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

(a) 25. (b) 30. (c) 608. (d) 609. (e) 865.

Short answer

The number of observations required is the closest to$609$.

Step-by-step solution

introduction

We know, confidence level = $90\%$

Interval is to have a margin of error E=$2$

The standard deviation of the amount spent will be P =$30$

explanation

Calculating the sample size $\mathbf{S}=\frac{{\mathrm{Z}}^2\times \mathrm{P}\times \mathrm{Q}}{{\mathrm{E}}^2}$

Z=$1.645$at confidence level =$90\%$

$$\begin{gathered} ={\left(\frac{1.645\times 30}{2}\right)}^2 \\ =609 \end{gathered}$$