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In a study of academic procrastination, the authors of the paper "Correlates and Consequences of Behavioral Procrastination" (Procrastination, Current Issues and New Directions [2000]) reported that for a sample of 411 undergraduate students at a mid-size public university, the mean time spent studying for the final exam in an introductory psychology course was 7.74 hours and the standard deviation of study times was 3.40 hours. Assume that this sample is representative of students taking introductory psychology at this university. Construct a \(95 \%\) confidence interval estimate of \(\mu,\) the mean time spent studying for the introductory psychology final exam. (Hint: See Example 12.7)

Short Answer

Expert verified
The 95% confidence interval estimate for the mean study time is approximately from 7.41 to 8.07 hours.

Step by step solution

01

Identify the relevant variables

From the problem, we can identify these key statistics: the sample size \(n = 411\), sample mean \(x̄ = 7.74\) hours, and the sample standard deviation \(s = 3.4\) hours. The confidence level is \(95%\).
02

Find the Z value

For a \(95\%\) confidence level, the associated Z-value (which you can find in a standard normal distribution table) is approximately \(1.96\).
03

Plug in the values into the confidence interval formula

We substitute the values into the formula \(x̄ ± Z * (σ/√n)\). Therefore, the confidence interval becomes: \(7.74 ± 1.96 * (3.4/√411)\).
04

Calculate the confidence interval

The calculation yields a confidence interval of \(7.74 ± 0.33\) hours. So the \(95%\) confidence interval estimate for the mean study time is approximately from \(7.41\) to \(8.07\) hours.

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Most popular questions from this chapter

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