Confidence Interval Estimation

psychmike1

New Member
Hello everyone. I have a question that I am stuck with. I have no idea if I am on the right track here. I have a series of questions like the one I posted below and I want to make sure I am solving this one correctly before I proceed.

1. Compute a 95% confidence interval for the population mean, based on the sample 1, 2, 3, 4, 5, 6, and 30. Change the number from 30 to 10 and recalculate the confidence interval. Using the results, describe the effect of an outlier or extreme value on the confidence interval.

Sample A Sample B
1 1
2 2
3 3
4 4
5 5
6 6
30 10
mean 7.285714286 4.428571429
SD 9.406901505 2.770102776

For Sample A, the sampling distribution of the mean has a mean of 7.285714 and a standard deviation of 9.406902/the square root of the sample size 2.64575 = 3.555476
7.285714 - (1.96) (3.555476) = 0.316984
7.285714 + (1.96) (3.555476) = 14.25444

For Sample B, the sampling distribution of the mean has a mean of 4.428571 and a standard deviation of 2.770103/the square root of the sample size 2.64575 = 1.04700
4.428571 - (1.96) (1.04700) = 2.37645
4.428571 + (1.96) (1.04700) = 6.48069

quark

Hi Mike,

Welcome to the forum.

The - and +1.96 are for normal distributions. Since your sample size is small (n=7), you should obtain the values from the t table with df=7-1=6.

psychmike

New Member

Quark, thank you for pointing me in the right direction. How does this look? (There are 2 sets of data for the confidence interval question, part A is 1-6 and 30 and part B is 1-6 and 10.

Question #1
Sample Data A computation for both sample data Sample Data B
1 1
2 2
3 3
4 4
5 5
6 6
30 10
Sum of data 51 31
Sample Mean = 7.2857 4.4286
Std. Dev. σ 10.1606 2.9921

Sample size (n < 30) n = 7 7
Conf. Level = 95% 0.95 0.95
Number of degrees of freedom is the sample size minus 1 = 7 - 1 6 6
Square root of number of values sqrt(n) sqrt (7) 2.6458
1-conf (alpha) 1 - 0.95 0.05
t (value from the t-table based on 6 degrees of freedom) (1-level of conf., degree of freedom) 2.4469
2.6458 (2.646)
t*s/sqrt(n) 2.4469 * (10.161/2.646) = 9.3964 2.4469 * (2.9921/2.646) =
2.1107 < 7.2857< 16.6821 0.2251< 4.4286 < 5.7591
95% possibility the true population mean is between [ -2.11, 16.68] [0.23, 5.76]

Standard error SE = SD / Ön. **(*10.1606 / Ö15 ) ( 2.9921 / Ö15) 10.1606/2.6458 = 3.8403 4.4286/2.6458 =
Q. Using the results, describe the effect of an outlier or extreme value on the confidence interval.
A. The variables is sample data B is not truly random, this result affected the results in the sample mean and variance.
Generally, the confidence interval calculations assume there is a genuine random sample of the relevant population, dample B is not normally distributed (1,2,3,4,5,6,30)
It can be presumed that the results [.211, 16.68] from sample shows some flaw in the sampling procedure

t-Test: Paired Two Sample for Means

Variable 1 Variable 2
Mean 7.2857 4.4286
Variance 103.2381 8.9524
Observations 7.0000 7.0000
Pearson Correlation 0.9054
Hypothesized Mean Difference 0.0000
df 6.0000 6.0000
t Stat 1.0000
P(T<=t) one-tail 0.1780
t Critical one-tail 1.9432
P(T<=t) two-tail 0.3559
t Critical two-tail 2.4469 2.4469