Use the t-distribution to find a confidence interval for a mean mu given the relevant sample results. Give the best point estimate for mu, 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 mu using the sample results x-bar equals 76.4, s = 8.6, and n = 42. Point estimate = ?
Margin of error = ?

Point estimate = 76.4

Margin of Error = 2.680

Step-by-step explanation:

Given that distribution is approximately normal;

The point estimate = sample mean, xbar = 76.4

The margin of error = Zcritical * s/√n

Tcritical at 95%, df = 42 - 1 = 41

Tcritical(0.05, 41) = 2.0195

Margin of Error = 2.0195 * (8.6/√42)

Margin of Error = 2.0195 * 1.327

Margin of Error = 2.67989

Margin of Error = 2.680

i think is g(1)=-1.

hope that i can

answer: the answer is simply nothing.

step-by-step explanation: because there is no ending to this question

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