Answer:
B. (2.869, 3.411)
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Step-by-step explanation:
<em>The question is incomplete:</em>
<em>The number of hours per day that Americans spend on social networking is approximately normally distributed. A random sample of 20 Americans who use social networks had M = 3.14 hours and s = 0.58 hours. Find a 95% confidence interval for the actual mean number of hours that Americans spend on social networking.
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<em>
</em>
<em>A. (2.860, 3.420)
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<em>
B. (2.869, 3.411)
</em>
<em>C. (2.886, 3.394)
</em>
<em>D. Not appropriate</em>
We have to calculate a 95% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=3.14.
The sample size is N=20.
When σ is not known, s divided by the square root of N is used as an estimate of σM:

The degrees of freedom for this sample size are:

The t-value for a 95% confidence interval and 19 degrees of freedom is t=2.093.
The margin of error (MOE) can be calculated as:

Then, the lower and upper bounds of the confidence interval are:

The 95% confidence interval for the mean is (2.869, 3.411).