Solved on Oct 29, 2023

In a survey of 254 students, $86\%$ enjoyed their classes. Find the margin of error, rounded to the nearest tenth of a percent.

STEP 1

Assumptions1. The total number of students surveyed is254. .86% of the students said they enjoyed the classes they were taking.
3. We are looking for the margin of error for the survey results.

STEP 2

The margin of error for a proportion in a population can be calculated using the following formula $Margin\,of\,Error = Z \times \sqrt{\frac{p(1-p)}{n}}$ where- $Z$ is the z-score, which corresponds to the confidence level (In most common surveys, a95% confidence level is used, and the corresponding z-score is approximately1.96). - $p$ is the proportion of the population which has the characteristic of interest (in this case, the proportion of students who said they enjoyed the classes they were taking). - $n$ is the size of the population (in this case, the number of students surveyed).

STEP 3

First, we need to convert the percentage of students who said they enjoyed the classes they were taking into a decimal.
$p =86\% =0.86$

STEP 4

Now, we can plug in the values into the formula to calculate the margin of error.
$Margin\,of\,Error =1.96 \times \sqrt{\frac{0.86(1-0.86)}{254}}$

STEP 5

Perform the calculation inside the square root first.
$Margin\,of\,Error =1.96 \times \sqrt{\frac{0.86 \times0.14}{254}}$

STEP 6

Calculate the value under the square root.
$Margin\,of\,Error =1.96 \times \sqrt{0.0048}$

STEP 7

Calculate the square root.
$Margin\,of\,Error =1.96 \times0.0693$

STEP 8

Finally, calculate the margin of error.
$Margin\,of\,Error =0.1358$

STEP 9

Convert the decimal value to a percentage and round to the nearest tenth of a percent.
$Margin\,of\,Error =13.6\%$ The margin of error for the survey results is13.6%.

Was this helpful?

In a survey of 254 students, $86\%$ enjoyed their classes. Find the margin of error, rounded to the nearest tenth of a percent.

STEP 1

Assumptions1. The total number of students surveyed is254. .86% of the students said they enjoyed the classes they were taking.
3. We are looking for the margin of error for the survey results.

STEP 2

STEP 3

First, we need to convert the percentage of students who said they enjoyed the classes they were taking into a decimal.
$p =86\% =0.86$

STEP 4

Now, we can plug in the values into the formula to calculate the margin of error.
$Margin\,of\,Error =1.96 \times \sqrt{\frac{0.86(1-0.86)}{254}}$

STEP 5

Perform the calculation inside the square root first.
$Margin\,of\,Error =1.96 \times \sqrt{\frac{0.86 \times0.14}{254}}$

STEP 6

Calculate the value under the square root.
$Margin\,of\,Error =1.96 \times \sqrt{0.0048}$

STEP 7

Calculate the square root.
$Margin\,of\,Error =1.96 \times0.0693$

STEP 8

Finally, calculate the margin of error.
$Margin\,of\,Error =0.1358$

STEP 9

Convert the decimal value to a percentage and round to the nearest tenth of a percent.
$Margin\,of\,Error =13.6\%$ The margin of error for the survey results is13.6%.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

In a survey of 254 students, 86%86\%86% enjoyed their classes. Find the margin of error, rounded to the nearest tenth of a percent.

STEP 1

Assumptions1. The total number of students surveyed is254. .86% of the students said they enjoyed the classes they were taking.3. We are looking for the margin of error for the survey results.

STEP 2

STEP 3

First, we need to convert the percentage of students who said they enjoyed the classes they were taking into a decimal.p=86%=0.86p =86\% =0.86p=86%=0.86

STEP 4

Now, we can plug in the values into the formula to calculate the margin of error.Margin of Error=1.96×0.86(1−0.86)254Margin\,of\,Error =1.96 \times \sqrt{\frac{0.86(1-0.86)}{254}}MarginofError=1.96×2540.86(1−0.86)​​

STEP 5

Perform the calculation inside the square root first.Margin of Error=1.96×0.86×0.14254Margin\,of\,Error =1.96 \times \sqrt{\frac{0.86 \times0.14}{254}}MarginofError=1.96×2540.86×0.14​​

STEP 6

Calculate the value under the square root.Margin of Error=1.96×0.0048Margin\,of\,Error =1.96 \times \sqrt{0.0048}MarginofError=1.96×0.0048​

STEP 7

Calculate the square root.Margin of Error=1.96×0.0693Margin\,of\,Error =1.96 \times0.0693MarginofError=1.96×0.0693

STEP 8

Finally, calculate the margin of error.Margin of Error=0.1358Margin\,of\,Error =0.1358MarginofError=0.1358

STEP 9

Convert the decimal value to a percentage and round to the nearest tenth of a percent.Margin of Error=13.6%Margin\,of\,Error =13.6\%MarginofError=13.6%The margin of error for the survey results is13.6%.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

In a survey of 254 students, $86\%$ enjoyed their classes. Find the margin of error, rounded to the nearest tenth of a percent.

Assumptions1. The total number of students surveyed is254. .86% of the students said they enjoyed the classes they were taking.
3. We are looking for the margin of error for the survey results.

First, we need to convert the percentage of students who said they enjoyed the classes they were taking into a decimal.
$p =86\% =0.86$

Now, we can plug in the values into the formula to calculate the margin of error.
$Margin\,of\,Error =1.96 \times \sqrt{\frac{0.86(1-0.86)}{254}}$

Perform the calculation inside the square root first.
$Margin\,of\,Error =1.96 \times \sqrt{\frac{0.86 \times0.14}{254}}$

Calculate the value under the square root.
$Margin\,of\,Error =1.96 \times \sqrt{0.0048}$

Calculate the square root.
$Margin\,of\,Error =1.96 \times0.0693$

Finally, calculate the margin of error.
$Margin\,of\,Error =0.1358$

Convert the decimal value to a percentage and round to the nearest tenth of a percent.
$Margin\,of\,Error =13.6\%$ The margin of error for the survey results is13.6%.