In a survey of 110 shoppers at a grocery store, 23 said they use coupons regularly. Given a 95% level of confidence with a z-score of 1.96, what is the margin of error, rounded to the nearest tenth of a percent?(1 point)

The margin of error is
%

1 answer

To calculate the margin of error (MOE) for a proportion, you can use the formula:

\[ \text{MOE} = z \times \sqrt{\frac{p(1-p)}{n}} \]

Where:

  • \( z \) is the z-score (1.96 for a 95% confidence level),
  • \( p \) is the sample proportion,
  • \( n \) is the sample size.

In this case:

  • The sample size \( n = 110 \),
  • The number of shoppers using coupons regularly \( x = 23 \).

First, we calculate the sample proportion \( p \):

\[ p = \frac{x}{n} = \frac{23}{110} \approx 0.2091 \]

Now, calculate \( 1-p \):

\[ 1-p = 1 - 0.2091 \approx 0.7909 \]

Next, we can substitute the values into the MOE formula:

\[ \text{MOE} = 1.96 \times \sqrt{\frac{0.2091 \times 0.7909}{110}} \]

Calculating the component inside the square root:

\[ 0.2091 \times 0.7909 \approx 0.1653 \] \[ \frac{0.1653}{110} \approx 0.001503 \] \[ \sqrt{0.001503} \approx 0.0388 \]

Now, substitute back into the MOE formula:

\[ \text{MOE} = 1.96 \times 0.0388 \approx 0.0761 \]

Convert the margin of error from a proportion to a percentage by multiplying by 100:

\[ \text{MOE} \approx 0.0761 \times 100 \approx 7.61% \]

Rounding to the nearest tenth of a percent:

\[ \text{MOE} \approx 7.6% \]

Therefore, the margin of error is 7.6%.