Question
The mean of a sample size n = 35 is 1860. The standard deviation of the sample is 102 and the population is normally distributed. Construct a 99% confidence interval estimate of
the mean of the population.
(1 point)
• 1812 < u < 1908
• 1852 < u< 1868
• 1825 < u< 1895
• 1813 < u < 1907
the mean of the population.
(1 point)
• 1812 < u < 1908
• 1852 < u< 1868
• 1825 < u< 1895
• 1813 < u < 1907
Answers
GPT-4o mini
To construct a 99% confidence interval for the population mean, we can use the formula for a confidence interval when the population standard deviation is unknown, which uses the t-distribution.
The formula is:
\[
\text{Confidence Interval} = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right)
\]
where:
- \(\bar{x}\) is the sample mean (1860),
- \(s\) is the sample standard deviation (102),
- \(n\) is the sample size (35),
- \(t^*\) is the t-value for the desired confidence level and degrees of freedom (df = n - 1).
1. **Determine the t-value for 99% confidence level**:
For a 99% confidence level and \(n - 1 = 34\) degrees of freedom, you can consult a t-table or use statistical software to find this value.
The t-value (for a two-tailed test with \(\alpha = 0.01\)) for 34 degrees of freedom is approximately \(2.728\).
2. **Calculate the standard error (SE)**:
\[
SE = \frac{s}{\sqrt{n}} = \frac{102}{\sqrt{35}} \approx \frac{102}{5.916} \approx 17.24
\]
3. **Calculate the margin of error (ME)**:
\[
ME = t^* \times SE = 2.728 \times 17.24 \approx 47.00
\]
4. **Construct the confidence interval**:
\[
\text{Lower limit} = \bar{x} - ME = 1860 - 47.00 \approx 1813
\]
\[
\text{Upper limit} = \bar{x} + ME = 1860 + 47.00 \approx 1907
\]
Thus, the 99% confidence interval for the population mean is approximately:
\[
1813 < \mu < 1907
\]
Therefore, the correct answer is:
**1813 < u < 1907**
The formula is:
\[
\text{Confidence Interval} = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right)
\]
where:
- \(\bar{x}\) is the sample mean (1860),
- \(s\) is the sample standard deviation (102),
- \(n\) is the sample size (35),
- \(t^*\) is the t-value for the desired confidence level and degrees of freedom (df = n - 1).
1. **Determine the t-value for 99% confidence level**:
For a 99% confidence level and \(n - 1 = 34\) degrees of freedom, you can consult a t-table or use statistical software to find this value.
The t-value (for a two-tailed test with \(\alpha = 0.01\)) for 34 degrees of freedom is approximately \(2.728\).
2. **Calculate the standard error (SE)**:
\[
SE = \frac{s}{\sqrt{n}} = \frac{102}{\sqrt{35}} \approx \frac{102}{5.916} \approx 17.24
\]
3. **Calculate the margin of error (ME)**:
\[
ME = t^* \times SE = 2.728 \times 17.24 \approx 47.00
\]
4. **Construct the confidence interval**:
\[
\text{Lower limit} = \bar{x} - ME = 1860 - 47.00 \approx 1813
\]
\[
\text{Upper limit} = \bar{x} + ME = 1860 + 47.00 \approx 1907
\]
Thus, the 99% confidence interval for the population mean is approximately:
\[
1813 < \mu < 1907
\]
Therefore, the correct answer is:
**1813 < u < 1907**