Since the population standard deviation is unknown, the sample mean follows a t(n-1) distribution with mean "\\mu" and standard deviation"\\frac{s}{\\sqrt{n}}"
From t-table or =T.INV(0.975,29) Excel formula, "t(29, 0.05)=2.04523"
Upper 95% CI is given by "90+(2.04523\u00d7\\frac{18}{\\sqrt{30}})=90+6.72131=96.72131"
Similarly, the answer can be obtained from =90+CONFIDENCE.T(0.05,18,30) Excel formula. Thus, the correct answer is (1)
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