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

Null hypothesis:
`p=0.1`

Alternative hypothesis:
`p \\eq 0.1`

`z=\frac{0.087 -0.1}{\sqrt{(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.

Explanation:

Data given and notation n

n=367 represent the random sample taken

X=32 represent the orders that were not accurate

`\hat p=(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)

Alternative hypothesis:
`p \\eq 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{(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{(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% .