Question

The distribution of systolic blood pressures in a group of females with diabetes is approximately normal with unknown mean. Fortunately, the standard deviation is known: σ= 11.8 mm/Hg.

1. In a sample of 10 women from this group the mean systolic blood pressure was x̅= 130 mmHg. Calculate a two-sided 95% confidence interval for the true mean systolic blood pressure, μ, for the group of diabetic women
2. Repeat (a) but construct a 90% confidence interval instead.

How does the two confidence intervals compare

Answer 1

a)

sample mean, xbar = 130
sample standard deviation, σ = 11.8
sample size, n = 10

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 11.8/sqrt(10)
ME = 7.31

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (130 - 1.96 * 11.8/sqrt(10) , 130 + 1.96 * 11.8/sqrt(10))
CI = (122.69 , 137.31)

b)

sample mean, xbar = 130
sample standard deviation, σ = 11.8
sample size, n = 10

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

ME = zc * σ/sqrt(n)
ME = 1.64 * 11.8/sqrt(10)
ME = 6.12

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (130 - 1.64 * 11.8/sqrt(10) , 130 + 1.64 * 11.8/sqrt(10))
CI = (123.88 , 136.12)

as the confidenc einterval decreases the margin of error decreases and interval would be narrower