Assume that σ is unknown, the lower 100(1−α)% confidence bound on μ is: a) μ≤x―+tα,n−1sn b) x―−tα,n−1sn≤μ c) μ≤x―+tα2,n−1sn d) x―−tα2,n−1sn≤μ

Question

Assume that σ is unknown, the lower 100(1−α)% confidence bound on μ is:  a) μ≤x―+tα,n−1sn  b) x―−tα,n−1sn≤μ  c) μ≤x―+tα2,n−1sn  d) x―−tα2,n−1sn≤μ

Ian Adams · Accepted Answer

Step 1 To estimate μ, we draw a random sample of X1…..,Xn of size n from a normal population population N(μ,σ). Let X― be the mean of the random sample, then x―∼N(μ,σ2n) ⇒x―−μσn∼N(0,1)=Z Since σ is unknown, the standard deviation σ is replaced by the estimated standard deviation i.e. the standard error S because S2 is the unbiased estimator of σ2. Since the standard error is an estimate for the true value of the standard deviation, the distribution of the sample mean X― is no longer normal with mean μ and standard deviation σn. Instead, the sample mean follows the Student&#039;s t distribution with n−1 degrees of freedom and with mean μ and standard deviation Sn. Step 2 From the table of Student’s t distribution with n−1 degrees of freedom where α is the confidence coefficient, we can find T such that P(−tn−1,α2≤T≤tn−1,α2)=1−α ⇒P(−tn−1,α2≤X―−usn≤tn−1,α2)=1−α ⇒P(−tn−1,α2Sn≤X―−μ≤tn−1,α2Sn)=1−α ⇒P(−X―−tn−1,α2Sn≤−μ≤−X―+tn−1,α2Sn)=1−α ⇒P(X+tn−1,α2Sn≥μ≥X−tn−1,α2Sn)=1−α ⇒P(X−tn−1,α2Sn≤μ≤X―+tn−1,α2Sn)=1−α Thus the

Assume that \sigma is unknown, the lower 100(1-\alpha)\% confidence bound on \mu is. Give your answers to all of provided questions.

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