Answer :
To construct a 99% confidence interval for the mean body temperature of all healthy humans, we need to follow these steps:
1. Identify the given data:
- Sample size (\(n\)) = 110
- Sample mean (\(\bar{x}\)) = 98.0°F
- Sample standard deviation (\(s\)) = 0.74°F
- Confidence level = 99%
2. Determine the degrees of freedom (df):
- Degrees of freedom (\(df\)) = \(n - 1\) = 110 - 1 = 109
3. Find the t-critical value:
- For a 99% confidence level and 109 degrees of freedom. Based on tabulated values or computational tools, we find the t-critical value (\(t_{\alpha/2}\)) to be approximately 2.622.
4. Calculate the margin of error (ME):
- The formula for the margin of error is:
[tex]\[ \text{ME} = t_{\alpha/2} \times \left(\frac{s}{\sqrt{n}}\right) \][/tex]
- Plug in the values:
[tex]\[ \text{ME} = 2.622 \times \left(\frac{0.74}{\sqrt{110}}\right) \][/tex]
- Calculate the standard error (SE):
[tex]\[ \text{SE} = \frac{0.74}{\sqrt{110}} \approx 0.0704 \][/tex]
- Now, calculate the margin of error:
[tex]\[ \text{ME} \approx 2.622 \times 0.0704 \approx 0.185 \][/tex]
5. Construct the confidence interval:
- The lower limit of the confidence interval is:
[tex]\[ \text{Lower limit} = \bar{x} - \text{ME} = 98.0 - 0.185 \approx 97.815 \][/tex]
- The upper limit of the confidence interval is:
[tex]\[ \text{Upper limit} = \bar{x} + \text{ME} = 98.0 + 0.185 \approx 98.185 \][/tex]
6. State the confidence interval:
- The 99% confidence interval for the mean body temperature of all healthy humans is approximately:
[tex]\[ 97.815^\circ F < \mu < 98.185^\circ F \][/tex]
Interpretation about the use of \(98.6^\circ F\):
Considering the constructed confidence interval of [tex]\(97.815^\circ F < \mu < 98.185^\circ F\)[/tex], we observe that 98.6°F is not within this interval. This suggests that the traditional average body temperature of 98.6°F might not be an accurate estimate of the mean body temperature for healthy humans based on the provided sample.
1. Identify the given data:
- Sample size (\(n\)) = 110
- Sample mean (\(\bar{x}\)) = 98.0°F
- Sample standard deviation (\(s\)) = 0.74°F
- Confidence level = 99%
2. Determine the degrees of freedom (df):
- Degrees of freedom (\(df\)) = \(n - 1\) = 110 - 1 = 109
3. Find the t-critical value:
- For a 99% confidence level and 109 degrees of freedom. Based on tabulated values or computational tools, we find the t-critical value (\(t_{\alpha/2}\)) to be approximately 2.622.
4. Calculate the margin of error (ME):
- The formula for the margin of error is:
[tex]\[ \text{ME} = t_{\alpha/2} \times \left(\frac{s}{\sqrt{n}}\right) \][/tex]
- Plug in the values:
[tex]\[ \text{ME} = 2.622 \times \left(\frac{0.74}{\sqrt{110}}\right) \][/tex]
- Calculate the standard error (SE):
[tex]\[ \text{SE} = \frac{0.74}{\sqrt{110}} \approx 0.0704 \][/tex]
- Now, calculate the margin of error:
[tex]\[ \text{ME} \approx 2.622 \times 0.0704 \approx 0.185 \][/tex]
5. Construct the confidence interval:
- The lower limit of the confidence interval is:
[tex]\[ \text{Lower limit} = \bar{x} - \text{ME} = 98.0 - 0.185 \approx 97.815 \][/tex]
- The upper limit of the confidence interval is:
[tex]\[ \text{Upper limit} = \bar{x} + \text{ME} = 98.0 + 0.185 \approx 98.185 \][/tex]
6. State the confidence interval:
- The 99% confidence interval for the mean body temperature of all healthy humans is approximately:
[tex]\[ 97.815^\circ F < \mu < 98.185^\circ F \][/tex]
Interpretation about the use of \(98.6^\circ F\):
Considering the constructed confidence interval of [tex]\(97.815^\circ F < \mu < 98.185^\circ F\)[/tex], we observe that 98.6°F is not within this interval. This suggests that the traditional average body temperature of 98.6°F might not be an accurate estimate of the mean body temperature for healthy humans based on the provided sample.
