For assessment of degree of reliability of the value of the measured size received in the settlement way it is necessary to define a confidential interval. It is an interval in which borders there is its expected value.
It is required to you
- - Laplace's table.
Instruction
1. Search of a confidential interval – one of ways of assessment of an error of statistical calculations. Unlike a dot method which assumes calculation of concrete size of a deviation (expected value, a mean square deviation and so forth), the interval method allows to capture wider range of possible errors.
2. To define a confidential interval, it is necessarysearchborder within which the value of expected value fluctuates. It is necessary for their calculation that the considered random variable was distributed under the normal law around some average expected value.
3. So, let there is a random variable which selective values are a set of X, and their probabilities are distribution function elements. Let the mean square deviation σ be also known, then the confidential interval can be defined in the form of the following double inequality: m(x) – t • σ/√ n
Calculation of a confidential interval requires the table of values of function of Laplace which represent probabilities that the value of a random variable will get to this interval. Expressions of m (x) there is t • σ/√ n and m(x) + t • σ/√ n are called confidential limits.
Example: find a confidential interval if the sample of volume of 25 elements is given and it is known that a mean square deviation σ=8, selective average m(x) = 15, and the level of reliability of an interval 0.85 is set.
Decision. Calculate value of an argument of function of Laplace according to the table. For φ(t) = 0.85 it are equal to 1.44. Substitute all known sizes in the general formula: 15 – 1.44•8/5
Write down result: 12.696
4. Calculation of a confidential interval requires the table of values of function of Laplace which represent probabilities that the value of a random variable will get to this interval. Expressions of m (x) there is t • σ/√ n and m(x) + t • σ/√ n are called confidential limits.
5. Example: find a confidential interval if the sample of volume of 25 elements is given and it is known that a mean square deviation σ=8, selective average m(x) = 15, and the level of reliability of an interval 0.85 is set.
6. Decision. Calculate value of an argument of function of Laplace according to the table. For φ(t) = 0.85 it are equal to 1.44. Substitute all known sizes in the general formula: 15 – 1.44•8/5
Write down result: 12.696
7. Write down result: 12.696