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- We are trying to predict GPA (Y) using number of hours spent studying (X1) and gender (X2). Our indicator variable for gender with male as the base is: If = {1 if Female} {0 else Our model is Y = Bo + B1X1+ B2IF + 8 } What is the expected value of Y if gender is female? O HF = Bo + B1X1 O µF = 0 O HF = Bo + 1 O uF = Bo + B1X1 + B2A 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.
- Heights (cm) and weights (kg) are measured for 100 randomly selected adult males, and range from heights of 133 to 188 cm and weights of 40 to 150 kg. Let the predictor variable x be the first variable given. The 100 paired measurements yield x = 167.54 cm, y = 81.35 kg, r=0.186, P-value = 0.064, and y = - 109 + 1.12x. Find the best predicted value of ŷ (weight) given an adult male who is 180 cm tall. Use a 0.10 significance level. The best predicted value of y for an adult male who is 180 cm tall is (Round to two decimal places as needed.) kg.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?Suppose we obtain n observations from a sinusoid contaminated by Gaussian noise: X = A· sin(2 · T· f·k+¢) + Wk, k = 1, 2, ...,n, %3D where Wg are independent identically distributed (iid) Gaussian random variables with mean o and variance o?. The frequency f and phase o are known constants, while A is an unknown constant. Find the maximum likelihood estimator for A. Show all work.
- 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 137 to 192 cm and weights of 40 to 150 kg. Let the predictor variable x be the first variable given. The 100 paired measurements yield x = 167.80 cm, y = 81.45 kg, r=0.211, P-value = 0.035, and y = -103 +1.07x. Find the best predicted value of ŷ (weight) given an adult male who is 145 cm tall. Use a 0.01 significance level. The best predicted value of y for an adult male who is 145 cm tall is (Round to two decimal places as needed.) kg.ndependent random samples are selected from two populations and are used to test the hypothesis Hn: (H, - H2) = 0 against the alternative H,: (4, - H2) #0. An analysis of 232 observations from population 1 and 311 from population 2 yielded p-value of 0.113. Complete parts a and b below. a. Interpret the results of the computer analysis. Use as0.10. A. Since this p-value exceeds the given value of a, there is insufficient evidence to indicate that the population means are different. O B. Since the given a value exceeds this p-value, there is insufficient evidence to indicate that the population means are different. OC. Since the given a value exceeds this p-value, there sufficient evidence to indicate that the population means are different. O D. Since this p-value exceeds the given value of , there is sufficient evidence to indicate that the population means are different. b. If the alternative hypothesis had been Ha: (H1 - H2) <0, how would the p-value change? Interpret the p-value…