Respuesta :
Solution :
It is given that :
[tex]$H_0:p \geq 0.05$[/tex] (Null hypothesis, [tex]$H_0 : p = 0.05$[/tex] is also correct)
[tex]$H_a:p<0.05$[/tex] (Alternate hypothesis, also called [tex]$H_1$[/tex])
This is a lower tailed test,
Therefore,
Sample proportion, [tex]$\hat p = \frac{x}{n}$[/tex]
[tex]$=\frac{8}{200}$[/tex]
= 0.04
And claimed proportion, P = 0.05
Significance level, [tex]$\alpha$[/tex] = 0.01
Therefore, calculating the statistics, we get
Standard deviation of [tex]$\hat p, \sigma_{\hat p}=\sqrt{\frac{P\times (1-P)}{n}}$[/tex]
[tex]$=\sqrt{\frac{0.05\times (1-0.05)}{200}}$[/tex]
[tex]$\approx 0.0154$[/tex]
Test statistic,
[tex]$z_{observed}=\frac{\hat p -0.05}{\sigma_{\hat p}}$[/tex]
[tex]$=\frac{0.04 -0.05}{0.0154}$[/tex]
[tex]$\approx -0.65$[/tex]
Now since this is a lower tailed test, p-value = [tex]$P(Z < z_{observed})=P(Z < -0.65)=0.2578$[/tex]
Rejection Criteria : Reject [tex]$H_0$[/tex] if p-value < α .
Conclusion : Since the p-value [tex]$\geq \alpha$[/tex], we fail to reject the null hypothesis. There is insufficient evidence that p is less than 0.05
The proportion 'p-hat' for the proportion of pills that have failed to retain their potency is: 0.04
The given parameters are:
H0 : p = 0.05, Ha : p < 0.05
n = 200
x = 8
p-hat is then calculated as:
[tex]\mathbf{\^p = \frac{x}{n}}[/tex]
Substitute values for x and n
[tex]\mathbf{\^p = \frac{8}{200}}[/tex]
Divide
[tex]\mathbf{\^p = 0.04}[/tex]
Hence, the proportion is 0.04
Read more about proportion p-hat at:
https://brainly.com/question/22816678
