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kipiarov [429]
2 years ago
6

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

oportions
Mathematics
1 answer:
postnew [5]2 years ago
7 0

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:brainly.com/question/17421912

#SPJ4

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