# Testen

### 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 $0,36n- 1,96 \cdot \sqrt{0,2304n}$ $>10$ $|-0,36n$ $-1,96 \cdot \sqrt{0,2304n}$ $>10-0,36n$ $|()^2$ $0,8851n$ $<100-7,2n+0,13n^2$ $|-0,8851n$ $0$ $< 0,13n^2-8,0851n+100$ $|:0,13$ $0$ $ $n_1>45,38$ $n_2<17,002$

4.The upperbound of the CI has to be lower than 10 $0,36n+ 1,96 \cdot \sqrt{0,2304n}$ $<10$ $|-0,36n$ $1,96 \cdot \sqrt{0,2304n}$ $<10-0,36n$ $|()^2$ $3,84 \cdot 0,2304n$ $<100-7,2n+0,13n^2$ $0,8851n$ $<100-7,2n+0,13n^2$ $0$ $< 0,13n^2-8,0851n+100$ $|:0,13$ $0$ $ $n_1<17,002$ $n_2>45,38$

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.

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