To estimate the number of people needed to reduce the uncertainty (margin of error) to 0.03, we can use the formula for the sample size in proportion estimates:

$n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right)$

where:

• $n$ = required sample size
• $Z$ = Z-score corresponding to the desired confidence level (for example, 1.96 for 95% confidence)
• $p$ = estimated proportion of success (percentage of people who made a purchase)
• $E$ = margin of error (uncertainty)

From your dataTo estimate the number of people needed to reduce the uncertainty (margin of error) to 0.03, we can use the formula for the sample size in proportion estimates:

$n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right)$

where:

• $n$ = required sample size
• $Z$ = Z-score corresponding to the desired confidence level (for example, 1.96 for 95% confidence)
• $p$ = estimated proportion of success (percentage of people who made a purchase)
• $E$ = margin of error (uncertainty)

From your

• Sample size $n = 150$
• Number of people who made a purchase = 83
• Estimated proportion $p = \frac{83}{150} \approx 0.5533$
• Desired margin of error $E = 0.03$

Assuming a 95% confidence level (Z = 1.96):

1. Calculate the required sample size:

$n = \left( \frac{(1.96)^2 \cdot 0.5533 \cdot (1 - 0.5533)}{(0.03)^2} \right)$

$n = \left( \frac{(1.96)^2 \cdot 0.5533 \cdot 0.4467}{(0.03)^2} \right)$

Calculating this gives:

$n \approx \left( \frac{3.8416 \cdot 0.5533 \cdot 0.4467}{0.0009} \right)$

$n \approx \left( \frac{0.9547}{0.0009} \right) \approx 1050.78$

Since sample size must be a whole number, round up:

$n \approx 1051$

Thus, you need to sample approximately 1051 people to achieve the desired level of certainty.

Summary:

The estimated number of people that must be sampled to reduce the uncertainty to 0.03 is 1051.

Would you like more details or have any questions about this calculation? Here are some related questions you might consider:

1. How do you interpret the Z-score in this context?
2. What happens if the estimated proportion $p$ changes?
3. How would the required sample size differ for a 90% confidence level?
4. What is the importance of the margin of error in survey sampling?
5. Can you explain how to calculate a confidence interval for the proportion?

Tip: Always round up your sample size to ensure you meet the desired precision in your estimates.