Question

In: Statistics and Probability

A recent survey of 1000 adults between 18 and 40 showed that 34% said they had...

A recent survey of 1000 adults between 18 and 40 showed that 34% said they had no credit cards. Find the 99% confidence interval for the population proportion.

Expert Solution

Solution :

Given that,

n = 1000

Point estimate = sample proportion = =0.34

1 - = 1-0.34 =0.66

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* ( $\sqrt$ ((0.34*0.66) /1000 )

E = 0.0386

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.34-0.0386 < p < 0.34+0.0386

0.3014< p < 0.3786

The 99% confidence interval for the population proportion p is : 0.3014, 0.3786

orchestra answered 1 year ago

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