May someone help me with this? if gladly appreciate it :)

Null hypothesis:  

Alternative hypothesis:  

The p value obtained was a very high value and using the significance level given  we have  so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of orders that were not accurate is not significant different from 0.1 or 10% .  

and the accuracy of the test yes is acceptable since the p value obtained is large enough to fail to reject the null hypothesis

Step-by-step explanation:

Data given and notation n  

n=367 represent the random sample taken

X=32 represent the orders that were not accurate

estimated proportion of orders that were not accurate

is the value that we want to test

represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

Alternative hypothesis:  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion  is significantly different from a hypothesized value .

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

Statistical decision  

It's important to refresh the p value method or p value approach . "This methos is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided . The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

So the p value obtained was a very high value and using the significance level given  we have  so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of orders that were not accurate is not significant different from 0.1 or 10% .