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Sagot :
Let's go through the detailed, step-by-step solution for the given problem.
1. Calculate the Sample Mean ([tex]\(\bar{x}\)[/tex]):
The sample provided is: 66, 68, 72, 72, 75, 74.
The sample mean ([tex]\(\bar{x}\)[/tex]) is calculated as follows:
[tex]\[ \bar{x} = \frac{\sum_{i=1}^{n} X_i}{n} = \frac{66 + 68 + 72 + 72 + 75 + 74}{6} = 71.16666666666667 \][/tex]
2. Determine the Population Standard Deviation ([tex]\(\sigma\)[/tex]) and Sample Size ([tex]\(n\)[/tex]):
The population standard deviation ([tex]\(\sigma\)[/tex]) is given as 2 inches, and the sample size ([tex]\(n\)[/tex]) is 6.
3. Calculate the Standard Error of the Mean (SE):
The standard error (SE) is calculated using the formula:
[tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{6}} = 0.8164965809277261 \][/tex]
4. Find the Z-score for a 99% Confidence Level:
The Z-score corresponding to a 99% confidence level is approximately 2.5758 (This can be found using Z-tables or statistical software).
5. Calculate the Margin of Error (ME):
The margin of error (ME) is calculated using the formula:
[tex]\[ \text{ME} = Z \times \text{SE} = 2.5758 \times 0.8164965809277261 = 2.103155819401123 \][/tex]
6. Determine the Confidence Interval:
The 99% confidence interval is given by:
[tex]\[ \left(\bar{x} - \text{ME}, \bar{x} + \text{ME}\right) = \left(71.16666666666667 - 2.103155819401123, 71.16666666666667 + 2.103155819401123\right) = \left(69.06351084726555, 73.2698224860678\right) \][/tex]
Summarizing:
- The sample mean ([tex]\(\bar{x}\)[/tex]) is [tex]\(71.16666666666667\)[/tex].
- The margin of error at the 99% confidence level is [tex]\(2.103155819401123\)[/tex].
- The 99% confidence interval for the mean height is [tex]\([69.06351084726555, 73.2698224860678]\)[/tex].
Thus:
Sample Mean ([tex]\(\bar{x}\)[/tex]):
[tex]\[\bar{x} = 71.16666666666667\][/tex]
Margin of error at 99% confidence level:
[tex]\[2.103155819401123\][/tex]
99% confidence interval:
[tex]\[[69.06351084726555, 73.2698224860678]\][/tex]
1. Calculate the Sample Mean ([tex]\(\bar{x}\)[/tex]):
The sample provided is: 66, 68, 72, 72, 75, 74.
The sample mean ([tex]\(\bar{x}\)[/tex]) is calculated as follows:
[tex]\[ \bar{x} = \frac{\sum_{i=1}^{n} X_i}{n} = \frac{66 + 68 + 72 + 72 + 75 + 74}{6} = 71.16666666666667 \][/tex]
2. Determine the Population Standard Deviation ([tex]\(\sigma\)[/tex]) and Sample Size ([tex]\(n\)[/tex]):
The population standard deviation ([tex]\(\sigma\)[/tex]) is given as 2 inches, and the sample size ([tex]\(n\)[/tex]) is 6.
3. Calculate the Standard Error of the Mean (SE):
The standard error (SE) is calculated using the formula:
[tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{6}} = 0.8164965809277261 \][/tex]
4. Find the Z-score for a 99% Confidence Level:
The Z-score corresponding to a 99% confidence level is approximately 2.5758 (This can be found using Z-tables or statistical software).
5. Calculate the Margin of Error (ME):
The margin of error (ME) is calculated using the formula:
[tex]\[ \text{ME} = Z \times \text{SE} = 2.5758 \times 0.8164965809277261 = 2.103155819401123 \][/tex]
6. Determine the Confidence Interval:
The 99% confidence interval is given by:
[tex]\[ \left(\bar{x} - \text{ME}, \bar{x} + \text{ME}\right) = \left(71.16666666666667 - 2.103155819401123, 71.16666666666667 + 2.103155819401123\right) = \left(69.06351084726555, 73.2698224860678\right) \][/tex]
Summarizing:
- The sample mean ([tex]\(\bar{x}\)[/tex]) is [tex]\(71.16666666666667\)[/tex].
- The margin of error at the 99% confidence level is [tex]\(2.103155819401123\)[/tex].
- The 99% confidence interval for the mean height is [tex]\([69.06351084726555, 73.2698224860678]\)[/tex].
Thus:
Sample Mean ([tex]\(\bar{x}\)[/tex]):
[tex]\[\bar{x} = 71.16666666666667\][/tex]
Margin of error at 99% confidence level:
[tex]\[2.103155819401123\][/tex]
99% confidence interval:
[tex]\[[69.06351084726555, 73.2698224860678]\][/tex]
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