5.05b Unbiased estimates: of population mean and variance

4 In a survey a random sample of 150 households in Nantville were asked to fill in a questionnaire about household budgeting.

1. The results showed that 33 households owned more than one car. Find an approximate \(99 \%\) confidence interval for the proportion of all households in Nantville with more than one car. [4]
2. The results also included the weekly expenditure on food, \(x\) dollars, of the households. These were summarised as follows. $$n = 150 \quad \Sigma x = 19035 \quad \Sigma x ^ { 2 } = 4054716$$ Find unbiased estimates of the mean and variance of the weekly expenditure on food of all households in Nantville.
3. The government has a list of all the households in Nantville numbered from 1 to 9526. Describe briefly how to use random numbers to select a sample of 150 households from this list.

3 Jagdeesh measured the lengths, \(x\) minutes, of 60 randomly chosen lectures. His results are summarised below. $$n = 60 \quad \Sigma x = 3420 \quad \Sigma x ^ { 2 } = 195200$$

1. Calculate unbiased estimates of the population mean and variance.
2. Calculate a \(98 \%\) confidence interval for the population mean.

7 The diameter, in cm, of pistons made in a certain factory is denoted by \(X\), where \(X\) is normally distributed with mean \(\mu\) and variance \(\sigma ^ { 2 }\). The diameters of a random sample of 100 pistons were measured, with the following results. $$n = 100 \quad \Sigma x = 208.7 \quad \Sigma x ^ { 2 } = 435.57$$

1. Calculate unbiased estimates of \(\mu\) and \(\sigma ^ { 2 }\). The pistons are designed to fit into cylinders. The internal diameter, in cm , of the cylinders is denoted by \(Y\), where \(Y\) has an independent normal distribution with mean 2.12 and variance 0.000144 . A piston will not fit into a cylinder if \(Y - X < 0.01\).
2. Using your answers to part (i), find the probability that a randomly chosen piston will not fit into a randomly chosen cylinder.

1 The weights, in kilograms, of a random sample of eight 16-year old males are given below. $$\begin{array} { l l l l l l l l } 58.9 & 63.5 & 62.7 & 59.4 & 66.9 & 68.0 & 60.4 & 68.2 \end{array}$$ Find unbiased estimates of the population mean and variance of the weights of all 16 -year old males.

5 It is claimed that \(30 \%\) of packets of Froogum contain a free gift. Andre thinks that the actual proportion is less than \(30 \%\) and he decides to carry out a hypothesis test at the \(5 \%\) significance level. He buys 20 packets of Froogum and notes the number of free gifts he obtains.

1. State null and alternative hypotheses for the test.
2. Use a binomial distribution to find the probability of a Type I error. Andre finds that 3 of the 20 packets contain free gifts.
3. Carry out the test.

4 The height of sweet pea plants grown in a nursery is a random variable. A random sample of 50 plants is measured and is found to have a mean height 1.72 m and variance \(0.0967 \mathrm {~m} ^ { 2 }\).

1. Calculate an unbiased estimate for the population variance of the heights of sweet pea plants.
2. Hence test, at the \(10 \%\) significance level, whether the mean height of sweet pea plants grown by the nursery is 1.8 m , stating your hypotheses clearly.

1 A random sample of observations of a random variable \(X\) is summarised by $$n = 100 , \quad \Sigma x = 4830.0 , \quad \Sigma x ^ { 2 } = 249 \text { 509.16. }$$

1. Obtain unbiased estimates of the mean and variance of \(X\).
2. The sample mean of 100 observations of \(X\) is denoted by \(\bar { X }\). Explain whether you would need any further information about the distribution of \(X\) in order to estimate \(\mathrm { P } ( \bar { X } > 60 )\). [You should not attempt to carry out the calculation.]

6 A company with a large fleet of cars compared two types of tyres, \(A\) and \(B\). They measured the stopping distances of cars when travelling at a fixed speed on a dry road. They selected 20 cars at random from the fleet and divided them randomly into two groups of 10 , one group being fitted with tyres of type \(A\) and the other group with tyres of type \(B\). One of the cars fitted with tyres of type \(A\) broke down so these tyres were tested on only 9 cars. The stopping distances, \(x\) metres, for the two samples are summarised by $$n _ { A } = 9 , \quad \bar { x } _ { A } = 17.30 , \quad s _ { A } ^ { 2 } = 0.7400 , \quad n _ { B } = 10 , \quad \bar { x } _ { B } = 14.74 , \quad s _ { B } ^ { 2 } = 0.8160 ,$$ where \(s _ { A } ^ { 2 }\) and \(s _ { B } ^ { 2 }\) are unbiased estimates of the two population variances.
It is given that the two populations have the same variance.

1. Show that an unbiased estimate of this variance is 0.780 , correct to 3 decimal places. The population mean stopping distances for cars with tyres of types \(A\) and \(B\) are denoted by \(\mu _ { A }\) metres and \(\mu _ { B }\) metres respectively.
2. Stating any further assumption you need to make, calculate a \(98 \%\) confidence interval for \(\mu _ { A } - \mu _ { B }\). The manufacturers of Type \(B\) tyres assert that \(\mu _ { B } < \mu _ { A } - 2\).
3. Carry out a significance test of this assertion at the \(5 \%\) significance level. \section*{[Question 7 is printed overleaf.]}

7 The continuous random variable \(X\) has a uniform distribution over the interval \([ 0 , \theta ]\) so that the probability density function is given by $$f ( x ) = \begin{cases} \frac { 1 } { \theta } & 0 \leqslant x \leqslant \theta \\ 0 & \text { otherwise } \end{cases}$$ where \(\theta\) is a positive constant. A sample of \(n\) independent observations of \(X\) is taken and the sample mean is denoted by \(\bar { X }\).

1. The estimator \(T _ { 1 }\) is defined by \(T _ { 1 } = 2 \bar { X }\). Show that \(T _ { 1 }\) is an unbiased estimator of \(\theta\). It is given that the probability density function of the largest value, \(U\), in the sample is $$g ( u ) = \begin{cases} \frac { n u ^ { n - 1 } } { \theta ^ { n } } & 0 \leqslant u \leqslant \theta \\ 0 & \text { otherwise } \end{cases}$$
2. Find \(\mathrm { E } ( U )\) and show that \(\operatorname { Var } ( U ) = \frac { n \theta ^ { 2 } } { ( n + 1 ) ^ { 2 } ( n + 2 ) }\).
3. The estimator \(T _ { 2 }\) is defined by \(T _ { 2 } = \frac { n + 1 } { n } U\). Given that \(T _ { 2 }\) is also an unbiased estimator of \(\theta\), show that \(T _ { 2 }\) is a more efficient estimator than \(T _ { 1 }\) for \(n > 1\).

6 The continuous random variable \(Y\) has cumulative distribution function given by $$\mathrm { F } ( y ) = \begin{cases} 0 & y < a , \\ 1 - \frac { a ^ { 3 } } { y ^ { 3 } } & y \geqslant a , \end{cases}$$ where \(a\) is a positive constant. A random sample of 3 observations, \(Y _ { 1 } , Y _ { 2 } , Y _ { 3 }\), is taken, and the smallest is denoted by \(S\).

1. Show that \(\mathrm { P } ( S > s ) = \left( \frac { a } { s } \right) ^ { 9 }\) and hence obtain the probability density function of \(S\).
2. Show that \(S\) is not an unbiased estimator of \(a\), and construct an unbiased estimator, \(T _ { 1 }\), based on \(S\). It is given that \(T _ { 2 }\), where \(T _ { 2 } = \frac { 2 } { 9 } \left( Y _ { 1 } + Y _ { 2 } + Y _ { 3 } \right)\), is another unbiased estimator of \(a\).
3. Given that \(\operatorname { Var } ( Y ) = \frac { 3 } { 4 } a ^ { 2 }\) and \(\operatorname { Var } ( S ) = \frac { 9 } { 448 } a ^ { 2 }\), determine which of \(T _ { 1 }\) and \(T _ { 2 }\) is the more efficient estimator.
4. The values of \(Y\) for a particular sample are 12.8, 4.5 and 7.0. Find the values of \(T _ { 1 }\) and \(T _ { 2 }\) for this sample, and give a reason, unrelated to efficiency, why \(T _ { 1 }\) gives a better estimate of \(a\) than \(T _ { 2 }\) in this case.

7 The continuous random variable \(U\) has unknown mean \(\mu\) and known variance \(\sigma ^ { 2 }\). In order to estimate \(\mu\), two random samples, one of 4 observations of \(U\) and the other of 6 observations of \(U\), are taken. The sample means are denoted by \(\bar { U } _ { 4 }\) and \(\bar { U } _ { 6 }\) respectively. One estimator \(S\), given by \(S = \frac { 1 } { 2 } \left( \bar { U } _ { 4 } + \bar { U } _ { 6 } \right)\), is proposed.

1. Show that \(S\) is unbiased and find \(\operatorname { Var } ( S )\) in terms of \(\sigma ^ { 2 }\). A second estimator \(T\) of the form \(a \bar { U } _ { 4 } + b \bar { U } _ { 6 }\) is proposed, where \(a\) and \(b\) are chosen such that \(T\) is an unbiased estimator for \(\mu\) with the smallest possible variance.
2. Find the values of \(a\) and \(b\) and the corresponding variance of \(T\).
3. State, giving a reason, which of \(S\) and \(T\) is the better estimator.
4. Compare the efficiencies of this preferred estimator and the mean of all 10 observations.

7 The continuous random variable \(X\) has probability density function given by $$f ( x ) = \begin{cases} \frac { 1 } { 4 } ( 1 + a x ) & - 2 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$ where \(a\) is a constant.

1. Show that \(| a | \leqslant \frac { 1 } { 2 }\).
2. Find \(\mathrm { E } ( X )\) in terms of \(a\).
3. Construct an unbiased estimator \(T _ { 1 }\) of \(a\) based on one observation \(X _ { 1 }\) of \(X\).
4. A second observation \(X _ { 2 }\) is taken. Show that \(T _ { 2 }\), where \(T _ { 2 } = \frac { 3 } { 8 } \left( X _ { 1 } + X _ { 2 } \right)\), is also an unbiased estimator of a.
5. Given that \(\operatorname { Var } ( X ) = \sigma ^ { 2 }\), determine which of \(T _ { 1 }\) and \(T _ { 2 }\) is the better estimator.

6 The continuous random variable \(X\) has mean \(\mu\) and variance \(\sigma ^ { 2 }\), and the independent continuous random variable \(Y\) has mean \(2 \mu\) and variance \(3 \sigma ^ { 2 }\). Two observations of \(X\) and three observations of \(Y\) are taken and are denoted by \(X _ { 1 } , X _ { 2 } , Y _ { 1 } , Y _ { 2 }\) and \(Y _ { 3 }\) respectively.

1. Find the expectation of the sum of these 5 observations and hence construct an unbiased estimator, \(T _ { 1 }\), of \(\mu\).
2. The estimator \(T _ { 2 }\), where \(T _ { 2 } = X _ { 1 } + X _ { 2 } + c \left( Y _ { 1 } + Y _ { 2 } + Y _ { 3 } \right)\), is an unbiased estimator of \(\mu\). Find the value of the constant \(c\).
3. Determine which of \(T _ { 1 }\) and \(T _ { 2 }\) is more efficient.
4. Find the values of the constants \(a\) and \(b\) for which $$a \left( X _ { 1 } ^ { 2 } + X _ { 2 } ^ { 2 } \right) + b \left( Y _ { 1 } ^ { 2 } + Y _ { 2 } ^ { 2 } + Y _ { 3 } ^ { 2 } \right)$$ is an unbiased estimator of \(\sigma ^ { 2 }\).

7 The continuous random variable \(X\) has probability density function $$f ( x ) = \left\{ \begin{array} { c l } \frac { k } { ( x + \theta ) ^ { 5 } } & \text { for } x \geqslant 0 \\ 0 & \text { otherwise } \end{array} \right.$$ where \(k\) is a positive constant and \(\theta\) is a parameter taking positive values.

1. Find an expression for \(k\) in terms of \(\theta\).
2. Show that \(\mathrm { E } ( X ) = \frac { 1 } { 3 } \theta\). You are given that \(\operatorname { Var } ( X ) = \frac { 2 } { 9 } \theta ^ { 2 }\). A random sample \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\) of \(n\) observations of \(X\) is obtained. The estimator \(T _ { 1 }\) is defined as \(T _ { 1 } = \frac { 3 } { n } \sum _ { i = 1 } ^ { n } X _ { i }\).
3. Show that \(T _ { 1 }\) is an unbiased estimator of \(\theta\), and find the variance of \(T _ { 1 }\).
4. A second unbiased estimator \(T _ { 2 }\) is defined by \(T _ { 2 } = \frac { 1 } { 3 } \left( X _ { 1 } + 3 X _ { 2 } + 5 X _ { 3 } \right)\). For the case \(n = 3\), which of \(T _ { 1 }\) and \(T _ { 2 }\) is more efficient? \section*{OCR}

1 A parcel is weighed, independently, on two scales. The weights are given by the random variables \(W _ { 1 }\) and \(W _ { 2 }\) which have underlying Normal distributions as follows. $$W _ { 1 } \sim \mathrm {~N} \left( \mu , \sigma _ { 1 } ^ { 2 } \right) , \quad W _ { 2 } \sim \mathrm {~N} \left( \mu , \sigma _ { 2 } ^ { 2 } \right) ,$$ where \(\mu\) is an unknown parameter and \(\sigma _ { 1 } ^ { 2 }\) and \(\sigma _ { 2 } ^ { 2 }\) are taken as known.

1. Show that the maximum likelihood estimator of \(\mu\) is $$\hat { \mu } = \frac { \sigma _ { 2 } ^ { 2 } } { \sigma _ { 1 } ^ { 2 } + \sigma _ { 2 } ^ { 2 } } W _ { 1 } + \frac { \sigma _ { 1 } ^ { 2 } } { \sigma _ { 1 } ^ { 2 } + \sigma _ { 2 } ^ { 2 } } W _ { 2 } .$$ [You may quote the probability density function of the general Normal distribution from page 9 in the MEI Examination Formulae and Tables Booklet (MF2).]
2. Show that \(\hat { \mu }\) is an unbiased estimator of \(\mu\).
3. Obtain the variance of \(\hat { \mu }\).
4. A simpler estimator \(T = \frac { 1 } { 2 } \left( W _ { 1 } + W _ { 2 } \right)\) is proposed. Write down the variance of \(T\) and hence show that the relative efficiency of \(T\) with respect to \(\hat { \mu }\) is $$y = \left( \frac { 2 \sigma _ { 1 } \sigma _ { 2 } } { \sigma _ { 1 } ^ { 2 } + \sigma _ { 2 } ^ { 2 } } \right) ^ { 2 }$$
5. Show that \(y \leqslant 1\) for all values of \(\sigma _ { 1 } ^ { 2 }\) and \(\sigma _ { 2 } ^ { 2 }\). Explain why this means that \(\hat { \mu }\) is preferable to \(T\) as an estimator of \(\mu\).

1 The random variable \(X\) has the continuous uniform distribution with probability density function $$\mathrm { f } ( x ) = \frac { 1 } { \theta } , \quad 0 \leqslant x \leqslant \theta$$ where \(\theta ( \theta > 0 )\) is an unknown parameter.
A random sample of \(n\) observations from \(X\) is denoted by \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\), with sample mean \(\bar { X } = \frac { 1 } { n } \sum _ { i = 1 } ^ { n } X _ { i }\).

1. Show that \(2 \bar { X }\) is an unbiased estimator of \(\theta\).
2. Evaluate \(2 \bar { X }\) for a case where, with \(n = 5\), the observed values of the random sample are \(0.4,0.2\), 1.0, 0.1, 0.6. Hence comment on a disadvantage of \(2 \bar { X }\) as an estimator of \(\theta\). For a general random sample of size \(n\), let \(Y\) represent the sample maximum, \(Y = \max \left( X _ { 1 } , X _ { 2 } , \ldots , X _ { n } \right)\). You are given that the probability density function of \(Y\) is $$g ( y ) = \frac { n y ^ { n - 1 } } { \theta ^ { n } } , \quad 0 \leqslant y \leqslant \theta$$
3. An estimator \(k Y\) is to be used to estimate \(\theta\), where \(k\) is a constant to be chosen. Show that the mean square error of \(k Y\) is $$k ^ { 2 } \mathrm { E } \left( Y ^ { 2 } \right) - 2 k \theta \mathrm { E } ( Y ) + \theta ^ { 2 }$$ and hence find the value of \(k\) for which the mean square error is minimised.
4. Comment on whether \(k Y\) with the value of \(k\) found in part (iii) suffers from the disadvantage identified in part (ii).

1 The random variable \(X\) has the Poisson distribution with parameter \(\theta\) so that its probability function is $$\mathrm { P } ( X = x ) = \frac { \mathrm { e } ^ { - \theta } \theta ^ { x } } { x ! } , \quad x = 0,1,2 , \ldots$$ where \(\theta ( \theta > 0 )\) is unknown. A random sample of \(n\) observations from \(X\) is denoted by \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\).

1. Find \(\hat { \theta }\), the maximum likelihood estimator of \(\theta\). The value of \(\mathrm { P } ( X = 0 )\) is denoted by \(\lambda\).
2. Write down an expression for \(\lambda\) in terms of \(\theta\).
3. Let \(R\) denote the number of observations in the sample with value zero. By considering the binomial distribution with parameters \(n\) and \(\mathrm { e } ^ { - \theta }\), write down \(\mathrm { E } ( R )\) and \(\operatorname { Var } ( R )\). Deduce that the observed proportion of observations in the sample with value zero, denoted by \(\tilde { \lambda }\), is an unbiased estimator of \(\lambda\) with variance \(\frac { \mathrm { e } ^ { - \theta } \left( 1 - \mathrm { e } ^ { - \theta } \right) } { n }\).
4. In large samples, the variance of the maximum likelihood estimator of \(\lambda\) may be taken as \(\frac { \theta \mathrm { e } ^ { - 2 \theta } } { n }\). Use this and the appropriate result from part (iii) to show that the relative efficiency of \(\tilde { \lambda }\) with respect to the maximum likelihood estimator is \(\frac { \theta } { \mathrm { e } ^ { \theta } - 1 }\). Show that this expression is always less than 1 . Show also that it is near 1 if \(\theta\) is small and near 0 if \(\theta\) is large.

1 The random variable \(X\) has probability density function $$\mathrm { f } ( x ) = \frac { x \mathrm { e } ^ { - x / \lambda } } { \lambda ^ { 2 } } \quad ( x > 0 )$$ where \(\lambda\) is a parameter \(( \lambda > 0 ) . X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\) are \(n\) independent observations on \(X\), and \(\bar { X } = \frac { 1 } { n } \sum _ { i = 1 } ^ { n } X _ { i }\) is their mean.

1. Obtain \(\mathrm { E } ( X )\) and deduce that \(\hat { \lambda } = \frac { 1 } { 2 } \bar { X }\) is an unbiased estimator of \(\lambda\).
2. \(\operatorname { Obtain } \operatorname { Var } ( \hat { \lambda } )\).
3. Explain why the results in parts (i) and (ii) indicate that \(\hat { \lambda }\) is a good estimator of \(\lambda\) in large samples.
4. Suppose that \(n = 3\) and consider the alternative estimator $$\tilde { \lambda } = \frac { 1 } { 8 } X _ { 1 } + \frac { 1 } { 4 } X _ { 2 } + \frac { 1 } { 8 } X _ { 3 } .$$ Show that \(\tilde { \lambda }\) is an unbiased estimator of \(\lambda\). Find the relative efficiency of \(\tilde { \lambda }\) compared with \(\hat { \lambda }\). Which estimator do you prefer in this case?

1 The random variable \(X\) has the following probability density function, in which \(a\) is a (positive) parameter. $$\mathrm { f } ( x ) = \frac { 2 } { a } x \mathrm { e } ^ { - x ^ { 2 } / a } , \quad x \geqslant 0 .$$

1. Verify that \(\int _ { 0 } ^ { \infty } \mathrm { f } ( x ) \mathrm { d } x = 1\).
2. Show that \(\mathrm { E } \left( X ^ { 2 } \right) = a\) and \(\mathrm { E } \left( X ^ { 4 } \right) = 2 a ^ { 2 }\). The parameter \(a\) is to be estimated by maximum likelihood based on an independent random sample from the distribution, \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\).
3. Show that the logarithm of the likelihood function is $$n \ln 2 - n \ln a + \sum _ { i = 1 } ^ { n } \ln X _ { i } - \frac { 1 } { a } \sum _ { i = 1 } ^ { n } X _ { i } ^ { 2 }$$ Hence obtain the maximum likelihood estimator, \(\hat { a }\), for \(a\).
   [0pt] [You are not required to verify that any turning point you find is a maximum.]
4. Using the results from part (ii), show that \(\hat { a }\) is unbiased for \(a\) and find the variance of \(\hat { a }\).
5. In a particular random sample from this distribution, \(n = 100\) and \(\sum x _ { i } ^ { 2 } = 147.1\). Obtain an approximate 95\% confidence interval for \(a\). (You may assume that the Central Limit Theorem holds in this case.) Option 2: Generating Functions

1 The random variable \(X\) has a Cauchy distribution centred on \(m\). Its probability density function ( pdf ) is \(\mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \frac { 1 } { \pi } \frac { 1 } { 1 + ( x - m ) ^ { 2 } } , \quad \text { for } - \infty < x < \infty$$

1. Sketch the pdf. Show that the mode and median are at \(x = m\).
2. A sample of size 1 , consisting of the observation \(x _ { 1 }\), is taken from this distribution. Show that the maximum likelihood estimate (MLE) of \(m\) is \(x _ { 1 }\).
3. Now suppose that a sample of size 2 , consisting of observations \(x _ { 1 }\) and \(x _ { 2 }\), is taken from the distribution. By considering the logarithm of the likelihood function or otherwise, show that the MLE, \(\hat { m }\), satisfies the cubic equation $$\left( 2 \hat { m } - \left( x _ { 1 } + x _ { 2 } \right) \right) \left( \hat { m } ^ { 2 } - \left( x _ { 1 } + x _ { 2 } \right) \hat { m } + 1 + x _ { 1 } x _ { 2 } \right) = 0$$
4. Obtain expressions for the three roots of this equation. Show that if \(\left| x _ { 1 } - x _ { 2 } \right| < 2\) then only one root is real. How do you know, without doing further calculations, that in this case the real root will be the MLE of \(m\) ?
5. Obtain the three possible values of \(\hat { m }\) in the case \(x _ { 1 } = - 2\) and \(x _ { 2 } = 2\). Evaluate the likelihood function for each value of \(\hat { m }\) and comment on your answer.

2 A random variable \(C\) has the distribution \(\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)\). A random sample of 10 observations of \(C\) is obtained, and the results are summarised as $$n = 10 , \Sigma c = 380 , \Sigma c ^ { 2 } = 14602 .$$

1. Calculate unbiased estimates of \(\mu\) and \(\sigma ^ { 2 }\).
2. Hence calculate an estimate of the probability that \(C > 40\).

7 A continuous random variable \(Y\) has cumulative distribution function $$\mathrm { F } ( y ) = \left\{ \begin{array} { c c } 0 & y < a \\ 1 - \frac { a ^ { 5 } } { y ^ { 5 } } & y \geqslant a \end{array} \right.$$ where \(a\) is a parameter.
Two independent observations of \(Y\) are denoted by \(Y _ { 1 }\) and \(Y _ { 2 }\). The smaller of them is denoted by S .

1. Show that \(P ( S > \mathrm { s } ) = \frac { a ^ { 10 } } { s ^ { 10 } }\) and hence find the probability density function of \(S\).
2. Show that \(S\) is not an unbiased estimator of \(a\), and construct an unbiased estimator of \(a , T _ { 1 }\) based on \(S\).
3. Construct another unbiased estimator of \(a , T _ { 2 }\), of the form \(k \left( Y _ { 1 } + Y _ { 2 } \right)\), where \(k\) is a constant to be found.
4. Without further calculation, explain how you would decide which of \(T _ { 1 }\) and \(T _ { 2 }\) is the more efficient estimator.

6 The continuous random variable \(Z\) has probability density function $$f ( z ) = \left\{ \begin{array} { c c } \frac { 4 z ^ { 3 } } { k ^ { 4 } } & 0 \leqslant z \leqslant k \\ 0 & \text { otherwise } \end{array} \right.$$ where \(k\) is a parameter whose value is to be estimated.

1. Show that \(\frac { 5 Z } { 4 }\) is an unbiased estimator of \(k\).
2. Find the variance of \(\frac { 5 Z } { 4 }\). The parameter \(k\) can also be estimated by making observations of a random variable \(X\) which has mean \(\frac { 1 } { 2 } k\) and variance \(\frac { 1 } { 12 } k ^ { 2 }\). Let \(Y = X _ { 1 } + X _ { 2 } + X _ { 3 }\) where \(X _ { 1 } , X _ { 2 }\) and \(X _ { 3 }\) are independent observations of \(X\).
3. \(c Y\) is also an unbiased estimator of \(k\). Find the value of \(c\).
4. For the value of \(c\) found in part (iii), determine which of \(\frac { 5 Z } { 4 }\) and \(c Y\) is the more efficient estimator of \(k\).

1 A random sample of nine observations of a random variable is obtained. The results are summarised as $$\Sigma x = 468 , \quad \Sigma x ^ { 2 } = 24820 .$$ Calculate unbiased estimates of the population mean and variance.

6 The continuous random variable \(R\) has the distribution \(\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)\). The results of 100 observations of \(R\) are summarised by $$\Sigma r = 3360.0 , \quad \Sigma r ^ { 2 } = 115782.84 .$$

1. Calculate an unbiased estimate of \(\mu\) and an unbiased estimate of \(\sigma ^ { 2 }\).
2. The mean of 9 observations of \(R\) is denoted by \(\bar { R }\). Calculate an estimate of \(\mathrm { P } ( \bar { R } > 32.0 )\).
3. Explain whether you need to use the Central Limit Theorem in your answer to part (ii).