The interval (l1, l2) which center is l assessment * and in which about probability an alpha the true value of parameter is concluded, is called the confidential interval corresponding to confidential probability an alpha. It should be noted that l * belongs to estimates dot, and a confidential interval – to interval.
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Instruction
1. It is necessary to tell several words about estimates. Let by results of selective values random variable Х {x1, x2..., xn } be required to define the unknown parameter l on which distribution depends. Receiving assessment of parameter l * is that to each sample some value of parameter is put in compliance, that is function of results of observation of Q which value is accepted equal to estimated value of parameter l * = by Q is created (x1, x2..., xn).
2. Any function of results of observations is called statistics. If at the same time it completely describes this parameter (phenomenon), then it is called sufficient statistics. As results of observations are accidental, l * also random variable. The problem of definition of statistics has to be solved taking into account its criteria of quality. At the same time it should be noted that the distribution law of assessment is quite defined if distribution of W (x, l) Is known (W – probability density).
3. Confidential probability is chosen the researcher and has to be rather big, that is such that in the conditions of the considered task it could be considered the probability of almost reliable event. The confidential interval can be calculated most simply if assessment distribution law is known. For an example it is possible to consider a confidential interval of assessment of expected value (average value of a random variable) of mx * = (1/n) (x1+x2 + … +xn). Such assessment is not displaced, that is its expected value (average value) to equally true value of parameter (M {mx*} = mx).
4. Besides, it is easy to establish that dispersion of assessment of expected value бх*^2=Dx/n. On the basis of the central limit theorem it is possible to draw a conclusion that the distribution law of this assessment Gaussian (normal). Therefore, for carrying out calculations it is possible to use integral of probabilities of F(z) (it is not necessary to confuse with F0(z) – one of integral forms). Then, having chosen length of a confidential interval as equal 2ld, it will turn out: alpha = P {mx-ld
5. From here the following technique of creation of a confidential interval of assessment of expected value follows: 1. Having set by confidential probability an alpha, find size (an alpha +1)/2.2. According to tables of integral of probability you will choose ld/sqrt (Dx/n).3 value. As true dispersion is unknown, instead of it it is possible to take its assessment: Dx *= (1/n) ((x1 - mx *)^2+ (x2 - mx *)^2+ …+ (xn - mx *)^2).4. Find ld. 5. Write down a confidential interval (mx*-ld, mx *+ld)