Questions: Suppose a marketing company wants to determine the current proportion of customers who click on ads on their smartphones. It was estimated that the current proportion of customers who click on ads on their smartphones is 0.68. How many customers should the company survey in order to be 97% confident that the margin of error is 0.03 for the confidence interval of true proportion of customers who click on ads on their smartphones? Answer: (Round up your answer to nearest whole number)

Solution Steps

We are tasked with determining the required sample size \( n \) for a marketing company to estimate the proportion of customers who click on ads on their smartphones. The following values are provided:

• Estimated population proportion \( p = 0.68 \)
• Desired margin of error \( ME = 0.29 \)
• Confidence level = 97%

For a confidence level of 97%, the corresponding Z-score is approximately \( Z = 2.17 \).

Using the formula for the margin of error for a population proportion:

\[ ME = Z \cdot \sqrt{\frac{p(1-p)}{n}} \]

We can rearrange this to solve for \( n \):

\[ n = \frac{Z^2 \cdot p(1-p)}{ME^2} \]

Substituting the known values:

\[ n = \frac{(2.17)^2 \cdot 0.68 \cdot (1 - 0.68)}{(0.29)^2} \]

Calculating this gives:

\[ n \approx 12.99 \]

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

\[ n = \lceil 12.99 \rceil = 13 \]

Final Answer