Question - Calculating 95% Confidence Interval for Mean Magnesium Concentration
Solution:
To calculate a 95% confidence interval for the mean magnesium concentration in this sample, we need to use the following formula for a confidence interval when the population standard deviation is unknown:\[ \bar{x} \pm t_{\frac{\alpha}{2}, n-1} \left( \frac{s}{\sqrt{n}} \right) \]where:- $$ \bar{x} $$ is the sample mean,- $$ t_{\frac{\alpha}{2}, n-1} $$ is the t-score from the t-distribution for a 95% confidence level with $$ n-1 $$ degrees of freedom (where $$ n $$ is the sample size),- $$ s $$ is the sample standard deviation, and- $$ n $$ is the sample size.First, we need to calculate the sample mean ($$ \bar{x} $$) and the sample standard deviation (s). The data we have is:\[ 175, 177, 175, 180, 138, 138 \]Now let's calculate the mean and the standard deviation.Mean ($$ \bar{x} $$):\[ \bar{x} = \frac{175 + 177 + 175 + 180 + 138 + 138}{6} = \frac{983}{6} \approx 163.83 \]To calculate the standard deviation, we need to follow these steps:1. Find the squared difference from the mean for each observation.2. Sum these squared differences.3. Divide by the number of observations minus 1 to get the variance.4. Take the square root of the variance to get the standard deviation.Let's do these calculations.1. Calculate the squared differences from the mean:\[ (175 - 163.83)^2 = 123.69 \]\[ (177 - 163.83)^2 = 173.69 \]\[ (175 - 163.83)^2 = 123.69 \]\[ (180 - 163.83)^2 = 262.89 \]\[ (138 - 163.83)^2 = 664.89 \]\[ (138 - 163.83)^2 = 664.89 \]2. Sum of squared differences:\[ 123.69 + 173.69 + 123.69 + 262.89 + 664.89 + 664.89 = 2014.74 \]3. Variance (s^2):\[ s^2 = \frac{2014.74}{6 - 1} = \frac{2014.74}{5} = 402.948 \]4. Standard deviation (s):\[ s = \sqrt{402.948} \approx 20.07 \]Next, we need to find the t-score for a 95% confidence level with 5 degrees of freedom (since $$ n = 6 $$). This value can be found in a t-distribution table or calculated using statistical software. Assuming we are using a t-distribution table or a calculator:\[ t_{\frac{\alpha}{2}, 5} \text{ at } \alpha = 0.05 \text{ is approximately } 2.571 \]Finally, calculate the margin of error (ME):\[ ME = t_{\frac{\alpha}{2}, 5} \left( \frac{s}{\sqrt{n}} \right) = 2.571 \left( \frac{20.07}{\sqrt{6}} \right) \approx 2.571 \left( 8.187 \right) \approx 21.04 \]Now we can calculate the 95% confidence interval:\[ \bar{x} \pm ME = 163.83 \pm 21.04 \]Lower limit:\[ 163.83 - 21.04 = 142.79 \]Upper limit:\[ 163.83 + 21.04 = 184.87 \]Therefore, the 95% confidence interval for the mean magnesium concentration is approximately (142.79, 184.87) parts per million.
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