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 95% confidence interval for μ using the sample results x̅=10.4, s=5.3, and n=30.

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 95% confidence interval is?

Answer #1

Point estimate for = =
**10.4**

margin of error = t * S / sqrt(n)

= 2.045 * 5.3 / sqrt(30)

= **1.98**

The 95% confidence interval for is

- E < < + E

10.4 - 1.98 < < 10.4 + 1.98

8.42 < < 12.38

95% CI is **( 8.42 , 12.38 )**

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