In a study of academic procrastination, the authors of a paper reported that for a sample of 451 undergraduate students at a midsize public university preparing for a final exam in an introductory psychology course, the mean time spent studying for the exam was 7.74 hours and the standard deviation of study times was 3.10 hours. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of students taking introductory psychology at this university. Required:
Construct a 95% confidence interval to estimate μ, the mean time spent studying for the final exam for students taking introductory psychology at this university.

The 95% confidence interval to estimate μ is between 7.45 hours and 8.03 hours.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

That is z with a pvalue of , so Z = 1.96.

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 7.74 - 0.29 = 7.45 hours

The upper end of the interval is the sample mean added to M. So it is 7.74 + 0.29 = 8.03 hours

The 95% confidence interval to estimate μ is between 7.45 hours and 8.03 hours.

30 people weigh 160 pounds or more

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.