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2 answers:

asambeis [7]3 years ago

8 0

Answer:

tyy

Explanation:

rewona [7]3 years ago

7 0

Answer:

thanks for free points

Explanation:

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Students in a statistics class are conducting a survey to estimate the mean number of units students at their college are enroll

Andrew [12]

Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).

<h3>What is a t-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

t is the critical value.

n is the sample size.

s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed <em>95% confidence interval</em>, with 49 - 1 = <em>48 df</em>, is t = 2.0106.

Hence:

$\overline{x} - t\frac{s}{\sqrt{n}} = 12.2 - 2.0106\frac{1.6}{\sqrt{49}} = 11.7$