Question

In a recent poll, 600 people were asked if they liked soccer, and 72% said they did. Based on this, construct a 99% confidence interval for the true population proportion of people who like soccer.
As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.


Give your answers as decimals, to 4 decimal places.

Out of 500 people sampled, 80 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.

As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.
Give your answers to three decimals

Give your answers as decimals, to three decimal places.

Answer 1

Answer:

(0.6728, 0.7672)

(0.1330, 0.1870)

Step-by-step explanation:

Given that in a recent poll, 600 people were asked if they liked soccer, and 72% said they did.

Std error = $\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.72(0.28)}{600} } \\=0.0183$

Margin of error for 99% we would use the value $z = 2.576$

Margin of error = $2.576*SE\\=2.576*0.0183\\=0.0472$

Confidence interval lower bound = $0.72-0.0472=0.6728$

Upper bound = $0.72+0.0472=0.7672$

99% confidence interval for the true population proportion of people who like soccer.=(0.6728, 0.7672)

b) n =500

Sample proportion p=$\frac{80}{500} =0.16$

Margin of error for 90% = $1.645*\sqrt{\frac{0.16*0.84}{500} } \\\\=0.0270$

90% confidence interval for the true population proportion of people with kids. =$(0.16-0.0270, 0.16+0.0270)\\=(0.1330, 0.1870)$