Solved on Mar 25, 2024

Find the margin of error for a poll of 130 people, where $64\%$ said they liked dogs, at a $90\%$ confidence level.

STEP 1

1. The percentage of people who said they liked dogs is a sample proportion from the poll.
2. The margin of error can be calculated using the formula for the margin of error of a proportion, which is $ME = z \cdot \sqrt{\frac{p(1-p)}{n}}$ , where $z$ is the z-score corresponding to the confidence level, $p$ is the sample proportion, and $n$ is the sample size.
3. The z-score for a $90\%$ confidence level is approximately $1.645$ , which corresponds to the critical value for a two-tailed test where the tails (each) contain $5\%$ of the distribution (since $100\% - 90\% = 10\%$ and $10\%/2 = 5\%$ ).

STEP 2

1. Calculate the sample proportion ( $p$ ).
2. Find the z-score corresponding to the $90\%$ confidence level.
3. Calculate the margin of error using the formula.

STEP 3

Calculate the sample proportion ( $p$ ) using the information given in the poll.
$p = \frac{\text{number of people who like dogs}}{\text{total number of people polled}} = \frac{64\% \text{ of } 130}{130} = 0.64$

STEP 4

Find the z-score corresponding to the $90\%$ confidence level.
The z-score for a $90\%$ confidence level is approximately $1.645$ .

STEP 5

Calculate the margin of error using the formula $ME = z \cdot \sqrt{\frac{p(1-p)}{n}}$ .
$ME = 1.645 \cdot \sqrt{\frac{0.64(1-0.64)}{130}}$

STEP 6

Simplify the calculation inside the square root.
$ME = 1.645 \cdot \sqrt{\frac{0.64 \cdot 0.36}{130}}$

STEP 7

Calculate the value inside the square root.
$\frac{0.64 \cdot 0.36}{130} \approx \frac{0.2304}{130} \approx 0.001773$

STEP 8

Take the square root of the calculated value.
$\sqrt{0.001773} \approx 0.0421$

STEP 9

Multiply the square root by the z-score to find the margin of error.
$ME = 1.645 \cdot 0.0421 \approx 0.0692$

STEP 10

Round the margin of error to three decimal places as requested.
$ME \approx 0.069$
The margin of error of this poll, at the $90\%$ confidence level, is approximately $0.069$ .

Was this helpful?

Find the margin of error for a poll of 130 people, where $64\%$ said they liked dogs, at a $90\%$ confidence level.

STEP 1

STEP 2

1. Calculate the sample proportion ( $p$ ).
2. Find the z-score corresponding to the $90\%$ confidence level.
3. Calculate the margin of error using the formula.

STEP 3

Calculate the sample proportion ( $p$ ) using the information given in the poll.
$p = \frac{\text{number of people who like dogs}}{\text{total number of people polled}} = \frac{64\% \text{ of } 130}{130} = 0.64$

STEP 4

Find the z-score corresponding to the $90\%$ confidence level.
The z-score for a $90\%$ confidence level is approximately $1.645$ .

STEP 5

Calculate the margin of error using the formula $ME = z \cdot \sqrt{\frac{p(1-p)}{n}}$ .
$ME = 1.645 \cdot \sqrt{\frac{0.64(1-0.64)}{130}}$

STEP 6

Simplify the calculation inside the square root.
$ME = 1.645 \cdot \sqrt{\frac{0.64 \cdot 0.36}{130}}$

STEP 7

Calculate the value inside the square root.
$\frac{0.64 \cdot 0.36}{130} \approx \frac{0.2304}{130} \approx 0.001773$

STEP 8

Take the square root of the calculated value.
$\sqrt{0.001773} \approx 0.0421$

STEP 9

Multiply the square root by the z-score to find the margin of error.
$ME = 1.645 \cdot 0.0421 \approx 0.0692$

STEP 10

Round the margin of error to three decimal places as requested.
$ME \approx 0.069$
The margin of error of this poll, at the $90\%$ confidence level, is approximately $0.069$ .

Start learning now

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

Find the margin of error for a poll of 130 people, where 64%64\%64% said they liked dogs, at a 90%90\%90% confidence level.

STEP 1

STEP 2

1. Calculate the sample proportion (ppp).2. Find the z-score corresponding to the 90%90\%90% confidence level.3. Calculate the margin of error using the formula.

STEP 3

STEP 4

Find the z-score corresponding to the 90%90\%90% confidence level.The z-score for a 90%90\%90% confidence level is approximately 1.6451.6451.645.

STEP 5

Calculate the margin of error using the formula ME=z⋅p(1−p)nME = z \cdot \sqrt{\frac{p(1-p)}{n}}ME=z⋅np(1−p)​​.ME=1.645⋅0.64(1−0.64)130 ME = 1.645 \cdot \sqrt{\frac{0.64(1-0.64)}{130}} ME=1.645⋅1300.64(1−0.64)​​

STEP 6

Simplify the calculation inside the square root.ME=1.645⋅0.64⋅0.36130 ME = 1.645 \cdot \sqrt{\frac{0.64 \cdot 0.36}{130}} ME=1.645⋅1300.64⋅0.36​​

STEP 7

Calculate the value inside the square root.0.64⋅0.36130≈0.2304130≈0.001773 \frac{0.64 \cdot 0.36}{130} \approx \frac{0.2304}{130} \approx 0.001773 1300.64⋅0.36​≈1300.2304​≈0.001773

STEP 8

Take the square root of the calculated value.0.001773≈0.0421 \sqrt{0.001773} \approx 0.0421 0.001773​≈0.0421

STEP 9

Multiply the square root by the z-score to find the margin of error.ME=1.645⋅0.0421≈0.0692 ME = 1.645 \cdot 0.0421 \approx 0.0692 ME=1.645⋅0.0421≈0.0692

STEP 10

Round the margin of error to three decimal places as requested.ME≈0.069 ME \approx 0.069 ME≈0.069The margin of error of this poll, at the 90%90\%90% confidence level, is approximately 0.0690.0690.069.

Start learning now

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

Find the margin of error for a poll of 130 people, where $64\%$ said they liked dogs, at a $90\%$ confidence level.

1. Calculate the sample proportion ( $p$ ).
2. Find the z-score corresponding to the $90\%$ confidence level.
3. Calculate the margin of error using the formula.

Find the z-score corresponding to the $90\%$ confidence level.
The z-score for a $90\%$ confidence level is approximately $1.645$ .

Calculate the margin of error using the formula $ME = z \cdot \sqrt{\frac{p(1-p)}{n}}$ .
$ME = 1.645 \cdot \sqrt{\frac{0.64(1-0.64)}{130}}$

Simplify the calculation inside the square root.
$ME = 1.645 \cdot \sqrt{\frac{0.64 \cdot 0.36}{130}}$

Calculate the value inside the square root.
$\frac{0.64 \cdot 0.36}{130} \approx \frac{0.2304}{130} \approx 0.001773$

Take the square root of the calculated value.
$\sqrt{0.001773} \approx 0.0421$

Multiply the square root by the z-score to find the margin of error.
$ME = 1.645 \cdot 0.0421 \approx 0.0692$

Round the margin of error to three decimal places as requested.
$ME \approx 0.069$
The margin of error of this poll, at the $90\%$ confidence level, is approximately $0.069$ .