The Democrat and Chronicle reported that 25% of the flights arriving at the San

Question

The Democrat and Chronicle reported that 25% of the flights arriving at the San Diego airport during the first five months of 2001 were late (Democrat and Chronicle, July 23, 2001). Assume the population proportion is p = .25.

Calculate () (sigma p-bar) with a sample size of 800 flights (to 4 decimals).

What is the probability that the sample proportion lie between 0.22 and 0.28 if a sample of size 800 is selected (to 4 decimals)?

What is the probability that the sample proportion will lie between 0.22 and 0.28 if a sample of size 400 is selected (to 4 decimals)?

Explanation / Answer

Here p=0.25

SE (p) for n=800 is sqrt(p(1-p)/n)=0.015

Now we need to find P(0.22<p<0.28)

As we know proportion is normally distributed with mean=p and SE=sqrt(p(1-p)/n)

Hence we will convert p to z

P(0.22-0.25/0.015<z<0.28-0.25/0.015)=P(-2<z<2)=P(0<z<2)-P(0<z<-2)=0.4772+0.4772=0.9545

Now we need to find P(0.22<p<0.28) for n=400

Same as above we will convert p to z

Now n=400, SE=0.02

So we get P(-1.5<z<1.5)=P(0<z<1.5)-P(0<z<-1.5)=0.4332+0.4332=0.8664

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.