A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a $90 \%$ confidence interval with an error of no more than 0.06 . A consultant has informed them that a previous study found the mean to be 3 fast food meals per week and found the variance to be 1.69 . What is the minimum sample size required to create the specified confidence interval? Round your answer up to the next integer.

Step 1 ：A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a 90% confidence interval with an error of no more than 0.06. A consultant has informed them that a previous study found the mean to be 3 fast food meals per week and found the variance to be 1.69. What is the minimum sample size required to create the specified confidence interval? Round your answer up to the next integer.

Step 2 ：First, we need to calculate the standard deviation, which is the square root of the variance. In this case, the variance is 1.69, so the standard deviation is \( \sqrt{1.69} = 1.3 \).

Step 3 ：Next, we use the Z-score for a 90% confidence level, which is 1.645. The margin of error is 0.06.

Step 4 ：We can calculate the sample size using the formula \( n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \), where Z is the Z-score, \(\sigma\) is the standard deviation, and E is the margin of error.

Step 5 ：Substituting the given values into the formula, we get \( n = \left( \frac{1.645 \cdot 1.3}{0.06} \right)^2 \).

Step 6 ：Calculating this gives us a sample size of approximately 1270.5.

Step 7 ：However, since we can't have a fraction of a person, we round this up to the next whole number, giving us a minimum sample size of 1271.

Step 8 ：Final Answer: The minimum sample size required to create the specified confidence interval is \(\boxed{1271}\).