Question

Estimate Confidence Interval for Population Mean based on sample data and using Appendix Table for t-Distribution

(Population standard deviation is not known, but sample standard deviation is given).

Read material from our e-textbook (link eText) pages 314-330.

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 (see page 328)

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.

Here is your DB-5 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

Solution:

Here, we have to find the 90% confidence interval for the population mean.

C = 90% = 0.90, ? = 1 – C = 1 – 0.90 = 0.10

We are not given a population standard deviation.

So, we have to use t confidence interval.

For sample size, we have to choose a value between 4 and 10.

Suppose, we select sample size = n = 9.

For sample mean Xbar, we have to choose a value between 10 and 80.

Suppose, we select sample mean = Xbar = 50

For sample standard deviation S, we have to choose a value between 3 and 20.

Suppose, we select sample standard deviation = S = 12.

Now, we have

Xbar = 50, S = 12, n = 9

C = 90% = 0.90, ? = 1 – C = 1 – 0.90 = 0.10, ?/2 = 0.10/2 = 0.05

df = n – 1 = 9 – 1 = 8

t-value = 1.8595

(By using t-table)

Confidence Interval = Sample Mean ± Margin of Error

Margin of error = t-value * S/sqrt(n)

Margin of error = 1.8595*12/sqrt(9)

Margin of error =1.8595*12/3

Margin of error =1.8595*4

Margin of error = 7.438192

Confidence Interval = Sample Mean ± Margin of Error

Confidence Interval = 50 ± 7.438192

Lower limit = 50 - 7.438192 = 42.561808

Upper limit = 50 + 7.438192 = 57.438192

Confidence interval = (42.561808, 57.438192)

We are 90% confident that the population mean will be lies between 42.561808 and 57.438192.