We wish to estimate the average height of Oompa Loompa’s with a 98% ?confidence intprval. Suppose we know that the population standard deviation of heights is equal to 3cm. ?What sample size should you use to ensure that the margin of error is no larger than 2.7cm ?
The correct answer and explanation is :
To determine the required sample size (( n )) for estimating the average height of Oompa Loompas with a 98% confidence interval and a margin of error (E) of no more than 2.7 cm, we use the formula for the margin of error in a confidence interval for a population mean:
[
E = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}
]
where:
- ( E = 2.7 ) cm (the margin of error)
- ( Z_{\alpha/2} ) is the critical value for a 98% confidence level
- ( \sigma = 3 ) cm (the population standard deviation)
- ( n ) is the required sample size
Step 1: Find the Critical Value (( Z_{\alpha/2} ))
For a 98% confidence interval, the critical value (( Z_{\alpha/2} )) is found from the standard normal (Z) table. The area in each tail is:
[
\frac{1 – 0.98}{2} = 0.01
]
Looking up the Z-score that leaves 0.01 in the upper tail, we get:
[
Z_{\alpha/2} = 2.33
]
Step 2: Solve for Sample Size (( n ))
Rearranging the margin of error formula:
[
n = \left( \frac{Z_{\alpha/2} \times \sigma}{E} \right)^2
]
Substituting the values:
[
n = \left( \frac{2.33 \times 3}{2.7} \right)^2
]
[
n = \left( \frac{6.99}{2.7} \right)^2
]
[
n = \left( 2.59 \right)^2
]
[
n \approx 6.72
]
Since sample size must be a whole number, we round up to 7.
Explanation:
A larger sample size reduces variability and ensures the estimate is precise. We rounded up to ensure the margin of error does not exceed 2.7 cm. This calculation guarantees that with 98% confidence, our estimate of the average Oompa Loompa height will be within ±2.7 cm of the true mean.