A food snack manu. samples 15 bags of pretzels off the assembly line and weighed their contents. If the sample mean is 10.0 and the sample standard deviation is .15, find the 95% confidence interval estimate for the true mean.
95% = mean ± 1.96 SEm
SEm = SD/√n
To find the 95% confidence interval estimate for the true mean, we can use the formula:
CI = x̄ ± Z * (σ / sqrt(n))
Where:
- CI is the confidence interval
- x̄ is the sample mean
- Z is the Z-score for the desired confidence level (95% confidence level corresponds to a Z-score of approximately 1.96)
- σ is the population standard deviation (unknown, so we'll use the sample standard deviation)
- n is the sample size
Given:
x̄ = 10.0 (sample mean)
σ (sample standard deviation) = 0.15
n (sample size) = 15
Now let's calculate the confidence interval:
CI = 10.0 ± 1.96 * (0.15 / sqrt(15))
To calculate the square root of 15, we need to approximate it as a decimal:
sqrt(15) ≈ 3.87
CI = 10.0 ± 1.96 * (0.15 / 3.87)
Now we can calculate the values:
CI = 10.0 ± 1.96 * 0.0387
CI ≈ 10.0 ± 0.0758
The 95% confidence interval estimate for the true mean is approximately 9.9242 to 10.0758.