To calculate a 90% confidence interval for the mean mass of all such dog food, we can follow these steps:
Step 1: Calculate the sample mean (\(\bar{x}\)) and sample standard deviation (s)
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Sample Mean (\(\bar{x}\)): \[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \] where \(n\) is the number of samples (6 in this case).
\[ \bar{x} = \frac{9.8 + 10.4 + 9.8 + 10 + 10.2 + 9.6}{6} = \frac{60.8}{6} \approx 10.13 \]
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Sample Standard Deviation (s): \[ s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n - 1}} \]
First, we need to calculate each \((x_i - \bar{x})^2\):
- \((9.8 - 10.13)^2 \approx 0.1089\)
- \((10.4 - 10.13)^2 \approx 0.0729\)
- \((9.8 - 10.13)^2 \approx 0.1089\)
- \((10 - 10.13)^2 \approx 0.0169\)
- \((10.2 - 10.13)^2 \approx 0.0049\)
- \((9.6 - 10.13)^2 \approx 0.2809\)
Now sum these squared deviations: \[ \sum (x_i - \bar{x})^2 \approx 0.1089 + 0.0729 + 0.1089 + 0.0169 + 0.0049 + 0.2809 \approx 0.5924 \]
Now calculate the sample standard deviation: \[ s = \sqrt{\frac{0.5924}{6 - 1}} = \sqrt{\frac{0.5924}{5}} \approx \sqrt{0.11848} \approx 0.344 \]
Step 2: Determine the critical value
For a 90% confidence interval and \(n-1 = 5\) degrees of freedom, we look up the t-distribution:
- The critical t-value (\(t_{0.05, 5}\)) for a 90% confidence interval with 5 degrees of freedom is approximately \(t_{0.05, 5} = 2.015\).
Step 3: Calculate the margin of error (E)
The margin of error is calculated as: \[ E = t \cdot \left(\frac{s}{\sqrt{n}}\right) \] where \(s\) is the sample standard deviation, and \(n\) is the sample size.
\[ E = 2.015 \cdot \left(\frac{0.344}{\sqrt{6}}\right) \approx 2.015 \cdot \left(\frac{0.344}{2.449}\right) \approx 2.015 \cdot 0.140 = 0.2821 \]
Step 4: Construct the confidence interval
The confidence interval is given by: \[ \bar{x} \pm E \] \[ CI = [10.13 - 0.2821, 10.13 + 0.2821] \] \[ CI \approx [9.8479, 10.4121] \]
Conclusion
The 90% confidence interval for the mean mass of all such dog food is approximately:
\[ (9.85, 10.41) \] (rounded to two decimal places).