Question

Jackson sampled 101 students and calculated an average of 6.5
hours of sleep each night with a standard deviation of 2.14. Using
a 95% confidence level, he also found that t* = 1.984.
confidence interval = X±t* s/√n
A 95% confidence interval calculates that the average number
of hours of sleep for working college students is between
hours. Answer choices are rounded to the
hundredths place.

Answer 1

Using the t-distribution, the 95% confidence interval for the average number of hours of sleep for working college students is between 6.03 and 6.97 hours.

What is a t-distribution confidence interval?

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.

For this problem, the parameters are given as follows:

$\overline{x} = 6.5, s = 2.4, t = 1.984, n = 101$

Hence the bounds of the interval are:

• $\overline{x} - t\frac{s}{\sqrt{n}} = 6.5 - 1.984\frac{2.4}{\sqrt{101}} = 6.03$

• $\overline{x} + t\frac{s}{\sqrt{n}} = 6.5 + 1.984\frac{2.4}{\sqrt{101}} = 6.97$

The 95% confidence interval for the average number of hours of sleep for working college students is between 6.03 and 6.97 hours.

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

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