Question

Using graphite furnace atomic absorption spectrometry, the mean cadmium (cd2 ) content value in a homogenized sample of 40 zebra mussles was found to be 7.50 ppb cd2 with a standard deviation (s) of 0.52 ppb cd2 (assume that s is a good approximation of σ). calculate the 90% and 95% confidence intervals (ci) for: (a) two analyses and (b) five analyses of this homogenized sample.

Answer 1

Part A1:

Given that the mean cadmium (cd2 ) content value in a homogenized sample of 40 zebra mussles was found to be 7.50 ppb cd2 with a standard deviation (s) of 0.52 ppb cd2

The 90% confidence interval for a sample of size n, with sample mean $(\bar{x})$ and a standard deviation of s is given by:

$90\%\ C.I.=\bar{x}\pm1.645\left( \frac{s}{ \sqrt{n} } \right)$

Thus, the 90% confidence intervals (ci) for two analyses is given by:

$7.50\pm1.645\left( \frac{0.52}{ \sqrt{2} } \right)=7.50\pm1.645(0.3677) \\ \\ =7.50\pm0.6049=(6.90, \ 8.1)$


Part A2:

Given that the mean cadmium (cd2 ) content value in a homogenized sample of 40 zebra mussles was found to be 7.50 ppb cd2 with a standard deviation (s) of 0.52 ppb cd2

The 95% confidence interval for a sample of size n, with sample mean $(\bar{x})$ and a standard deviation of s is given by:

$95\%\ C.I.=\bar{x}\pm1.96\left( \frac{s}{ \sqrt{n} } \right)$

Thus, the 95% confidence intervals (ci) for two analyses is given by:

$7.50\pm1.96\left( \frac{0.52}{ \sqrt{2} } \right)=7.50\pm1.96(0.3677) \\ \\ =7.50\pm0.7207=(6.78, \ 8.22)$



Part B1:

Given that the mean cadmium (cd2 ) content value in a homogenized sample of 40 zebra mussles was found to be 7.50 ppb cd2 with a standard deviation (s) of 0.52 ppb cd2

The 90% confidence interval for a sample of size n, with sample mean $(\bar{x})$ and a standard deviation of s is given by:

$90\%\ C.I.=\bar{x}\pm1.645\left( \frac{s}{ \sqrt{n} } \right)$

Thus, the 90% confidence intervals (ci) for five analyses is given by:

$7.50\pm1.645\left( \frac{0.52}{ \sqrt{5} } \right)=7.50\pm1.645(0.2326) \\ \\ =7.50\pm0.3825=(7.12, \ 7.88)$



Part B2:

Given that the mean cadmium (cd2 ) content value in a homogenized sample of 40 zebra mussles was found to be 7.50 ppb cd2 with a standard deviation (s) of 0.52 ppb cd2

The 95% confidence interval for a sample of size n, with sample mean $(\bar{x})$ and a standard deviation of s is given by:

$95\%\ C.I.=\bar{x}\pm1.96\left( \frac{s}{ \sqrt{n} } \right)$

Thus, the 95% confidence intervals (ci) for five analyses is given by:

$7.50\pm1.96\left( \frac{0.52}{ \sqrt{5} } \right)=7.50\pm1.96(0.2326) \\ \\ =7.50\pm0.4559=(7.04, \ 7.96)$