In: Statistics and Probability
study of the accuracy of fast food drive through orders, Restaurant A had 339 accurate orders and 57 that were not accurate. construct a 90% confidence interval estimate of the percentage of orders not accurate. compare the results from part (a) to the 90% confidence interval for the percentage of orders that are not accurate at Restaurant B: 0.134<p<0.184. what do you conclude?
Answer;
Given that:
construct a 90% confidence interval estimate of the percentage of orders not accurate.The proportion of restaurants that are not accurate here is computed as:
p = 57 / ( 57 + 339 )
= 57 / 396
= 0.1439
Now from standard normal tables, we get:
P( -1.645 < Z < 1.645 ) = 0.9
Therefore the confidence interval for the proportion here is obtained as:
( 11.488 \% , 17.291 \%) ( In terms of percentage terms )
This is the required 90% confidence interval here.
compare the results from part (a) to the 90% confidence interval for the percentage of orders that are not accurate at Restaurant B: 0.134<p<0.184. what do you conclude?
As the confidence interval above has intersection points with the proportion confidence interval for restaurant B, therefore we cannot conclude anything here as to which restaurant has got a higher proportion of inaccurate orders.