To calculate a 95% confidence interval for the mean mass of the Chow dog food containers, we can follow these steps:
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Calculate the sample mean (\( \bar{x} \)): \[ \bar{x} = \frac{\sum x_i}{n} = \frac{9.5 + 10.4 + 9 + 11 + 10.5 + 9.6}{6} \] \[ \bar{x} = \frac{60.5}{6} \approx 10.0833 \]
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Calculate the sample standard deviation (\( s \)): The formula for the sample standard deviation is: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \] First, calculate each deviation from the mean, square it, and sum it: \[ (9.5 - 10.0833)^2 \approx 0.3422 \] \[ (10.4 - 10.0833)^2 \approx 0.1014 \] \[ (9 - 10.0833)^2 \approx 1.1669 \] \[ (11 - 10.0833)^2 \approx 0.8361 \] \[ (10.5 - 10.0833)^2 \approx 0.1736 \] \[ (9.6 - 10.0833)^2 \approx 0.2322 \]
Now sum these squared deviations: \[ 0.3422 + 0.1014 + 1.1669 + 0.8361 + 0.1736 + 0.2322 \approx 2.8524 \]
Now, divide by \( n - 1 = 6 - 1 = 5 \): \[ s^2 = \frac{2.8524}{5} \approx 0.5705 \] Therefore, \[ s = \sqrt{0.5705} \approx 0.7567 \]
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Determine the critical value (\( t \)) for 95% confidence: For a sample size of 6, the degrees of freedom \( df = n - 1 = 5 \). Using a t-table or calculator, the critical value \( t \) for a 95% confidence level with 5 degrees of freedom is approximately \( t \approx 2.571 \) (the exact value can vary slightly depending on the source).
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Calculate the standard error (\( SE \)): \[ SE = \frac{s}{\sqrt{n}} = \frac{0.7567}{\sqrt{6}} \approx 0.3082 \]
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Calculate the margin of error (\( ME \)): \[ ME = t \times SE \approx 2.571 \times 0.3082 \approx 0.7938 \]
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Calculate the confidence interval: \[ \text{Lower bound} = \bar{x} - ME \approx 10.0833 - 0.7938 \approx 9.2895 \] \[ \text{Upper bound} = \bar{x} + ME \approx 10.0833 + 0.7938 \approx 10.8771 \]
Thus, the 95% confidence interval for the mean mass of all such dog food containers is approximately: \[ (9.29, 10.88) \]