Respuesta :
Answer:
There is statistical evidence to support the claim that the goal of reducing the rate of undercutting to less than 5% has been met.
P-value=0.01923.
Step-by-step explanation:
We have to test the hypothesis that the proportion of defective circuits is under 5%.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.05\\\\H_a:\pi<0.05[/tex]
We will assume a level of significance of 0.05.
The sample, of size n=1000, has 35 defecteive circuits, so the sample proportion is:
[tex]p=35/1000=0.035[/tex]
The standard error is calculated as if the null hypothesis is true, so it is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.05*0.95}{1000}}}=\sqrt{0.0000475}=0.007[/tex]
The z-statistic can be calculated as:
[tex]z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.035-0.050+0.5/1000}{0.007}=\dfrac{-0.0145}{0.007}= -2.07[/tex]
For this one-tailed test, the P-value is:
[tex]P-value=P(z<-2.07)=0.01923[/tex]
As the P-value is smaller than the significance level, the effect is significant and the null hypothesis is rejected. There is statistical evidence to support the claim that the goal of reducing the rate of undercutting to less than 5% has been met.
