Suppose we wish to calculate a 90% confidence interval for the average amount spent on books by freshmen in their first year at a major university.The interval is to have a margin of error of $2.Assume that the amount spent on books by freshmen has a Normal distribution with a standard deviation of $\sigma$ = $30.How many observations are required to achieve this margin of error?
A)25
B)608
C)609
D)865

Question

Quizplus · Accepted Answer

The margin of error is given by:
Margin of Error = z-value * (standard deviation / sqrt(sample size))
We are given that the margin of error is $2, and we want a 90% confidence interval. From the Normal distribution, we know that a 90% confidence interval corresponds to a z-value of 1.645.
Plugging in the given values, we get:
2 = 1.645 * (30 / sqrt(n))
Solving for n, we get:
n = (1.645^2 * 30^2) / 2^2 = 608.71
Since we cannot have a fractional number of observations, we must round up to the nearest whole number. Therefore, we need at least n = 609 observations. Therefore, the correct answer is C.

Suppose We Wish to Calculate a 90% Confidence Interval for the Average