Question

Suppose we want 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 at most \(2. Based on last year’s book sales, we estimate that the standard deviation of the amount spent will be close to \)30. The number of observations required is closest to which of the following?

a. $25$

b.$30$

c. $225$

d. $609$

e. $865$

Short answer

The required number of observations is $25$

The correct option is$\left(a\right)$

Step-by-step solution

Given information

Standard deviation $\left(\sigma \right)=30$

Confidence level $=90\%$

Margin of error$\left(E\right)=2$

The objective is to find out the number of observations.

We know,

The formula to compute the sample size is:

$n={\left(\frac{z_{{\displaystyle \frac{\alpha }{2}}}\times \sigma }{E}\right)}^2$

The standard normal table yields the value of $z_{\frac{a}{2}}$corresponding to a $90\%$confidence level of$1.645$.

The total number of observations can be calculated as follows:

$$\begin{gathered} n={\left(\frac{z_{{\displaystyle \frac{\alpha }{2}}}\times \sigma }{E}\right)}^2 \\ ={\left(\frac{1.645\times 30}{2}\right)}^2 \\ =24.675 \\ \approx 25 \end{gathered}$$

The sample size is $25$

Therefore, the correct option is$\left(a\right)$