Question

Researchers in a populous country contacted more than​ 25,000 inhabitants aged 25 years to see if they had finished high​ school; 81.3 % of the 12 comma 759 males and 80.4​% of the 12 comma 873 females indicated that they had high school diplomas. ​a) What assumptions are necessary to satisfy the conditions necessary for​ inference? ​b) Create a 90​% confidence interval for the difference in graduation rates between males and​ females, p Subscript males Baseline minus p Subscript females. ​c) Interpret your confidence interval. ​d) Is there evidence that boys are more likely than girls to complete high​ school? Create a 90​% confidence interval for the difference in graduation rates between males and​ females, p Subscript males Baseline minus p Subscript females.

Answer 1

MALE	FEMALE
P1= 81.3 % = 0.813	P2 =80.4% = 0.804
n1=12,759	n2 = 12,873

true population = P1 - P2

= 0.813-0.804 = 0.009

a)

Assumption:the samples are independent arepresentative of all recent graduates.

b)   

90​% confidence interval  

Alpha = 0.10

Z(alpha/2) = 1.645

S.E = sqrtof (P1(1- P1))/n1 + (P2(1 - P2)) / n2

= sqrt of ( 0.814 (1 - 0.814)/12,759 + (0.804 (1-8040))/12,873

= sqrt of (0.1514/12,759 + 0.1575/12,873)

= sqrt of (1.186*10-5 + 1.223*10-5)

= sqrt (2.409*10-5)

=4.908* 10-3

C.I = 0.009 +/- (1.645)(4.908* 10-3)

= 0.009+/- 8.07*10-3

= 0.0170 = 0.00093

C.I = (0.00093, 0.0170)

C)

90% intervals are narrower. But this makes perfect sense. If you want more confidence that C.I = (0.00093, 0.0170) an interval contains the true parameter, then the intervals will be wider.

D)

yes, boys are more likely than girls.because of the evidence ishere we observed the confidence interval doesnot contain 0.at Alpha = 0.10

so we conclude that true mean diffference is not zero.so we conclude that the true mean difference is greater than 0.