Use the t-distribution to find a confidence interval
for a mean μ given the relevant sample results. Give the best point
estimate for μ, the margin of error, and the confidence interval.
Assume the results come from a random sample from a population that
is approximately normally distributed.
A 90% confidence interval for μ using the sample results x¯=144.2, s=55.7, and n=50
Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places.
point estimate =
margin of error =
The 90% confidence interval is
Sample Mean: x̅ = 144.2
Sample Standard Deviation: s = 55.7
Sample Size= 50
We want to find 90% Confidence interval for population mean ‘µ’
Sample is come from the normal distribution.
1) Point estimate for µ: Since sample is from normal distribution with sample size n =50 (>30).
By using Central Limit Theorem, The Sample mean is the point estimate for the population
Point estimate for µ = Sample Mean x̅= 144.2
2) Margin of error
Margin of error = Z * s/√n
Z: The chosen Z-value for 90% confidence of 0.10 level of significance = 1.645
s: The standard deviation = 55.7
n: The number of observations = 50
Margin of error = 1.645 * 55.7/√50
Margin of error =12.96
90% Confidence interval for population mean µ = X̅ ± Z * s/√n
= 144.2 ± 12.96
90% Confidence interval for population mean µ = (131.24, 157.16)
Get Answers For Free
Most questions answered within 1 hours.