Questions: Suppose the number of dollars spent per week on groceries is normally distributed. If the population standard deviation is 7 dollars, what minimum sample size is needed to be 90% confident that the sample mean is within 3 dollars of the true population mean? Use the table above for the z-score, and be sure to round up to the nearest integer. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.57

Solution Steps

We are given the following values:

• Population standard deviation \( \sigma = 7 \) dollars
• Margin of error \( E = 3 \) dollars
• Z-score for 90% confidence level \( z = 1.645 \)

The formula for the margin of error \( E \) is given by:

\[ E = z \times \frac{\sigma}{\sqrt{n}} \]

Rearranging this formula to solve for the sample size \( n \):

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

Substituting the known values into the formula:

\[ n = \left( \frac{1.645 \times 7}{3} \right)^2 \]

Calculating the expression step-by-step:

1. Calculate \( z \times \sigma \):

\[ 1.645 \times 7 = 11.515 \]

2. Divide by \( E \):

\[ \frac{11.515}{3} = 3.8383 \]

3. Square the result:

\[ n = (3.8383)^2 \approx 14.7077 \]

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

\[ n = \lceil 14.7077 \rceil = 15 \]

Final Answer