Question

A severely damaged manuscript by a long-deceased scholar was excavated. The scholar measured the heights of 40 people living on a remote Pacic island and reported that x = 158.7. Assuming the standard deviation was known, the scholar presented a 95% confidence interval (for the mean height of the island inhabitants) with 155.8 as its lower limit. Provide the following information.

a) the standard deviation that the scholar used in writing the manuscript;

b) the (missing) upper limit of his 95% confidence interval;

c) a 99% confidence interval for the mean height of the island inhabitants.

d) the precise p-value (assume that the scholar was interested in testing u= 164 vs. u is not= 164).

Answer 1

x̅ = 158.7, n = 40

At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) = 1.960

95% confidence interval Lower limit = 155.8

a) Margin of error = x̅ -Lower limit = 158.7 - 155.8 = 2.9

Margin of error , E = z*σ/√n

σ = E*√n / z = 2.9*√40 / 1.96 = 9.3578

Standard deviation = 9.3578

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b) Upper Bound = x̅ + z_c*σ/√n = 158.7 + 1.96 * 9.3578/√40 = 161.6

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c)

99% Confidence interval :

At α = 0.01 two tailed critical value, z_c = ABS(NORM.S.INV(0.01/2)) = 2.576

Lower Bound = x̅ - z_c*σ/√n = 158.7 - 2.576 * 9.3578/√40 = 154.889

Upper Bound = x̅ + z_c*σ/√n = 158.7 + 2.576 * 9.3578/√40 = 162.511

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d)

Null and Alternative hypothesis:

Ho : µ = 164

H1 : µ ≠ 164

Test statistic:

z = (x̅- µ)/(σ/√n) = (158.7 - 164)/(9.3578/√40) = -3.5821

p-value :

p-value = 2*(1-NORM.S.DIST(ABS(-3.5821, 1) = 0.0003