Let p be the population proportion of orders are delivered within 10 minutes of the time the order is placed.
Then according to the claim we have ,
[tex]H_0:p=0.90\\\\ H_a:p\neq0.90[/tex]
Since the alternative hypothesis is two-tailed so the hypothesis test is a two-tailed test.
For sample ,
n = 90
Proportion of orders are delivered within 10 minutes of the time the order is placed=[tex]\hat{p}=\dfrac{80}{90}\approx0.89[/tex]
Test statistics for population proportion :-
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\=\dfrac{0.89-0.90}{\sqrt{\dfrac{0.90(0.10)}{90}}}\approx-0.32[/tex]
The p-value : [tex]2(z>-0.32)=0.7489683\approx0.75[/tex] [By using standard normal distribution table]
Since the p-value is greater that the significance level (0.01), so we do not reject the null hypothesis.
Hence, we conclude that we have enough evidence to support the claim that 90% of its orders are delivered within 10 minutes of the time the order is placed.
