Question

In a study of the accuracy of fast food drive-through orders, one restaurant had 32 orders that were not accurate among 367 orders observed. Use a 0.05 significance level to test the claim that the rate of inaccurate orders is equal to 10%. Does the accuracy rate appear to be acceptable?

Identify the rest statistic for this hypothesis test. Round to two decimal places.

Identify the P-value for this hypothesis test. Round to two decimal places.

Identify the conclusion for this hypothesis tes.

Does the accuracy rate appear to be acceptable?

Answer 1

Answer:

Null hypothesis:$p=0.1$  

Alternative hypothesis:$p \neq 0.1$  

$z=\frac{0.087 -0.1}{\sqrt{\frac{0.1(1-0.1)}{367}}}=-0.83$  

$p_v =2*P(z<-0.83)=0.41$  

The p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ 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

$\hat p=\frac{32}{367}=0.087$ estimated proportion of orders that were not accurate

$p_o=0.1$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the rate of inaccurate orders is equal to 10%:  

Null hypothesis:[tex]p=0.1$  

Alternative hypothesis:$p \neq 0.1$  

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

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

Calculate the statistic  

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

$z=\frac{0.087 -0.1}{\sqrt{\frac{0.1(1-0.1)}{367}}}=-0.83$  

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 $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(z<-0.83)=0.41$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ 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% .