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Math  /  Data & Statistics

QuestionWe wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 468 had kids. Based on this, construct a 90%90 \% confidence interval for the proportion p of adult residents who are parents in this county:
Express your answer in tri-inequality form. Give your answers as decimals, to three places. □\square <p<<\mathrm{p}< □\square Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places. p=p= □\square ±\pm □\square

Studdy Solution

STEP 1

1. The sample size is 600 adult residents.
2. Out of the sample, 468 are parents.
3. We are constructing a 90%90\% confidence interval for the proportion pp.
4. The sample proportion p^\hat{p} is an unbiased estimator of the population proportion pp.

STEP 2

1. Calculate the sample proportion p^\hat{p}.
2. Determine the standard error of the sample proportion.
3. Find the critical value for a 90%90\% confidence interval.
4. Calculate the margin of error.
5. Construct the confidence interval in tri-inequality form.
6. Express the confidence interval using the point estimate and margin of error.

STEP 3

Calculate the sample proportion p^\hat{p}:
p^=468600=0.78\hat{p} = \frac{468}{600} = 0.78

STEP 4

Determine the standard error (SE) of the sample proportion:
SE=p^(1−p^)n=0.78×0.22600SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.78 \times 0.22}{600}}
SE≈0.1716600≈0.000286≈0.0169SE \approx \sqrt{\frac{0.1716}{600}} \approx \sqrt{0.000286} \approx 0.0169

STEP 5

Find the critical value z∗z^* for a 90%90\% confidence interval. For a 90%90\% confidence level, z∗≈1.645z^* \approx 1.645.

STEP 6

Calculate the margin of error (ME):
ME=z∗×SE=1.645×0.0169≈0.0278ME = z^* \times SE = 1.645 \times 0.0169 \approx 0.0278

STEP 7

Construct the confidence interval in tri-inequality form:
0.78−0.0278<p<0.78+0.02780.78 - 0.0278 < p < 0.78 + 0.0278
0.752<p<0.8080.752 < p < 0.808

STEP 8

Express the confidence interval using the point estimate and margin of error:
p=0.78±0.028p = 0.78 \pm 0.028
The 90%90\% confidence interval for the proportion pp is:
0.752<p<0.8080.752 < p < 0.808
Expressed with the point estimate and margin of error:
p=0.78±0.028p = 0.78 \pm 0.028

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