Answer:
The number of observations required is closest to 609 ⇒ b
Step-by-step explanation:
The formula of the sample size is n = [([tex]z_{\frac{\alpha}{2} }[/tex] . σ)/E]² , where
∵ 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 [tex]\frac{\alpha }{2}[/tex]
∴ [tex]\frac{\alpha }{2}[/tex] = [tex]\frac{0.10}{2}[/tex] = 0.05
- Look to the normal distribution table to find z-score for this
value with opposite sign
∴ [tex]z_{\frac{\alpha}{2} }[/tex] = 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
The number of observations required based on the confidence interval is B. 609.
From the information given, the standard deviation is $30,the critical value is 1.645 for 90% confidence level.
Therefore, the sample size will be:
= [(30/2) × 1.645]²
= 609
In conclusion, the sample size is 609.
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