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[3] Question 8 A study is conducted to determine the mean age of students who take STTN111. A sample of 50 randomly chosen students from the 1200 students that take STTN111 showed a mean age of 19 with a standard deviation 2.8. If the mean age of all STTN111 students must be calculated accurately to the nearest year with \( 95 \% \) certainty, what is the smallest possible sample size that needs to be drawn?

Ask by Lawson Paul. in South Africa
Feb 25,2025

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The smallest sample size needed is 31 students.

Solución

To determine the smallest possible sample size needed to estimate the mean age of students taking STTN111 with \( 95\% \) certainty, we can use the formula for the sample size in estimating a population mean: \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] Where: - \( n \) = required sample size - \( Z \) = Z-value corresponding to the desired confidence level - \( \sigma \) = standard deviation of the population - \( E \) = margin of error (the precision we want in our estimate) ### Step 1: Identify the values 1. **Confidence Level**: \( 95\% \) - The Z-value for \( 95\% \) confidence is approximately \( 1.96 \). 2. **Standard Deviation**: \( \sigma = 2.8 \) 3. **Margin of Error**: Since we want to calculate the mean age to the nearest year, we set \( E = 1 \). ### Step 2: Plug the values into the formula Now we can substitute the values into the formula: \[ n = \left( \frac{1.96 \cdot 2.8}{1} \right)^2 \] ### Step 3: Calculate the sample size Let's perform the calculation to find \( n \). \[ n = \left( 1.96 \cdot 2.8 \right)^2 \] Calculating \( 1.96 \cdot 2.8 \): \[ 1.96 \cdot 2.8 = 5.488 \] Now squaring that value: \[ n = (5.488)^2 \] Now, let's calculate \( (5.488)^2 \). Calculate the value by following steps: - step0: Calculate: \(5.488^{2}\) - step1: Convert the expressions: \(\left(\frac{686}{125}\right)^{2}\) - step2: Use the properties of exponents: \(\frac{686^{2}}{125^{2}}\) - step3: Evaluate the power: \(\frac{686^{2}}{15625}\) Calculating \( (5.488)^2 \) gives us approximately \( 30.118144 \). ### Step 4: Round up to the nearest whole number Since the sample size must be a whole number, we round \( 30.118144 \) up to the nearest whole number: \[ n \approx 31 \] ### Conclusion The smallest possible sample size that needs to be drawn to estimate the mean age of students taking STTN111 with \( 95\% \) certainty, to the nearest year, is \( \boxed{31} \).

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To calculate the necessary sample size for estimating the mean age with 95% certainty, you can use the formula for the required sample size \( n \) when estimating the mean: \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] where: - \( Z \) is the z-score for the desired confidence level (1.96 for 95% confidence), - \( \sigma \) is the known population standard deviation (2.8 years), - \( E \) is the margin of error (1 year in this case, as you want to round to the nearest year). Plugging in the values gives: \[ n = \left( \frac{1.96 \cdot 2.8}{1} \right)^2 \] Calculating this, we find: \[ n = (5.488)^2 \approx 30.19 \] Since you can't take a fraction of a sample size, you would round up to 31. Therefore, you'd need a minimum sample size of 31 students to estimate the mean age with the desired level of confidence. Moreover, this process highlights the importance of sample size in research. Larger sample sizes generally yield more accurate and reliable results, minimizing the margin of error and ensuring better representation of the population. Fun fact: the concept of the margin of error can be likened to a game of darts—aiming for the bullseye (true population mean) means you want your darts (sample estimates) as close as possible, and a bigger dartboard (larger sample size) gives you a better chance!

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