**How about this?**

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