Question

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

Answer 1

Given,

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

Where,

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

=(131.24, 157.16)

90% Confidence interval for population mean µ = (131.24, 157.16)