Information on a sample of 362 college students is collected. The complete dataset is available at StudentSurvey. We see that 27 of the 193 males in the sample smoke while 16 of the 169 females in the sample smoke. Find a 99% confidence interval for the difference in proportions. Round your answers to three decimal places. Use 8000 samples.

For males and females C.I. at 99 % 596.24±0.8932 and 353.7±0.7517 respectively.

Step-by-step explanation:

Given:

362 students samples of college students.

193 are males and 169 females.

out of which 27 males and 16 females smokes respectively.

To Find:

99% Confidence interval with 8000 samples.

Solution:

For confidence interval we should know mean ,standard deviation and number of samples .

First calculate the percentage of males and females so

For males,

=193/362

=53.31% males are presented.

For females,

=1-53.31

=46.69% females are presented.

Hence for 8000 samples No of males and females will be ,

Males=8000*0.5331

=4265

Females=8000-4265

=3725

Now consider we know probability for smoking related to male sand females,

For males,

Probability of success =p=27/193=0.1398

probability of failure =q=1-0.1398=0.8602

For mean of males

Mean= no of samples*porbability of success

=n*p

=4265*0.1398

=596.24

Standard deviation=Sqrt(n*p*q)

=Sqrt(4265*0.1398*0.8602)

=Sqrt(512.89)

=22.647

Hence C.I at 99% Z=2.567

C.I=mean±Z[standard deviation/Sqrt(No.of males)]

=596.24±2.567[22.647/Sqrt(4265)]

=596.24±2.567[22.647/65.307]

=596.24±0.8932

For Females,

Probability of success=p=16/169=0.0947

Probability of failure=q=1-0.0947=0.9053

Mean=No of females*Probability of success

=0.0947*3735

Mean=353.7

Standard deviation=Sqrt(n*p*q)

=Sqrt(3735*0.0947*0.9053)

S.D.=17.87

For C.I. at 99% Z=2.567

C.I.=mean±Z[standard deviation/Sqrt(No.of females)]

=353.7±2.567[17.87/Sqrt(3725)]

=353.7±0.7517

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