Math  /  Data & Statistics

Question22. Let XX denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of XX is f(x,θ)={(θ+1)xθ0x10 otherwise f(x, \theta)=\left\{\begin{array}{cl} (\theta+1) x^{\theta} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. where 1<θ-1<\theta. A random sample of ten students yields data x1=.92,x2=.79,x3=.90,x4=.65x_{1}=.92, x_{2}=.79, x_{3}=.90, x_{4}=.65, x5=.86,x6=.47,x7=.73,x8=.97,x9=.94x_{5}=.86, x_{6}=.47, x_{7}=.73, x_{8}=.97, x_{9}=.94, x10=.77x_{10}=.77. a. Use the method of moments to obtain an estimator of θ\theta and then compute the estimate for this data. b. Obtain the maximum likelihood estimator of θ\theta and then compute the estimate for the given data.

Studdy Solution
Solve for θ\theta:
10θ+1=i=110lnxi \frac{10}{\theta + 1} = -\sum_{i=1}^{10} \ln x_i
θ+1=10i=110lnxi \theta + 1 = -\frac{10}{\sum_{i=1}^{10} \ln x_i}
θ=10i=110lnxi1 \theta = -\frac{10}{\sum_{i=1}^{10} \ln x_i} - 1
Calculate i=110lnxi\sum_{i=1}^{10} \ln x_i:
i=110lnxi=ln(0.92)+ln(0.79)+ln(0.90)+ln(0.65)+ln(0.86)+ln(0.47)+ln(0.73)+ln(0.97)+ln(0.94)+ln(0.77) \sum_{i=1}^{10} \ln x_i = \ln(0.92) + \ln(0.79) + \ln(0.90) + \ln(0.65) + \ln(0.86) + \ln(0.47) + \ln(0.73) + \ln(0.97) + \ln(0.94) + \ln(0.77)
0.08340.23630.10540.43080.15080.75500.31470.03050.06190.2614 \approx -0.0834 -0.2363 -0.1054 -0.4308 -0.1508 -0.7550 -0.3147 -0.0305 -0.0619 -0.2614
2.4302 \approx -2.4302
Substitute back to find θ\theta:
θ=102.43021 \theta = -\frac{10}{-2.4302} - 1
θ3.11 \theta \approx 3.11
The method of moments estimate of θ\theta is 3\boxed{3}.
The maximum likelihood estimate of θ\theta is 3.11\boxed{3.11}.

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