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Sagot :
The option that describes the relationship between the two confidence intervals is; The width of I₈₁ will be¹/₉ the width of I₉
How to calculate width of the confidence interval with normal distribution?
The width of the confidence interval will be calculated from the formula;
W = 2Z(σ/√n)
where;
z is the z-score at given confidence level
σ is standard deviation
n is sample size
For first sample;
W_i9 = 2Z(σ/√9)
W_i9 = ²/₃Zσ
For second Sample;
W_i81 = 2Z(σ/√81)
W_i81 = ²/₉Zσ
Thus, we will have;
W_i9/W_i81 = (²/₃Zσ)/(²/₉Zσ)
W_i9/W_i81 = 9
W_i81 = ¹/₉W_i9
Read more about width of confidence interval at; https://brainly.com/question/15934877
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