Respuesta :
Solution :
Given the number of the edges are = 11
Therefore we take [tex]$\frac{1}{11}=0.091$[/tex]
And the number of the vials total is = 100
Thus we can multiply = [tex]$100 \times 0.091 = 9.1$[/tex]
From the table provided below, we can obtained :
1. Null hypothesis and alternate hypothesis :
The following null and alternate hypothesis needs to be tested :
[tex]$H_0:p_1=0.091, \ p_2 = 0.091, p_3=0.091, \p_4=0.091, \p_5=0.091, \p_6=0.091, \p_7=0.091,$[/tex] [tex]$p_8=0.091, \ p_9=0.091, \ p_{10}=0.091, \ p_{11}=0.091 $[/tex]
[tex]$H_a :$[/tex] some of the population differ from the values stated in the null hypothesis. It corresponds the Chi-square test that is for the Goodness of the Fit.
2. Rejection region :
It is given --
significance level, α = 0.05
number of degrees of freedom, df = 11 - 1
= 10
Then the rejection region of this test is :
[tex]$R=\{ x^2:x^2 >18.307\}$[/tex]
3. Test statistics
The Chi-Squared statistics is computed as :
[tex]$x^2=\sum_{i=1}^n \frac{(O_i-E_i)^2}{E_i}$[/tex]
[tex]$=7.21+4.089+0.133+0.089+1.056+3.825+2.638+0.924+1.671+0.001+0.001$[/tex]
= 21.638
4. Decision about null hypothesis
Since we see that : [tex]$x^2=21.638>x^2_c = 18.307$[/tex], we can concluded that we can reject the null hypothesis.
5. Conclusion :
It is concluded that the null hypothesis [tex]$H_0$[/tex] is rejected. So there is enough evidence for us to claim that the population proportions differ from that stated in null hypothesis at α = 0.05 significance level.
