Question

Suppose we want 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) 865 b) 609 c) 608 d) 30​

Answer 1

Answer:

The number of observations required is closest to 609 ⇒ b

Step-by-step explanation:

The formula of the sample size is n = [($z_{\frac{\alpha}{2} }$ . σ)/E]² , where

• σ is the standard deviation
• E is the margin of error
• $z_{\frac{\alpha}{2} }$ is the z-score, where 1 - α = confidence interval

∵ We want 90% confidence interval for the average amount

   spent on books by freshmen in their first year at a major

   university

∴ 1 - α = 90%

∵ 90% = 90 ÷ 100 = 0.90

∴ 1 - α = 0.90

- Subtract 1 from both sides

∴ - α = -0.10

- Divide both sides by -1

∴ α = 0.10

- Find the value of $\frac{\alpha }{2}$

∴  $\frac{\alpha }{2}$ = $\frac{0.10}{2}$ = 0.05

- Look to the normal distribution table to find z-score for this

   value with opposite sign

∴ $z_{\frac{\alpha}{2} }$  = 1.645

∵ The interval is to have a margin of error of $2

∴ E = 2

∵ We estimate that the standard deviation of the amount spent

   will be close to $30

∴ σ = 30

- Substitute these values in the formula above

∴ n = [(1.645)(30) ÷ 2]²

∴ n = 608.855

- Round it to the whole number

∴ n = 609

The number of observations required is closest to 609