needed.)For 50 randomly selected speed dates, attractiveness ratings by males of their female date partners (x) are recorded along with the attractiveness ratings by females of their male date partners (y); the ratings range from 1 to 10. The 50 paired ratings yield x=6.4, y= 6.0, r= -0.124, P-value = 0.391, and y = 6.95-0.146x. Find the best predicted value of y (attractiveness rating by female of male) for a date in which the attractiveness rating by the male of the female is x = 7. Use a 0.10 significance level. The best predicted value of y when x 7 is. (Round to one decimal place as needed.)
