Question

part: 1 / 4
part 2 of 4
(b) find the 95% confidence interval of the mean of these pounds. round intermediate answers to at least three decimal places. round your final answers to one decimal place.
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Explanation:

To find the 95% confidence interval for the population mean \(\mu\), we typically use the formula for a confidence interval when the population standard deviation \(\sigma\) is unknown (which is common in practice):

$$ \bar{x} - t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} < \mu < \bar{x} + t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} $$

where:

• \(\bar{x}\) is the sample mean,
• \(s\) is the sample standard deviation,
• \(n\) is the sample size,
• \(t_{\alpha/2, n-1}\) is the critical value from the t-distribution with \(n-1\) degrees of freedom and \(\alpha/2\) significance level (for a 95% confidence interval, \(\alpha = 0.05\), so \(\alpha/2 = 0.025\)).

However, since the problem does not provide the sample data (the values of the pounds), we cannot compute the sample mean \(\bar{x}\), sample standard deviation \(s\), or the sample size \(n\).

To proceed, we need the following information:

1. The sample data (or the sample mean \(\bar{x}\), sample standard deviation \(s\), and sample size \(n\)).

If you provide the sample data (e.g., a list of the pounds), we can calculate the sample mean, sample standard deviation, and then find the 95% confidence interval.

For example, suppose we have a sample of weights: \(x_1, x_2, \dots, x_n\). We would:

1. Calculate the sample mean \(\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i\).
2. Calculate the sample standard deviation \(s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}\).
3. Determine the degrees of freedom \(df = n - 1\).
4. Find the critical value \(t_{\alpha/2, df}\) (for \(\alpha = 0.05\), this is the t-value that leaves 2.5% in the upper tail).
5. Compute the margin of error \(E = t_{\alpha/2, df} \cdot \frac{s}{\sqrt{n}}\).
6. The confidence interval is \(\bar{x} - E < \mu < \bar{x} + E\).

Please provide the sample data (the values of the pounds) so that we can complete the calculation.

Answer:

To find the 95% confidence interval for the population mean \(\mu\), we typically use the formula for a confidence interval when the population standard deviation \(\sigma\) is unknown (which is common in practice):

$$ \bar{x} - t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} < \mu < \bar{x} + t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} $$

where:

• \(\bar{x}\) is the sample mean,
• \(s\) is the sample standard deviation,
• \(n\) is the sample size,
• \(t_{\alpha/2, n-1}\) is the critical value from the t-distribution with \(n-1\) degrees of freedom and \(\alpha/2\) significance level (for a 95% confidence interval, \(\alpha = 0.05\), so \(\alpha/2 = 0.025\)).

However, since the problem does not provide the sample data (the values of the pounds), we cannot compute the sample mean \(\bar{x}\), sample standard deviation \(s\), or the sample size \(n\).

To proceed, we need the following information:

1. The sample data (or the sample mean \(\bar{x}\), sample standard deviation \(s\), and sample size \(n\)).

If you provide the sample data (e.g., a list of the pounds), we can calculate the sample mean, sample standard deviation, and then find the 95% confidence interval.

For example, suppose we have a sample of weights: \(x_1, x_2, \dots, x_n\). We would:

1. Calculate the sample mean \(\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i\).
2. Calculate the sample standard deviation \(s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}\).
3. Determine the degrees of freedom \(df = n - 1\).
4. Find the critical value \(t_{\alpha/2, df}\) (for \(\alpha = 0.05\), this is the t-value that leaves 2.5% in the upper tail).
5. Compute the margin of error \(E = t_{\alpha/2, df} \cdot \frac{s}{\sqrt{n}}\).
6. The confidence interval is \(\bar{x} - E < \mu < \bar{x} + E\).

Please provide the sample data (the values of the pounds) so that we can complete the calculation.