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Answer :
To construct a 99.8% confidence interval for the mean number of hours it takes for a student to meet the course objectives, we'll follow these steps:
1. Identify the Given Information:
- Sample size ([tex]\(n\)[/tex]) = 43
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 80.8 hours
- Sample standard deviation ([tex]\(s\)[/tex]) = 51.2 hours
- Confidence level = 99.8%
2. Determine the Critical Value:
- We use a normal distribution approximation because the sample size is large enough (n = 43).
- For a 99.8% confidence level, the critical value ([tex]\(z\)[/tex]) corresponds to the area in the tails of the standard normal distribution. The remaining area under the curve (for the tails) is 0.002 (since 1 - 0.998 = 0.002). Thus, we divide this by 2 to find the area in one tail, which is 0.001.
3. Calculate the Margin of Error:
[tex]\[
\text{Margin of Error} = z \times \left(\frac{s}{\sqrt{n}}\right)
\][/tex]
- The critical [tex]\(z\)[/tex]-score for the 99.8% confidence interval is approximately determined using statistical tables or software.
- Compute the margin of error using the formula above.
4. Construct the Confidence Interval:
[tex]\[
\text{Lower bound} = \bar{x} - \text{Margin of Error}
\][/tex]
[tex]\[
\text{Upper bound} = \bar{x} + \text{Margin of Error}
\][/tex]
5. Final Result:
- After performing the calculations, the 99.8% confidence interval for the mean number of hours is from 56.7 hours to 104.9 hours.
Thus, the 99.8% confidence interval for the mean number of hours it takes for a student to meet course objectives is [tex]\(56.7 < \mu < 104.9\)[/tex].
1. Identify the Given Information:
- Sample size ([tex]\(n\)[/tex]) = 43
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 80.8 hours
- Sample standard deviation ([tex]\(s\)[/tex]) = 51.2 hours
- Confidence level = 99.8%
2. Determine the Critical Value:
- We use a normal distribution approximation because the sample size is large enough (n = 43).
- For a 99.8% confidence level, the critical value ([tex]\(z\)[/tex]) corresponds to the area in the tails of the standard normal distribution. The remaining area under the curve (for the tails) is 0.002 (since 1 - 0.998 = 0.002). Thus, we divide this by 2 to find the area in one tail, which is 0.001.
3. Calculate the Margin of Error:
[tex]\[
\text{Margin of Error} = z \times \left(\frac{s}{\sqrt{n}}\right)
\][/tex]
- The critical [tex]\(z\)[/tex]-score for the 99.8% confidence interval is approximately determined using statistical tables or software.
- Compute the margin of error using the formula above.
4. Construct the Confidence Interval:
[tex]\[
\text{Lower bound} = \bar{x} - \text{Margin of Error}
\][/tex]
[tex]\[
\text{Upper bound} = \bar{x} + \text{Margin of Error}
\][/tex]
5. Final Result:
- After performing the calculations, the 99.8% confidence interval for the mean number of hours is from 56.7 hours to 104.9 hours.
Thus, the 99.8% confidence interval for the mean number of hours it takes for a student to meet course objectives is [tex]\(56.7 < \mu < 104.9\)[/tex].
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