Question

In a study of the accuracy of fast food​ drive-through orders, one restaurant had 34 orders that were not accurate among 391 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?

Answer 1

Answer: Yes , the accuracy rate appear to be​ acceptable .

Step-by-step explanation:

Let p be the population proportion of the orders that were not accurate .

Then according to the claim we have ,

$H_0:p=0.10\\\\ H_a:p\neq0.10$

Since the alternative hypothesis is two-tailed so the hypothesis test is a  two-tailed test.

For sample ,

n = 391

Proportion of  the orders that were not accurate =$\hat{p}=\dfrac{34}{391}\approx0.087$

Test statistics for population proportion :-

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\=\dfrac{0.087-0.10}{\sqrt{\dfrac{0.10(0.90)}{391}}}\approx-0.86$

By using the standard normal distribution table,

The p-value : $2(z>-0.86)=0.389789\approx0.39$

Since the p-value is greater that the significance level (0.05), so we do not reject the null hypothesis.

Hence, we conclude that the accuracy rate appear to be​ acceptable.