Question

Suppose the following five data values were samples from a normal distribution with a standard deviation of 2.5.

2,3,5,6 and 9

a.) Compute the mean and standard error

b.) Compute the 95% confidence interval

Answer 1

a.) Compute the mean and standard error

Mean = sum of all observations / number of observations

Sum of all observations = 2+3+5+6+9 = 25

Number of observations = n = 5

Mean = 25/5 = 5

Mean = Xbar = 5

Standard deviation = S = 2.5

Standard error = S/sqrt(n) = 2.5/sqrt(5) = 1.118034

Standard error = 1.118034

b.) Compute the 95% confidence interval

Confidence level = 95%

df = n - 1 = 5 - 1 = 4

Critical t value = 2.7764

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 5 ± 2.7764*2.5/sqrt(5)

Confidence interval = 5 ± 2.7764*1.118034

Confidence interval = 5 ± 3.1042

Lower limit = 5 - 3.1042 = 1.90

Upper limit = 5 + 3.1042 =8.10

Confidence interval = (1.90, 8.10)