Question

In a recent poll, 600 people were asked if they liked soccer, and 10% said they...

In a recent poll, 600 people were asked if they liked soccer, and 10% said they did. Based on this, construct a 90% 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.

[________, ________]

Math   Statistics-And-Probability   
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Homework Answers

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = 0.10

1 - = 1 - 0.10 = 0.90

Z/2 = Z0.05 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 (((0.10 * 0.90) / 600)

= 0.0201

A 90% confidence interval for population proportion p is ,

± E   

= 0.10  ± 0.0201

= ( 0.0799, 0.1201 )

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