Question

You know that y is a normally distributed variable with a variance of 9. You do not know its mean. You collect some data. For each sample below, form the 95% confidence interval and test the null hypothesis of the mean equaling 2.

a. (8,1,5)
b. (8,1,5,-4,-8)
c. (8,1,5,-4,-8,4,8,5)

Answer 1

a)

CI = ( 1.2718, 8.0614 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀

b)

CI = ( -2.2296, 3.0296 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀

c)

CI = ( 0.2962, 4.4538 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀

Explanation:

Given that;

Variance σ² = 9

so standard deviation σ = √9 = 3

with 95% confidence interval in each case. thus, The Z value for 95% confidence is Z = 1.96

Null Hypotheses H₀ : μ = 2

Now

a) (8,1,5)

sample mean x" = (8+1+5) / 3 = 14/3 = 4.6666

n = 3

Confidence Interval CI = x" ± ( z × σ/√n )

CI = 4.6666 ± ( 1.96 × 3/√3 )

CI = 4.6666 ± 3.3948

so CI = ( 1.2718, 8.0614 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀

b) (8,1,5,-4,-8)

sample mean X" = (8 +1 + 5 - 4 - 8) / 5 = 2 / 5 = 0.4

n = 5

Confidence Interval CI = x" ± ( z × σ/√n )

CI = 0.4 ± ( 1.96 × 3/√5 )

CI = 0.4 ± 2.6296

so CI = ( -2.2296, 3.0296 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀

c) (8,1,5,-4,-8,4,8,5)

sample mean x" = (8 + 1 + 5 - 4 - 8 + 4 + 8 + 5) / 8 = 19 / 8 = 2.375

n = 8

Confidence Interval CI = x" ± ( z × σ/√n )

CI = 2.375 ± ( 1.96 × 3/√8 )

CI = 2.375 ± 2.0788

so CI = ( 0.2962, 4.4538 )

Since value 2 is contained within Confidence Interval, fail to Reject H₀