Roderick_April_Unit_8_Project_MM_207-05

# Roderick_April_Unit_8_Project_MM_207-05 - 9.1#8 a Ho μ...

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Unformatted text preview: 9.1 #8 a. Ho; μ =8.7 sec. b. H1; μ >8.7 sec. c.H1: μ < 8.7 sec. d. b. It would be right of the mean, because the test is for μ > 8.7 d. c. It would be left of the mean, because the test is for μ < 8.7 9.1 #10 a. a = 0.05 H0: µ = 85 H1: µ > 85 Since H1: µ > 85 mg/100 ml, this is a right-tailed test b. Since the x distribution is normal and σ is known, use the standard normal distribution. z =(93.8 – 85) / (12.5 / 2.83) = 8.8 / 4.42 = 1.99 c. P(xb < 85) = 1 - P(z > 1.99) = 1 - 0.9767 P-value = 0.0233 d. P-value <= alpha is false. We do not reject H0 The data is not statistically significant. e. p-value less than the significance level we reject the null hypothesis and conclude the alternate hypothesis is true. 9.1 #14 a = 0.01 H : µ = 28 H 1 : µ ≠ 28 H 1 : µ ≠ 28, this is a two-tailed test b. z = (32.7– 28) / (4.75/ 2.65) = 4.7 / 1.79 = 2.63 c P(z > 2.63) = 2( 0.0043) P-value = 0.0086 d....
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Roderick_April_Unit_8_Project_MM_207-05 - 9.1#8 a Ho μ...

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