A toy manufacturer wants to know how many new toys children buy each year. A sample of 1417 children was taken to study their purchasing habits. Construct the $$ 99 \% $$ confidence interval for the mean number of toys purchased each year if the sample mean was found to be 7.7 . Assume that the population standard deviation is 1.8. Round your answers to one decimal place.

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1. The confidence interval for the mean is given by  
   $$
   \bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}},
   $$  
   where $$ \bar{x} $$ is the sample mean, $$ \sigma $$ is the population standard deviation, $$ n $$ is the sample size, and $$ z_{\alpha/2} $$ is the critical value.

2. Here, we have:
   - $$ \bar{x} = 7.7 $$
   - $$ \sigma = 1.8 $$
   - $$ n = 1417 $$
   - For a $$ 99\% $$ confidence level, $$ z_{\alpha/2} \approx 2.576 $$.

3. Calculate the standard error:
   $$
   \frac{\sigma}{\sqrt{n}} = \frac{1.8}{\sqrt{1417}}.
   $$
   Approximating,
   $$
   \sqrt{1417} \approx 37.63,
   $$
   so
   $$
   \frac{1.8}{37.63} \approx 0.0478.
   $$

4. Compute the margin of error:
   $$
   \text{Margin of Error} = 2.576 \times 0.0478 \approx 0.1233.
   $$

5. Determine the lower and upper bounds:
   $$
   \text{Lower bound} = 7.7 - 0.1233 \approx 7.5767,
   $$
   $$
   \text{Upper bound} = 7.7 + 0.1233 \approx 7.8233.
   $$

6. Rounding these to one decimal place gives:
   $$
   \text{Lower bound} \approx 7.6,\quad \text{Upper bound} \approx 7.8.
   $$

Thus, the $$ 99\% $$ confidence interval for the mean number of toys purchased each year is  
$$
(7.6,\, 7.8).
$$
The 99% confidence interval for the mean number of toys purchased each year is (7.6, 7.8).

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