Final answer:
When comparing samples of n=50 and n=100 with the same sample proportion, the margin of error for the sample of n=100 would be smaller, leading to a more precise estimate for the restaurant chain's proportion of new customers.
Step-by-step explanation:
The restaurant chain is dealing with a proportion problem because it wants to determine what proportion of their customers are new. The sample proportion is a categorical variable that represents a success (a new customer) or a failure (not a new customer). Since the central limit theorem for proportions applies here, the distribution of the sample proportions (P') is approximately normal with a mean equal to the population proportion (p) and a standard deviation equal to √p(1-p)/n.
Using the provided information, we know that when increasing the sample size from n=50 to n=100, the margin of error will decrease. This is because the margin of error is inversely proportional to the square root of the sample size (√n). Thus, the larger the sample size, the smaller the margin of error, assuming the sample proportion remains the same. This is essential in designing the confidence interval for a proportion, where the desired precision is to be met without unnecessarily inflating the sample size.
Therefore, in this case, the margin of error for a sample of n=100 would be smaller than that of a sample of n=50, resulting in a more precise estimate for a one-sample z interval for a proportion.