A random sample of size displaystyle{n}={25} from a normal distribution with mean displaystylemu and variance 36 has sample mean displaystyleoverline{

FobelloE

FobelloE

Answered question

2020-11-29

A random sample of size n=25 from a normal distribution with mean μ and variance 36 has sample mean X=16.3
a)
Calculate confidence intervals for μ at three levels of confidence: 80. How to the widths of the confidence intervals change?
b) How would the CI width change if n is increased to 100?

Answer & Explanation

Lacey-May Snyder

Lacey-May Snyder

Skilled2020-11-30Added 88 answers

Step 1
a)
Computation of 80% confidence interval is obtained below:
It is assumed that the population is approximately normal.
Here, α is 0.20(=10.80).
Degrees of freedom is 24=(n1).
Using EXCEL formula =T.INV.2T(0.2,24)", the critical value is obtained as 1.3178. That is, tα2=1.3178.
The 80% confidence interval is:
CI=16.3±(1.3178)625
=(16.31.581416.3+1.5814)
=(14.7186,17.8814)
Step 2
Computation of 90% confidence interval is obtained below:
It is assumed that the population is approximately normal.
Here, α is 0.10(=10.90).
Degrees of freedom is 24(=n1).
Using EXCEL formula =T.INV.2T(0.10,24)", the critical value is obtained as 1.7108. That is, tα2=1.7108.
The 90% confidence interval is:
CI=16.3±(1.7108)625
=(16.31.710816.3+1.7108)
=(14.589,18.011)
Step 3
Computation of 99% confidence interval is obtained below:
It is assumed that the population is approximately normal.
Here, α is 0.01(=10.99).
Degrees of freedom is 24(=n1)
Using EXCEL formula =T.INV.2T(0.01,24)", the critical value is obtained as 2.7969. That is, tα2=2.7969.
The 99% confidence interval is:
CI=16.3±(1.7108)625
=(16.32.05316.3+2.053)
=(14.247,18.353)
Step 4
The confidence level is increased, and then the length of the confidence interval gets wider.
Therefore, the confidence interval gets wider.

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