Obtain maximum likelihood estimate of 0 in f(x, 0) = (1 + 0) x®, Example 0
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- Consider a random sample drawn from the normal distribution with mean μ and variance o². The maximum likelihood estimator maximizes the value of the mean and variance of the normal distribution in order to estimate the unknown population mean. True FalseSuppose X₁,, Xn from a normal distribution N(μ, o²) where both μ and o are unknown. We wish to test the hypotheses Ho: o² = o vs. Ha: 0² # of at the level a. Show that the likelihood ratio test is equivalent to the x² test.A random variable X is binomially distributed with mean 15 and variance 14.7. Determine the values of n and p. Then, P (X<10) [Hint: use Poisson approximation to Binomial Distribution]
- Use a t-test to test the claim about the population mean y at the given level of significance a using the given sample statistics. Assume the population is normally distrbuted Claim: jI * 25; a = 0.01 Sample statistics: × = 25.1, s = 4.2, n= 11For each of the following distributions, compute the maximum likelihood estimator based on 7 i.i.d. observations X₁,..., X and the Fisher information, if defined. If it is not, enter DNE in each applicable input box. (a) (Enter barX_n for the sample average X Maximum likelihood estimator = X~ Ber (p), p = (0,1) Xn = 1-7 n T X₁. Hint: Use the definition of Fisher information that leads to the shorter computation. (If the Fisher information is not defined, enter DNE.) Fisher information I (p) = Use Fisher Information to find the asymptotic variance V (p) of the MLE >>. V (B) = 9 SaveRegression Statistics Multiple R R Square Adjusted R Square Standard Error Observations ANOVA Regression Residual Total Intercept DISTANCE PAX SWA=YES VACATION=YES HI 0.8567 0.7339 0.7318 39.3716 638 df 5 632 637 86.67589 0.07751 -0.00011 -61.42324 -52.21338 0.00735 t Stat P-value 7.08935 12.2262 5.2751E-31 0.00269 28.8351 2.4522E-117 0.00012 -0.9381 0.3485 3.54152 -17.3438 2.0809E-55 3.57526 -14.6041 7.8163E-42 0.00099 7.3950 4.5121E-13 a) What percent of total variation in FARE does the model overall explain/determine SS MS 2701807.711 540361.542 979676.999 1550.122 3681484.709 Coefficients Standard Error F 348.5929 Significance F 5.5794E-179 Upper 95% 100.5974 72.7544 0.0722 0.0828 -0.0004 0.0001 -68.3778 -54.4687 -59.2342 0.0054 -45.1926 0.0093 Lower 95% b) Does the model overall fit the dat? NO YES c) Should one interpret the estimated value for the intercept? NO 0.25 pt) d) Interpret the value for the estimated coefficient for DISTANCE e) Which explanatory variable has the least…
- What is the optimal time for a scuba diver to be on the bottom of the ocean? That depends on the depth of the dive. The U.S. Navy has done a lot of research on this topic. The Navy defines the "optimal time" to be the time at each depth for the best balance between length of work period and decompression time after surfacing. Let x = depth of dive in meters, and let y optimal time in hours. A random sample of divers gave the following data. 15.1 23.3 32.2 38.3 51.3 20.5 22.7 2.48 2.38 1.48 1.03 0.75 2.38 2.20 (a) Find Ex, Ey, Ex², Ey², Exy, and r. (Roundr to three decimal places.) Ex = 203.4 Ey = 12.7 Ex2 = 0.001 Ey? = 0.002 Exy = 931.638 r= -0.9655 (b) Use a 1% level of significance to test the claim that p < 0. (Round your answers to two decimal places.) t = -8.29 critical t = -3.36 Conclusion O Reject the null hypothesis. There is insufficient evidence that p < 0. Reject the null hypothesis. There is sufficient evidence that p < 0. Fail to reject the null hypothesis. There is…please include all the steps. thank you(4) Consider n i.i.d. samples of X ~ N(µ,0²). Find the maximum likelihood estimate of o?.
- A new method has been developed in the treatment of a disease. 12 randomly selected patients were treated with this method and the time until recovery was calculated as in the picture. Establish a confidence interval for the mass mean µ. (α=0.05)A simple random sample of size n =66, is obtained from a population that is skewed left with =33 and =3. . Does the population need to be normally distributed for the sampling distribution of x to be approximately normally distributed? Why? What is the sampling distribution of x? Does the population need to be normally distributed for the sampling distribution of x to be approximately normally distributed? Why?(A) Yes. The central limit theorem states that the sampling variability of nonnormal populations will increase as the sample size increases. (B) Yes. The central limit theorem states that only for underlying populations that are normal is the shape of the sampling distribution of x normal, regardless of the sample size, n. (C)No. The central limit theorem states that only if the shape of the underlying population is normal or uniform does the sampling distribution of x, become approximately normal as the sample size, n, increases. (D) No. The central limit theorem…