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?
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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