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Ü15, Exercise 8

In this exercise, we search n such that under α=0,05 10 ∉ A

n=?; H_0:p=0,36; \alpha=0,05; k=10

1.Calculation of necessary parameters

\mu=0,36n

\sigma= \sqrt{n \cdot 0,36 \cdot 0,64}= \sqrt{0,2304n}

2.Calculation of the confidence interval

A=[\mu-1,96\sigma;\mu+1,96\sigma] \approx 95%

A=[0,36n-1,96 \cdot \sqrt{0,2304n};0,36n+ 1,96 \cdot \sqrt{0,2304n}

3.The lowerbound of the CI has to be bigger 10

0,36n- 1,96 \cdot \sqrt{0,2304n} >10 |-0,36n
-1,96 \cdot \sqrt{0,2304n} >10-0,36n |()^2
0,8851n <100-7,2n+0,13n^2 |-0,8851n
0 < 0,13n^2-8,0851n+100 |:0,13
0 <n^2-62,19n+769,23

n_1>45,38

n_2<17,002

4.The upperbound of the CI has to be lower than 10

0,36n+ 1,96 \cdot \sqrt{0,2304n} <10 |-0,36n
1,96 \cdot \sqrt{0,2304n} <10-0,36n |()^2
3,84 \cdot 0,2304n <100-7,2n+0,13n^2
0,8851n <100-7,2n+0,13n^2
0 < 0,13n^2-8,0851n+100 |:0,13
0 <n^2-62,19n+769,23

n_1<17,002

n_2>45,38

Ü15 8

The values with our calculation of the right intervall will be accepted.

We can see it also so:

n<17, when 10< µ+1,96σ --> right intervall

n>46, when 10> µ-1,96σ --> left intervall


n \le 17 V n  \ge 46

4.Check with n ≤ 16

n=16; \mu=5,76; \sigma=1,92; 1,96 \cdot \sigma=3,76


A=[1,99868;9,5232]

\approx A=[1;10)

If we need n=16


n=15; \mu=5,4; \sigma=1,86; 1,96 \cdot \sigma=3,64

A=[1,76;9,04]

\approx A=[1;10)


5.Check with n≥46

n=46; \mu=16,56 \sigma=3,255

A=[10,1792;22,9408]

\approx A=(10;23]

n=47; \mu=16,92; \sigma=3,255

A=[10,53922;23,30078]

\approx A=(10;24]


For n≤16 or n≥46 we can reject H0 at the 5% signifiance level.

Aufgabe8

To reject H0 at the 5% signifiance level, the following condition has to hold: n ∉ [18;45]