Questions: You would like to start a new business providing Internet service and need to estimate the average Internet usage of households during one week for your business plan. How many households must you select to be 95 percent sure that the sample mean is within one minute (E=1) of the population mean? Assume a previous survey of household usage has shown that sigma = 6.95 minutes.

Solution Steps

We are tasked with estimating the average Internet usage of households during one week. The following parameters are provided:

• Population standard deviation \( \sigma = 6.95 \) minutes
• Margin of error \( E = 1 \) minute
• Confidence level = 95%, which corresponds to a Z-score \( Z \approx 1.96 \)

To determine the required sample size \( n \), we use the formula:

\[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \]

Substituting the known values:

\[ n = \left( \frac{1.96 \cdot 6.95}{1} \right)^2 \]

Calculating the expression inside the parentheses:

\[ n = \left( 13.617 \right)^2 \]

Calculating \( n \):

\[ n \approx 185.56 \]

Since the sample size must be a whole number, we round \( n \) to the nearest whole number:

\[ n \approx 186 \]

Final Answer