3 years ago

Can someone help me please

SAT

1 answer:

Scrat [10]3 years ago

4 0

Help with which one? 2 or 3

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2 years ago

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:

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

$\overline{x} + t\frac{s}{\sqrt{n}} = 12.2 + 2.0106\frac{1.6}{\sqrt{49}} = 12.7$

The 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).

More can be learned about the t-distribution at brainly.com/question/16162795

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2 years ago