To solve this problem, we need to determine the minimum sample size $n$ required for the consultant to estimate the mean shopping time $\mu$ within 6 minutes of the true mean with 90% confidence.

Key Information:

• The standard deviation of the population: $\sigma = 26$ minutes
• Margin of error (E): $6$ minutes
• Confidence level: $90\%$

The formula to determine the required sample size is: $n = \left( \frac{z \cdot \sigma}{E} \right)^2$ where:

• $z$ is the critical value corresponding to the confidence level (from the standard normal distribution),
• $\sigma$ is the population standard deviation,
• $E$ is the margin of error.

Steps:

1. Find the critical value $z$ for a 90% confidence level.

   • A 90% confidence level corresponds to a $z$-value of approximately $1.645$.

2. Substitute the values into the formula: $n = \left( \frac{1.645 \cdot 26}{6} \right)^2$

3. Simplify:

   • Calculate $\frac{1.645 \cdot 26}{6}$: $\frac{1.645 \cdot 26}{6} = \frac{42.77}{6} = 7.128$
   • Square the result: $n = (7.128)^2 = 50.82$

4. Round up to the nearest whole number: Since $n$ must be an integer and meet the requirement, round up to $n = 51$.

Final Answer:

The minimum sample size required is 51.

Would you like further explanation or help with a related question? Here are some suggestions for expanding this topic:

1. Why do we need to round up the sample size in such problems?
2. How does changing the confidence level affect the required sample size?
3. What happens to the sample size if the margin of error is halved?
4. Explain the relationship between $\sigma$ (standard deviation) and the sample size.
5. What is the role of the critical $z$-value in determining sample size?

Tip: Always round up the sample size in these problems, even if the decimal portion is very small, to ensure the requirement is met.