The 90% confidence interval for the true difference between the population proportions is [ - 0.0908, - 0.0092]
Given,
[tex]N_{1}[/tex] = 731
[tex]N_{2}[/tex] = 644
[tex]P_{1}[/tex] = 0.33
[tex]P_{2}[/tex] = 0.28
z score for 90% confidence interval = 1.645
Here, the confidence interval formula :
( [tex]P_{1} -P_{2}[/tex]) ± z [tex]\sqrt{\frac{P_{1}(1-P_{1}) }{N_{1} }+ \frac{P_{2}(1-P_{2}) }{N_{2} } }[/tex]
Substituting the values, we get
(0.33 - 0.28) ± 1.645 [tex]\sqrt{\frac{0.33(1-0.33)}{731}+\frac{0.28(1-0.28)}{644} }[/tex]
= 0.05 ± 1.645 [tex]\sqrt{0.0003024624+0.0003130435}[/tex]
= 0.05 ± 1.645 [tex]\sqrt{0.0006155059}[/tex]
= 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
#SPJ4
