For $$ C=99 \% $$ If $$ \sigma=6.69 $$ and $$ \mathrm{n}=40 $$ then $$ E=2.22 $$

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To solve the problem, we need to verify the conditions given and calculate the margin of error $$ E $$ using the provided values.

### Known Conditions:
- Confidence Level $$ C = 99\% $$
- Standard Deviation $$ \sigma = 6.69 $$
- Sample Size $$ n = 40 $$
- Margin of Error $$ E = 2.22 $$

### Step 1: Find the Z-score for the Confidence Level
For a confidence level of $$ 99\% $$, we need to find the Z-score that corresponds to the tails of the normal distribution. The Z-score for $$ 99\% $$ confidence level is approximately $$ 2.576 $$.

### Step 2: Calculate the Standard Error (SE)
The standard error (SE) is calculated using the formula:
$$
SE = \frac{\sigma}{\sqrt{n}}
$$
Substituting the known values:
$$
SE = \frac{6.69}{\sqrt{40}}
$$

### Step 3: Calculate the Margin of Error (E)
The margin of error $$ E $$ can be calculated using the formula:
$$
E = Z \times SE
$$
Substituting the Z-score and the calculated SE:
$$
E = 2.576 \times SE
$$

Now, let's perform the calculations for $$ SE $$ and then for $$ E $$.

I'll calculate $$ SE $$ first.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{6.69}{\sqrt{40}}$$
- step1: Simplify the root:
  $$\frac{6.69}{2\sqrt{10}}$$
- step2: Convert the expressions:
  $$\frac{\frac{669}{100}}{2\sqrt{10}}$$
- step3: Multiply by the reciprocal:
  $$\frac{669}{100}\times \frac{1}{2\sqrt{10}}$$
- step4: Multiply the fractions:
  $$\frac{669}{100\times 2\sqrt{10}}$$
- step5: Multiply:
  $$\frac{669}{200\sqrt{10}}$$
- step6: Multiply by the Conjugate:
  $$\frac{669\sqrt{10}}{200\sqrt{10}\times \sqrt{10}}$$
- step7: Multiply the numbers:
  $$\frac{669\sqrt{10}}{2000}$$
Calculate or simplify the expression $$ 2.576*1.057782 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$2.576\times 1.057782$$
- step1: Multiply the numbers:
  $$2.724846432$$
### Step 2: Calculate the Standard Error (SE)
The standard error (SE) is calculated as:
$$
SE \approx 1.0578
$$

### Step 3: Calculate the Margin of Error (E)
Using the Z-score for $$ 99\% $$ confidence level:
$$
E = 2.576 \times 1.0578 \approx 2.7248
$$

### Conclusion
The calculated margin of error $$ E \approx 2.7248 $$ does not match the provided margin of error $$ E = 2.22 $$.

This indicates that either the provided values or the margin of error is incorrect based on the calculations. If you have any further questions or need additional assistance, feel free to ask!
The margin of error $$ E $$ is approximately $$ 2.72 $$, which does not match the given $$ E = 2.22 $$.

For \( C=99 \% \) If \( \sigma=6.69 \) and \( \mathrm{n}=40 \) then \( E=2.22 \)

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