2. You have a CAN estimator \hat{\theta} for parameter \theta, with point estimate \hat{\theta} = 0.45 and the asymptotic standard error aSE(\hat{\theta}) = 0.22. You are interested in the parameter \beta = exp(\theta). (a) Find \hat{\beta}. Using the Delta method, find an asymptotic standard error aSE(\hat{\beta}) for \hat{\beta}. (b) Using the above, calculate a 95\% asymptotic confidence interval for \beta. (c) Using a t-test, find the P-value of the null hypothesis H_0: \beta = 1 against the alternative H_1: \beta \neq 1. (d) Using a t-test, find the P-value of the null hypothesis H_0: \beta \leq 1 against the alternative H_1: \beta > 1. (e) Calculate a 95\% asymptotic confidence interval [L, U] for the original parameter \theta. Calculate a 95\% asymptotic interval for \beta as [exp(L), exp(U)]. Compare this interval with your answer in (c).
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Suppose that $X_{1}, X_{2}, \ldots, X_{n_{1}}, Y_{1}, Y_{2}, \ldots, Y_{n_{1}},$ and $W_{1}, W_{2}, \ldots, W_{n_{1}}$ are independent random samples from normal distributions with respective unknown means $\mu_{1}, \mu_{2},$ and $\mu_{3}$ and common variances $\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma_{3}^{2}=\sigma^{2} .$ Suppose that we want to estimate a linear function of the means: a. What is the standard error of the estimator $\hat{\theta}$ ? b. What is the distribution of the estimator $\hat{\theta}$ ? c. If the sample variances are given by $S_{1}^{2}, S_{2}^{2},$ and $S_{3}^{2},$ respectively, consider $$S_{p}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}+\left(n_{3}-1\right) S_{3}^{2}}{n_{1}+n_{2}+n_{3}-3}$$ i. What is the distribution of $\left(n_{1}+n_{2}+n_{3}-3\right) S_{p}^{2} / \sigma^{2} ?$ ii. What is the distribution of $$T=\frac{\hat{\theta}-\theta}{S_{p} \sqrt{\frac{a_{1}^{2}}{n_{1}}+\frac{a_{2}^{2}}{n_{2}}+\frac{a_{3}^{2}}{n_{3}}} ?} ?$$ d. Give a confidence interval for $\theta$ with confidence coefficient $1-\alpha$ e. Develop a test for $H_{0}: \theta=\theta_{0}$ versus $H_{a}: \theta \neq \theta_{0}$ $\theta=a_{1} \mu_{1}+a_{2} \mu_{2}+a_{3} \mu_{3}$. Because the maximum-likelihood estimator (MLE) of a function of parameters is the function of the MLEs of the parameters, the MLE of $\theta$ is $\hat{\theta}=a_{1} X+a_{2} Y+a_{3} W$
Hypothesis Testing
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A random sample of n = 36 observations has a mean x = 25.3 and a standard deviation s = 3.9. 1. Give the point estimate of the population mean μ. Find the 95% margin of error for your estimate. (Round your answer to four decimal places.) 2. Find a 90% confidence interval for μ. (Round your answers to three decimal places.) ____ to _____. 3. What does "90% confident" mean? a) In repeated sampling, 90% of all intervals constructed in this manner will enclose the population mean. b) There is a 90% chance that an individual sample mean will fall within the interval. c) In repeated sampling, 90% of all intervals constructed in this manner will enclose the population mean. d) 90% of all values will fall within the interval limits. e) There is a 10% chance that an individual sample mean will fall within the interval limits. 4. Find a 90% lower confidence bound for the population mean μ. (Round your answer to two decimal places.) 5. Why is this bound different from the lower confidence limit in part (b)? a) This bound is calculated using n, while the lower confidence limit in part (b) is calculated using square root n. b) This bound is calculated using zα, while the lower confidence limit in part (b) is calculated using zα/2. c) This bound is calculated using square root n, while the lower confidence limit in part (b) is calculated using n. The lower bounds are based on different values of x. d) This bound is calculated using zα/2, while the lower confidence limit in part (b) is calculated using zα. 6. How many observations do you need to estimate μ to within 0.6, with probability equal to 0.95? (Round your answer up to the nearest whole number.)
Keondre P.
In the following confidence interval of proportion, find the point estimate and the margin of error: .39 < p < .45 14. Given the confidence interval .39 < p < .45, would the interval become wider, narrower, or stay the same size if the confidence level increased? 15. Given the confidence interval .39 < p < .45, would the interval become wider, narrower, or stay the same size if the sample size increased. 16. If an informal survey indicates that around 10% of people in a given area have coronavirus, find the sample size needed to construct a 99% confidence interval estimate of the true population proportion, with a margin of error of 1%. 17. In a survey of 867 respondents, 265 of them believe the coronavirus is "not a grave concern". Find a 95% confidence interval estimate for the true population proportion that believes this. 18. In the same survey as the previous problem, construct a 90% confidence interval estimate for the true population proportion that believes this. Did the interval get wider or narrrower? Explain why. 19. In the same survey as the previous problem, construct a 99.9% confidence interval estimate for the true population proportion that believes this. How useful is this interval? Explain your answer. 20. In the same survey as the previous problem, find the sample size needed to have a 99.9% confidence interval estimate with a margin of error of 0.5%. Is it realistic to have a survey with that large of a sample size? Explain why or why not.
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