I discussed the following question with a friend, but we got different results. What do you think about it?

Let (X1,X2,…,Xn)(X_1,X_2,…,X_n) be a random sample of nn observations from a normal population having an unknwon mean μ\mu and a known variance σ2\sigma^2. Suppose we calculate confidence intervals as follows:

CIn=[ˉxn−σ√n,ˉxn+σ√n]CI_n=\bigg{[}\bar{x}_n – \tfrac{\sigma}{\sqrt{n}}, \bar{x}_n + \tfrac{\sigma}{\sqrt{n}}\bigg{]}

What is the probability that the confidence interval covers the value 1 (false coverage probability) when the true mean μ=0.5,\mu=0.5, σ2=4\sigma^2 = 4 and n=36n=36?

My approach is as follows:

:σ√n=26=13\tfrac{\sigma}{\sqrt{n}} = \tfrac{2}{6} = \tfrac{1}{3}

Pr(1∈[ˉxn−13,ˉxn+13])=Pr(23≤ˉxn≤43)=Pr(ˉxn≤43)−Pr(ˉxn≤23)Pr(1 \in [\bar{x}_n – \tfrac{1}{3}, \bar{x}_n+\tfrac{1}{3}]) = Pr(\tfrac{2}{3} \leq \bar{x}_n \leq \tfrac{4}{3}) = Pr(\bar{x}_n \leq \tfrac{4}{3}) – Pr(\bar{x}_n \leq \tfrac{2}{3})

Standardize: z=3⋅(ˉxn−0.5)z=3\cdot (\bar{x}_n – 0.5).

Pr(ˉxn≤43)=Pr(ˉxn−0.513≤43−0.513)=Pr(ˉxn−0.513≤2.5)=0.99379Pr(\bar{x}_n \leq \tfrac{4}{3}) = Pr\bigg{(}\dfrac{\bar{x}_n – 0.5}{\tfrac{1}{3}} \leq \dfrac{\tfrac{4}{3} – 0.5}{\tfrac{1}{3}}\bigg{)} = Pr\bigg{(}\dfrac{\bar{x}_n – 0.5}{\tfrac{1}{3}} \leq 2.5 \bigg{)} = 0.99379 Pr(ˉxn≤23)=Pr(ˉxn−0.513≤0.5)=0.69146Pr(\bar{x}_n \leq \tfrac{2}{3})=Pr\bigg{(}\dfrac{\bar{x}_n – 0.5}{\tfrac{1}{3}} \leq 0.5\bigg{)} = 0.69146

Coverage probability =0.3023=30.23%__=\underline{\underline{0.3023 = 30.23\%}}.

Is this correct?

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