Question

In a study of the accuracy of fast food drive-through orders, McDonald's had 33. orders that were not accurate among 362 orders observed (based on data from QSR magazine). Construct a 95% confidence interval for the proportion of orders that are not accurate. What would be the sample sizes needed to get a 95% confidence interval of plus or minus 3% given that the initial estimate of the population proportion is either 1%, 25%, 50%, 75% or 99% (calculate the five intervals). What do you notice that is interesting?

What would be the sample sizes needed to get a 95% confidence interval of

plus or minus 3% given that the initial estimate of the population proportion is either 1%, 25%, 50%,

75% or 99% (calculate the five intervals). What do you notice that is interesting?

Answer 1

The sample proportion here is computed as:
p = x/n = 33/362 = 0.0912

From standard normal tables, we have:
P(-1.96 < Z < 1.96) = 0.95

Therefore the confidence interval here is obtained as:

This is the required 95% confidence interval for the proportion of inaccurate orders.

b) The sample size from margin of error is computed as:

Therefore, for different prior proportion values, the sample sizes required is computed here as:

The thing we notice here is that:

• The sample size require for a value of p is same as that required for (1-p), which is obvious given the fact that the sample size is directly proportional to p(1-p)
• Also as proportion increases till 0.5, the sample size required increases, after which it again reduces as it tends to 1.