Question

N1=731pˆ1=0. 33 n2=644pˆ2=0. 28 use this data to find the 90onfidence interval for the true difference between the population proportions

Answer 1

The 90% confidence interval for the true difference between the population proportions is [ - 0.0908, - 0.0092]

Given,

$N_{1}$ = 731

$N_{2}$ = 644

$P_{1}$ = 0.33

$P_{2}$ = 0.28

z score for 90% confidence interval = 1.645

Here, the confidence interval formula :

( $P_{1} -P_{2}$) ± z $\sqrt{\frac{P_{1}(1-P_{1}) }{N_{1} }+ \frac{P_{2}(1-P_{2}) }{N_{2} } }$

Substituting the values, we get

    (0.33 - 0.28) ± 1.645 $\sqrt{\frac{0.33(1-0.33)}{731}+\frac{0.28(1-0.28)}{644} }$

=   0.05 ± 1.645 $\sqrt{0.0003024624+0.0003130435}$

=   0.05 ± 1.645 $\sqrt{0.0006155059}$

=    0.05 ± 1.645 × 0.0248093914

=    0. 05 ± 0.0408114489

Confidence interval:

- 0.05 - 0.0408114489 = - 0.0908114489 ≈ - 0.0908

- 0.05 + 0.0408114489 = - 0.0091885511 ≈ - 0.0092

The confidence interval for the true difference between the population proportions is [ - 0.0908, - 0.0092]

Learn more about confidence interval here:https://brainly.com/question/17421912

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