Respuesta :
Given :
Possible chances, x = 92
Sample size, n = 150
Success rate, [tex]$p=\frac{x}{n}$[/tex]
[tex]$=\frac{92}{150}=0.6133$[/tex]
Success probability, [tex]$p_o$[/tex] = 0.647
Failure probability, [tex]$q_o$[/tex] = 0.353
The null hypothesis, [tex]$H_o:p=0.647$[/tex]
The alternate hypothesis, [tex]$H_1:p!=0.647$[/tex]
Level of significance, α = 0.01
Therefore from the standard table, the two tailed z = α/2 = 2.576
Since the test is a two tailed test,
we reject the null hypothesis, i.e. [tex]$H_0$[/tex], if [tex]$z_0$[/tex] < - 2.576 or if [tex]$z_0$[/tex] > 2.576
We use the test statistics z proportion = [tex]$\frac{p-p_0}{\sqrt{\frac{p_oq_o}{n}}}$[/tex]
[tex]$z_o=\frac{0.61333-0.647}{\sqrt{\frac{0.228391}{150}}}$[/tex]
[tex]$z_o=-0.8628$[/tex]
[tex]$|z_o|=0.8628$[/tex]
The critical value,
The value of | [tex]$z_{\alpha}$[/tex] | at los 0.01% is 2.576
So we got, [tex]$|z_o|=0.8628$[/tex] and | [tex]$z_{\alpha}$[/tex] | = 2.576
Conclusion :
Therefore, the value of [tex]$|z_o|<|z_{\alpha}|$[/tex] and here we do not reject the [tex]$H_0$[/tex]
The p-value : two tailed - [tex]$H_a$[/tex] : (p! = -0.86279) = 0.38825
Hence the value of p(0.01) < 0.3883, so here we do not reject the [tex]$H_0$[/tex].
