Alzheimer's patients is tested to assess the amount of time in stage IV sleep

Full Answer Section

• Confidence Level (α) = 1 - 0.95 = 0.05 (5%) for a 95% confidence interval
• Degrees of freedom (df) = n - 1 = 61 - 1 = 60
• Sample mean (x̄) = 48 minutes
• Sample standard deviation (s) = 14 minutes

b) Critical t-value:

We need the two-tailed critical t-value for a confidence level of 95% and 60 degrees of freedom. You can find this value using a t-distribution table or calculator. It's approximately ± 1.992.

c) Confidence Interval Calculation:

Confidence Interval = x̄ ± (critical t-value) * (standard error)

Standard Error (SE) = s / √n = 14 minutes / √61 ≈ 2.23 minutes

Lower Bound: 48 minutes - (1.992) * (2.23 minutes) ≈ 43.3 minutes Upper Bound: 48 minutes + (1.992) * (2.23 minutes) ≈ 52.7 minutes

Interpretation:

We can be 95% confident that the true average stage IV sleep duration for the population of Alzheimer's patients falls between 43.3 and 52.7 minutes per night.

Note: This confidence interval applies to the population mean, not a specific individual. We cannot determine the exact stage IV sleep duration for any single patient with this information.

2. World Events Knowledge of Students

We can compute a 99% confidence interval for the average world events knowledge score of the university's students. Here's the breakdown:

a) Formula and Values:

Similar to the first case, we'll use a one-sample t-distribution for the confidence interval.

• Confidence Level (α) = 1 - 0.99 = 0.01 (1%) for a 99% confidence interval
• Degrees of freedom (df) = n - 1 = 30 - 1 = 29
• Sample mean (x̄) = 58 points
• Standard Error (SE) = 3.2 points (provided)

b) Critical t-value:

Find the two-tailed critical t-value for a confidence level of 99% and 29 degrees of freedom. This value is approximately ± 2.756.

c) Confidence Interval Calculation:

Confidence Interval = x̄ ± (critical t-value) * (standard error)

Lower Bound: 58 points - (2.756) * (3.2 points) ≈ 49.6 points Upper Bound: 58 points + (2.756) * (3.2 points) ≈ 66.4 points

Interpretation:

We can be 99% confident that the true average world events knowledge score for all students at the university falls between 49.6 and 66.4 points.

Comparison with National Sample:

The national reported mean score is 65 points. The confidence interval for the university students (49.6 - 66.4) includes the national average. However, the lower bound is slightly lower than the national mean, suggesting the university students might score slightly lower on average compared to the national sample.

Note: This analysis is based on a sample and doesn't confirm a definitive difference. A larger sample size or a hypothesis test would provide stronger evidence.

Sample Answer

Confidence Intervals for Sleep and World Events Knowledge

1. Stage IV Sleep in Alzheimer's Patients

We can compute a 95% confidence interval for the average stage IV sleep duration in this scenario. Here's how:

a) Formula and Values:

We'll use the one-sample t-distribution for the confidence interval as the population standard deviation (σ) is unknown and the sample size (n) is less than 30.