Homework Answers
15. Probability of rejecting a true null hypothesis is 5% or 0.05.
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16. x̅ = 84, s = 5, n = 16
98% Confidence interval :
At α = 0.02 and df = n-1 = 15, two tailed critical value, t-crit = T.INV.2T(0.02, 15) = 2.602
Lower Bound = x̅ - t-crit*s/√n = 84 - 2.602 * 5/√16 = 80.7469
Upper Bound = x̅ + t-crit*s/√n = 84 + 2.602 * 5/√16 = 87.2531
80.7469 < µ < 87.2531
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17. Null and Alternative hypothesis:
Ho : µ = 3000
Ha : µ > 3000
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18. x̅ = 94, σ = 21, n = 49
a) Test statistic:
z = (x̅- µ)/(σ/√n) = (94 - 100)/(21/√49) = -2
b) The standard Normal Table
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19. Critical value at 0.05 for right tailed test :
Right tailed critical value, z crit = ABS(NORM.S.INV(0.05) = + 1.645
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