In a survey, 22 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $\$ 36$ and standard deviation of $\$ 5$. Find the margin of error at a $90 \%$ confidence level. Give your answer to two decimal places. \[ \text { MOE }= \]

Step 1 ：In a survey, 22 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $36 and standard deviation of $5. We are asked to find the margin of error at a 90% confidence level.

Step 2 ：The margin of error (MOE) for a confidence interval can be calculated using the formula: \[\text{MOE} = Z \times \frac{\sigma}{\sqrt{n}}\] where Z is the Z-score, which depends on the confidence level, \(\sigma\) is the standard deviation, and n is the sample size.

Step 3 ：For a 90% confidence level, the Z-score is approximately 1.645. The standard deviation (\(\sigma\)) is $5 and the sample size (n) is 22.

Step 4 ：Substituting these values into the formula, we get: \[\text{MOE} = 1.645 \times \frac{5}{\sqrt{22}}\]

Step 5 ：Solving the above expression, we find that the margin of error (MOE) is approximately 1.75.

Step 6 ：Final Answer: The margin of error at a 90% confidence level is approximately $1.75. So, the final answer is \(\boxed{1.75}\).