Fast food Accuracy| Confidence Interval and Hypothesis Testing for Population Proportion| Statistics

In a recent study of drive-through orders at Burger King, they found out that 365 were accurate and 71 were not accurate.
a) Construct a 95% confidence interval for the population percentage of their drive-through order that were not accurate.
b) If Burger King claims to have an accuracy of 85% on all drive-through orders, test the claim at the 0.05 levels of significance. 


Part A

Compute for the value of the point estimate. There are x=71 successes (orders that were not accurate) out of n=436 drive-through orders in the study, so


Because we want a 95% confidence interval, we have \alpha =1-0.95=0.05, so z_{\alpha /2}=z_{0.05/2}=z_{0.025}=1.96

So, we have everything we need to solve for the lower and upper bounds of the confidence interval. 

The lower bound is:

\hat{p}-z_{\alpha /2}\cdot \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}

=\frac{71}{436}-1.96\cdot \sqrt{\frac{\frac{71}{436}\left(1-\frac{71}{436}\right)}{436}}



The upper bound is:

\hat{p}+z_{\alpha /2}\cdot \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}

=\frac{71}{436}+1.96\cdot \sqrt{\frac{\frac{71}{436}\left(1-\frac{71}{436}\right)}{436}}




Therefore, we are 95% confident that the percentage of Burger King’s drive-through orders that were not accurate is between 12.8% and 19.8%. 

Part B

The null and alternative hypothese are:


H_1:\:p\ne 0.85

The sample proportion is 


The test statistic is




Because this is a two-tailed test, we determine the critical value at \alpha /2=0.025 level of significance to be z_{0.025}=\pm 1.96. The critical region is shown in the figure below.


Since the test statistic \left(z_0=-0.75\right) is not in the rejection region, we FAIL TO REJECT the null hypothesis. 

There is sufficient evidence at the \alpha =0.05 level of significance to conclude that Burger King’s accuracy in their drive-through orders is 85%.