An article in Engineering Horizons (Spring 1990, p. 26) reported that 117 of 484

Question

An article in Engineering Horizons (Spring 1990, p. 26) reported that 117 of 484 new engineering graduates were planning to continue studying for an advanced degree. Consider this as a random sample of the 1990 graduating class. Round your answers to 3 decimal places. Find a 95% confidence interval on the proportion of such graduates planning to continue their education. ______________ lessthanorequalto p lessthanorequalto ____________ Find a 99% confidence interval on the proportion of such graduates planning to continue their education. ______________ lessthanorequalto p lessthanorequalto __________________ The 95% confidence interval is _______________________ the 99% confidence interval.

Explanation / Answer

a.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=117
Sample Size(n)=484
Sample proportion = x/n =0.242
Confidence Interval = [ 0.242 ±Z a/2 ( Sqrt ( 0.242*0.758) /484)]
= [ 0.242 - 1.96* Sqrt(0) , 0.242 + 1.96* Sqrt(0) ]
= [ 0.204,0.28]
b.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=117
Sample Size(n)=484
Sample proportion = x/n =0.242
Confidence Interval = [ 0.242 ±Z a/2 ( Sqrt ( 0.242*0.758) /484)]
= [ 0.242 - 2.576* Sqrt(0) , 0.242 + 2.58* Sqrt(0) ]
= [ 0.192,0.292]
c.
narrower

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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