This Quiz: 8 pts po 3 of 8 (1 complete) This Question 1 pt A simple random sampl

Question

This Quiz: 8 pts po 3 of 8 (1 complete) This Question 1 pt A simple random sample of size n is drawn from a population that is normally distributed The sample mean, is found to bo 106, and the samplo standard deviatic s, is found to be 10. (a) Construct a 95% confidence interval about w if the sample size, n, is 20. (b) Construct a 95% confidence interval about w if the sample size, n is 24. (c) Construct a 99% confidence interval about wif the sample size, n, is 20 (d) Could we have computed the confidence intervals in parts (a (c) ir the population had not been normally distributed? (c) Construct a 99% confidence interval about w if the sample size, n, is 20 Lower bound Upper bound (use ascending order Round to one decimal place asneeded) compare the results to those obtained in part (a) How does increasina the level e affect the size of the margin of error, E? click to select your answer(s).

Explanation / Answer

a.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=106
Standard deviation( sd )=10
Sample Size(n)=20
Confidence Interval = [ 106 ± t a/2 ( 10/ Sqrt ( 20) ) ]
= [ 106 - 2.093 * (2.236) , 106 + 2.093 * (2.236) ]
= [ 101.32,110.68 ]
b.
when n = 24
Confidence Interval = [ 106 ± t a/2 ( 10/ Sqrt ( 24) ) ]
= [ 106 - 2.069 * (2.041) , 106 + 2.069 * (2.041) ]
= [ 101.777,110.223 ]
c.
sample size = 20
Confidence Interval = [ 106 ± t a/2 ( 10/ Sqrt ( 20) ) ]
= [ 106 - 2.861 * (2.236) , 106 + 2.861 * (2.236) ]
= [ 99.603,112.397 ]

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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