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)**

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