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A survey of 1,026 people asked about what they would do with an unexpected cash gift. Forty-seven percent responded that they would pay off debts. a. With $95 \%$ confidence, what is the margin of error?
Note: Round final answer to 3 decimal places. Margin of error $\square$ b. Construct the $95 \%$ confidence interval for the population proportion of people who would pay off debts with an unexpected cash gift. Note: Round final answers to 3 decimal places. \begin{tabular}{|l|l|} \hline Lower limit & \\ \hline Upper limit & \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. The sample size $n = 1026$ .
2. The sample proportion $\hat{p} = 0.47$ .
3. We are using a $95\%$ confidence level.

STEP 2

1. Calculate the margin of error.
2. Construct the confidence interval.

STEP 3

Determine the critical value for a $95\%$ confidence level. For a normal distribution, this is typically $z = 1.96$ .

STEP 4

Calculate the standard error (SE) using the formula:
$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.47 \times (1-0.47)}{1026}}$

STEP 5

Calculate the margin of error (ME) using the formula:
$ME = z \times SE = 1.96 \times \sqrt{\frac{0.47 \times 0.53}{1026}}$
Calculate the value and round to 3 decimal places.

STEP 6

Calculate the lower limit of the confidence interval using:
$\text{Lower limit} = \hat{p} - ME = 0.47 - ME$

STEP 7

Calculate the upper limit of the confidence interval using:
$\text{Upper limit} = \hat{p} + ME = 0.47 + ME$

STEP 8

Round both the lower and upper limits to 3 decimal places.
The margin of error is approximately:
$\boxed{0.030}$
The $95\%$ confidence interval for the population proportion is approximately:
\begin{tabular}{|l|l|} \hline \text{Lower limit} & \boxed{0.440} \\ \hline \text{Upper limit} & \boxed{0.500} \\ \hline \end{tabular}

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Studdy Solution

STEP 1

1. The sample size $n = 1026$ .
2. The sample proportion $\hat{p} = 0.47$ .
3. We are using a $95\%$ confidence level.

STEP 2

1. Calculate the margin of error.
2. Construct the confidence interval.

STEP 3

Determine the critical value for a $95\%$ confidence level. For a normal distribution, this is typically $z = 1.96$ .

STEP 4

Calculate the standard error (SE) using the formula:
$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.47 \times (1-0.47)}{1026}}$

STEP 5

Calculate the margin of error (ME) using the formula:
$ME = z \times SE = 1.96 \times \sqrt{\frac{0.47 \times 0.53}{1026}}$
Calculate the value and round to 3 decimal places.

STEP 6

Calculate the lower limit of the confidence interval using:
$\text{Lower limit} = \hat{p} - ME = 0.47 - ME$

STEP 7

Calculate the upper limit of the confidence interval using:
$\text{Upper limit} = \hat{p} + ME = 0.47 + ME$

STEP 8

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Studdy Solution

STEP 1

1. The sample size n=1026 n = 1026 n=1026.2. The sample proportion p^=0.47 \hat{p} = 0.47 p^​=0.47.3. We are using a 95% 95\% 95% confidence level.

STEP 2

1. Calculate the margin of error.2. Construct the confidence interval.

STEP 3

Determine the critical value for a 95% 95\% 95% confidence level. For a normal distribution, this is typically z=1.96 z = 1.96 z=1.96.

STEP 4

Calculate the standard error (SE) using the formula:SE=p^(1−p^)n=0.47×(1−0.47)1026SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.47 \times (1-0.47)}{1026}}SE=np^​(1−p^​)​​=10260.47×(1−0.47)​​

STEP 5

Calculate the margin of error (ME) using the formula:ME=z×SE=1.96×0.47×0.531026ME = z \times SE = 1.96 \times \sqrt{\frac{0.47 \times 0.53}{1026}}ME=z×SE=1.96×10260.47×0.53​​Calculate the value and round to 3 decimal places.

STEP 6

Calculate the lower limit of the confidence interval using:Lower limit=p^−ME=0.47−ME\text{Lower limit} = \hat{p} - ME = 0.47 - MELower limit=p^​−ME=0.47−ME

STEP 7

Calculate the upper limit of the confidence interval using:Upper limit=p^+ME=0.47+ME\text{Upper limit} = \hat{p} + ME = 0.47 + MEUpper limit=p^​+ME=0.47+ME

STEP 8

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1. The sample size $n = 1026$ .
2. The sample proportion $\hat{p} = 0.47$ .
3. We are using a $95\%$ confidence level.

1. Calculate the margin of error.
2. Construct the confidence interval.

Determine the critical value for a $95\%$ confidence level. For a normal distribution, this is typically $z = 1.96$ .

Calculate the standard error (SE) using the formula:
$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.47 \times (1-0.47)}{1026}}$

Calculate the margin of error (ME) using the formula:
$ME = z \times SE = 1.96 \times \sqrt{\frac{0.47 \times 0.53}{1026}}$
Calculate the value and round to 3 decimal places.

Calculate the lower limit of the confidence interval using:
$\text{Lower limit} = \hat{p} - ME = 0.47 - ME$

Calculate the upper limit of the confidence interval using:
$\text{Upper limit} = \hat{p} + ME = 0.47 + ME$