Suppose that the population of the number of hours slept by all college students the night before finals is approximately normally distributed. A report claimed that the mean of this population is 8.29 hours. As a student wellness advocate, you want to test this claim, so you select a random sample of 17 college students and record the number of hours each slept the night before finals.

Question

Follow the steps below to construct a 90% confidence interval for the population mean of all the numbers of hours slept by college students the night before finals. Then state whether the confidence interval you construct contradicts the report's claim. (If necessary, consult a list of formulas.)
Number of students= 17
sample mean= 6.875
sample standard deviation= 2.012
what is the:
sample size?
point estimate?
sample standard deviation?
critical value?

Answer 1

To construct a 90% confidence interval for the population mean of all the numbers of hours slept by college students the night before finals, let's go through the required steps.

Given Data:

• Sample size (n) = 17
• Sample mean (\(\bar{x}\)) = 6.875 hours
• Sample standard deviation (s) = 2.012 hours

1. Sample Size:

The sample size (n) is 17.

2. Point Estimate:

The point estimate for the population mean is the sample mean, which is 6.875 hours.

3. Sample Standard Deviation:

The sample standard deviation is given as 2.012 hours.

4. Critical Value:

To find the critical value for a 90% confidence interval, we can use the t-distribution because the sample size is small (n < 30) and we are estimating the population mean.

• The degrees of freedom (df) for this case: \[ df = n - 1 = 17 - 1 = 16 \]

• For a 90% confidence interval, the significance level (\(\alpha\)) is 0.10, and the area in each tail is \( \alpha/2 = 0.05 \).

Using a t-table or calculator to find the critical t-value for 16 degrees of freedom at a significance level of 0.05 (for one tail):

• The critical t-value (\(t^*\)) for a 90% confidence interval and 16 degrees of freedom is approximately 1.745.

Summary of Values:

• Sample Size (n): 17
• Point Estimate: 6.875 hours
• Sample Standard Deviation (s): 2.012 hours
• Critical Value (t)*: 1.745

5. Constructing the Confidence Interval:

To construct the confidence interval, we use the formula:

\[ CI = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) \]

Calculating the margin of error:

1. Calculate \( \frac{s}{\sqrt{n}} \): \[ \frac{s}{\sqrt{n}} = \frac{2.012}{\sqrt{17}} \approx \frac{2.012}{4.123} \approx 0.488 \]

2. Calculate the margin of error (ME): \[ ME = t^* \left( \frac{s}{\sqrt{n}} \right) = 1.745 \times 0.488 \approx 0.852 \]

3. Now calculate the confidence interval: \[ CI = \bar{x} \pm ME = 6.875 \pm 0.852 \]

   • Lower Bound: \( 6.875 - 0.852 \approx 6.023 \)
   • Upper Bound: \( 6.875 + 0.852 \approx 7.727 \)

Confidence Interval:

The 90% confidence interval for the population mean hours slept is approximately: \[ (6.023 , \text{hours}, 7.727 , \text{hours}) \]

Conclusion:

The reported mean of 8.29 hours is outside of this confidence interval. Therefore, the confidence interval does contradict the report's claim that the mean is 8.29 hours.