In a survey of 110 shoppers at a grocery store 23 said they use coupons regularly give 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 answer

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

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

Where:

  • \(z\) is the z-score (1.96 for a 95% confidence level),
  • \(p\) is the sample proportion of shoppers who use coupons regularly,
  • \(n\) is the total number of shoppers surveyed.

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

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

Now we can compute \(1 - p\):

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

Next, we can substitute these values into the margin of error formula:

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

Calculating the values under the square root:

\[ 0.2091 \times 0.7909 \approx 0.1654 \] \[ \frac{0.1654}{110} \approx 0.0015036 \] \[ \sqrt{0.0015036} \approx 0.0388 \]

Now substitute this back into the margin of error formula:

\[ ME = 1.96 \times 0.0388 \approx 0.0761 \]

To convert the margin of error into a percentage:

\[ ME \approx 0.0761 \times 100 \approx 7.61% \]

Rounded to the nearest tenth of a percent, the margin of error is:

\[ \boxed{7.6%} \]