Suppose we want a 90% confidence interval for the average amount spent on entertainment (movies, concerts, dates, etc.) by freshman in their first semester at a large university. the interval is to have a maximum bound on the error (or simply margin of error) of $2, and the amount spent has a normal distribution with a known standard deviation $30. the number of observations required is at least

609

Step-by-step explanation:

Standard deviation = = $30

Margin of error = E = $2

Confidence level = 90%

Since the distribution is said to be normal, we will use z scores to solve this problem.

The z score for 90% confidence level = z = 1.645

Sample size= n = ?

The formula to calculate the margin of error is:

Using the values in above equation, we get:

This means, the minimum number of observations required is 609

answer: it is d

step-by-step explanation: