Respuesta :
Answer:
There is strong evidence that less than 87% of the orders are delivered in less than 10 minutes.
Decision rule: Reject the null hypothesis if (P-value < level of significance)
Test statistic z=-2.70
Decision: Reject the null hypothesis (0.003<0.010)
Step-by-step explanation:
In this question we have to test an hypothesis.
The null and alternative hypothesis are:
[tex]H_0: \pi\geq0.87\\\\H_a:\pi<0.87[/tex]
The significance level is assumed to be 0.01.
The sample of size n=80 gives a proportion of p=61/80=0.7625.
The standard deviation is:
[tex]\sigma=\sqrt{\frac{\pi(1-\pi)}{N} }=\sqrt{\frac{0.87*0.13}{80} }=0.0376[/tex]
The statistic z is then
[tex]z=\frac{p-\pi+0.5/N}{\sigma}=\frac{0.7625-0.87+0.5/80}{0.0376} } =\frac{-10125}{0.0376} =-2.70[/tex]
The P-value is
[tex]P(z<-2.70)=0.00347[/tex]
The P-value (0.003) is smaller than the significance level (0.010), so the effect is significant. The null hypothesis is rejected.
There is strong evidence that less than 87% of the orders are delivered in less than 10 minutes.
Decision rule: Reject the null hypothesis if (P-value < level of significance)
Test statistic z=-2.70
Decision: Reject the null hypothesis (0.003<0.010)