Determine the margin of error for a confidence interval to estimate the populati

Question

Determine the margin of error for a confidence interval to estimate the population mean with n = 22 and s = 14.9 for the confidence levels below 80% 90% 99% The margin of error for an 80% confidence interval is. (Round to two decimal places as needed.) The margin of error for a 90% confidence interval is. (Round to two decimal places as needed.) The margin of error for a 99% confidence interval is. Round to two decimal places as needed.) Construct a 98% confidence interval to estimate the population mean when x = 131 and s = 33 for the sample sizes below. n = 20 n = 50 n = 100 The 98% confidence interval for the population mean when n = 20 is from a lower limit of to an upper limit of. (Round to two decimal places as needed.) The 98% confidence interval for the population mean when n a 50 is from a lower limit of to an upper limit of. (Round to two decimal places as needed.) The 98% confidence interval for the population mean when n 100 is from a lower limit

Explanation / Answer

7. Here sd=14. and n=22

a. t table value for df=21 at 80% CI is 1.323

Hence margin of error=t*sd/sqrt(n)=4.20

b. t table value for df=21 at 90% CI is 1.721

Hence margin of error=t*sd/sqrt(n)=5.47

c. t table value for df=21 at 99% CI is 2.831

Hence margin of error=t*sd/sqrt(n)=8.99

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

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