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An environmental scientist developed a new analytical method for the determination of cadmium (cd^2+) in mussels. To validate the method, the researcher measured the Cd^2+ concentration in standard reference material (SRM) 2976 that is known to contain 0.82 plusminus 0.16 ppm Cd^2+. Five replicate measurements of the SRM were obtained using the new method, giving values of 0.782, 0.762, 0.825, 0.838, and 0.761 ppm Cd^2+.

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The 95% confidence interval does not include the known concentration of 0.82 ppm Cd^2+. Therefore, at the 95% confidence level, the new method gives a result that differs from the known result of the SRM.

1. Calculate the Mean (
\bar{x}):

The mean (
\bar{x}) is calculated as the sum of all values divided by the number of values.


\bar{x} = (0.782 + 0.762 + 0.825 + 0.838 + 0.761) / 5 = 0.7936

2. Calculate the Standard Deviation (s_x):

The standard deviation (sx) is calculated using the formula:


\( s_x = √([(0.782-0.7936)^2 + (0.762-0.7936)^2 + \ldots + (0.761-0.7936)^2] / 4) \)

s_x ≈ 0.030

3. Calculate the 95% Confidence Interval:

Use the t-distribution table to find the t-value for a 95% confidence level with 4 degrees of freedom (n-1).


\( t_(\alpha/2) = t_(0.025,4) \)

Assuming a two-tailed test, let's say t_{0.025,4} is 2.776 (hypothetical value).

Calculate the margin of error (ME):


\( ME = t_(\alpha/2) * (s_x / √(n)) \)


\( ME \approx 2.776 * (0.030 / √(5)) \)

ME ≈ 0.034

Calculate the confidence interval:

Confidence Interval =
\bar{x} ± ME

Confidence Interval ≈ 0.7936 ± 0.034

Confidence Interval ≈ (0.7596, 0.8276)

Complete question:

An environmental scientist developed a new analytical method for the determination of cadmium (cd^2+) in mussels. To validate the method, the researcher measured the Cd^2+ concentration in standard reference material (SRM) 2976 that is known to contain 0.82 plusminus 0.16 ppm Cd^2+. Five replicate measurements of the SRM were obtained using the new method, giving values of 0.782, 0.762, 0.825, 0.838, and 0.761 ppm Cd^2+. Calculate the mean (Bar x), standard deviation (s_x), and the 95% confidence interval for this data set. At list of t values can be found in the student's t table. X Bar = s_x = 95% confidence interval = x Bar plusminus Does the new method give a result that differs from the known result of the SRM at the 95% confidence level?

User MylesK
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