For \( C=90 \% \) if \( \sigma=7.21 \) and \( E=2 \) then Sample Size

To determine the sample size \( n \), we can use the formula for the sample size for estimating a population mean: \[ n = \left( \frac{Z_{\alpha/2} \cdot \sigma}{E} \right)^2 \] Where: - \( Z_{\alpha/2} \) is the Z-value corresponding to the desired confidence level. - \( \sigma \) is the population standard deviation. - \( E \) is the margin of error. For a 90% confidence level, the Z-value \( Z_{0.05} \) is approximately 1.645. Plugging in the given values: \[ n = \left( \frac{1.645 \cdot 7.21}{2} \right)^2 \] Calculating this, we get: \[ n = \left( \frac{11.86645}{2} \right)^2 = \left( 5.933225 \right)^2 \approx 35.2 \] Since sample size must be a whole number, we round up to get \( n = 36 \). Thus, the required sample size is \( 36 \).