Question

what is the minimum sample size required for estimating μ for N(μ, σ2=96) to within ± 2.5 with confidence levels 95%, 90%, and 80%? a. Find the minimum sample size for the 95% confidence level. Enter an integer below. You must round up. Submit Answer Tries 0/5 b. Find the minimum sample size for the 90% confidence level Submit Answer Tries 0/5 C. Find the minimum sample size for the 80% confidence level

Answer 1

Sample size = Z² $\alpha$ /2 * $\sigma ^{2}$ / E² , Where E is margin of error.

a)

For 95% confidence level,

Sample size = 1.96² * 96 / 2.5²

= 59.01

Sample size = 60 (Rounded up to integer)

b)

For 90% confidence level,

Sample size = 1.6449² * 96 / 2.5²

= 41.56

Sample size = 42 (Rounded up to nearest integer)

c)

For 80% confidence level

Sample size = 1.2816² * 96 / 2.5²

= 25.23

Sample size = 26 (Rounded up to nearest integer)