Question

Here is an example with steps you can follow: sample size n=9, sample mean=80, sample standard deviation s=25 (population standard deviation is not known) Estimate confidence interval for population mean with confidence level 90%.

Confidence Interval = Sample Mean ± Margin of Error Margin of Error = (t-value)×s/√n

t-value should be taken from Appendix Table IV. For n=9 df=n-1=9-1=8 For Confidence Level 90% a = 1 - 0.90 = 0.10, a/2 = 0.10/2 = 0.05

So, we are looking for t-value in column t0.05 and row 8. t-value = 1.860

Margin of Error E = 1.860 * 25/√(9) = 15.5 ≈ 16 Confidence interval for population mean: 80 ± 16 or between 64 and 96.

Assignment: Keep Confidence Level 90%. For sample size n, choose a value for n between 4 and 10 For sample mean x , choose a value between 10 and 80 For sample standard deviation s, choose a value between 3 and 20 1. Use Appendix table IV and find t-value for your case; 2. Calculate Margin of Error; 3. Create Confidence Interval for Population Mean.

Answer 1

	CI
	xbar	60.000
	sd	15.000
	n	8
	alpha	0.1
critical
	2-tailed	1.895
	E	10.05
	lower	49.95
	upper	70.05

assuming sample mean = 60 ,sd = 15 , n = 8

critical value = 1.895

margin of error = 10.05

confidence interval is (49.95,70.05)

or between 50 to 70

Please rate