Confidence Interval

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1. Identify the sample mean, [pic 11]. While [pic 12] differs from [pic 13], population mean, they are still calculated the same way: [pic 14] .

2. Identify whether the standard deviation is known, [pic 15], or unknown, s.

- If standard deviation is known then z* is used as the critical value. This value is only dependent on the confidence level for the test. Typical two sided confidence levels are:

99%

2.576

98%

2.326

95%

1.96

90%

1.645

- If the standard deviation is unknown then t* is used as the critical value. This value is dependent on the confidence level (C) for the test and degrees of freedom. The degrees of freedom is found by subtracting one from the number of observations, n − 1. The critical value is found from the t-distribution Table. In this table the critical value is written as tα(r), where r is the degrees of freedom and .

3. Plug the found values into the appropriate equations:

- For a known standard deviation:

- For an unknown standard deviation:

4. The final step is to interpret the answer. Since the found answer is an interval with an upper and lower bound it is appropriate to state that based on the given data we are __ % (dependent on the confidence level) confident that the true mean of the population is between __ (lower bound) and __ (upper bound).

Example №1. When checking the validity Party tablets (250 pcs.), It was found that the average tablet weight of 0.3 g, and the standard deviation of weight 0.01 g Find the confidence interval, which with 90% probability adjudged to rate of weight of the tablet.

Solution.[pic 16]

We determine the value t kp tables Laplace function.

Then 2Ф(tkp) = 1 - γ

Ф(tkp) = γ/2 = (1- 0.05)/2 = 0.48

In the table find the Laplace function, at which the value tkp of Ф(tkp) = 0.48

tkp(γ) = Ф(0.48) = 2.06 [pic 17]

(0.3 - 0.206;0.3 + 0.206) = (0.094;0.51)

With probability 0.9 it can be argued that the average value of the haul larger volume will not exceed the found interval.

Example №2. On an area of 70 ha occupied by crop, is determined by the sampling method the proportion planting, affected by insects. How many samples must be taken in the sample with a probability of 0.997 to determine the desired value of up to 4% if the test sample shows that the proportion of the affected crop area amounts to 9%?

We seek a solution by the formula determining the sample size for re-selection.

[pic 18]

Ф(tkp) = γ/2 = 0.997/2 = 0,4985

and this value corresponds to the table by Laplace tkp =2.96. w = 9% = 0,09 Δ = 4% = 0,04 Result: n = 2.962*0,09(1-0,09)/0,042 = 448,4844 ≈ 449

Example 3. When checking the weight of imported cargo customs randomly re-sampling of 100 products were selected. As a result, the average weight of 5000 g of product with an average deviation of 40 was established with probability of 0.950 to determine the extent to which the average weight of the product is in the general population.

Solution. Because the n>30, In the table find the Laplace function, at which the value tkp .

then 2Ф(tkp) = γ

Ф(tkp) = γ/2 = 0.95/2 = 0.475

In the table find the Laplace function, at which the value tkp of Ф(tkp) = 0.475

tkp(γ) = (0.475) = 1.96

(5000 - 78.4;5000 + 78.4) = (4921.6;5078.4)

With a probability of 0.95 can be argued that the average value of the sample at a larger volume will not exceed the found interval.