Question

A researcher is interested in how lack of sleep impacts a participant’s focus. The researcher had a sample of 25 people stay awake all night and then measured their level of focus the next morning. For the sample, the researcher calculated the mean level of focus to be 30 (on a 100 point scale with higher numbers meaning greater focus) and the standard deviation to be 5.

A. Find a 95% confidence interval for the true mean level of focus for people who stay up all night. Make sure final answer is written using confidence interval notation.

B. Interpret the margin of error in context (i.e., what is the margin of error? What does it tell us?)

Answer 1

(A)

n = 25

= 30

s = 5

SE = s/

= 5/ = 1

= 0.05

ndf = 25 - 1 = 24

From Table, critical values of t = 2.0639

Confidence interval:

30 (2.0639 X1)

= ( 27.9361, 32.0639)

So, Answer is:

27.9361 < < 32.0639

(b)

Margin of Error = 2.0639

Interpretation:

The Margin of Error = MOE = 2.0639 with 95% level of confidence indicates that if the survey is conducted 100 times, the data will be within ( 27.9361, 32.0639) in 95 out of the 100 surveys.