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Homework7key - Homework 7 (15.2 pts) due Oct. 28 (1.2 pts.)...

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1 Homework 7 (15.2 pts) due Oct. 28 (1.2 pts.) 8.10 (p.354). For each of the given pairs of P-values and significance levels, state whether H 0 should be rejected: a) P-value = 0.084, = 0.05 Fail to reject H 0 , 0.084 > 0.05 b) P-value = 0.003., = 0.001 Fail to reject H 0 , 0.003 > 0.001 c) P-value = 0.048, = 0.05 Reject H 0 , 0.048 ≤ 0.05 d) P-value = 0.084, = 0.10 Reject H 0 , 0.084 ≤ 0.10 e) P-value = 0.039, = 0.01 Fail to reject H 0 , 0.039 > 0.01 f) P-value = 0.017, = 0.10 Reject H 0 , 0.017 ≤ 0.10 (2 pts.) 8.12 (you do not need to do the 7 steps, p.354). New purchased automobile tires of a certain type are supposed to be filled to a pressure of 34 psi. Let μ denote the true average pressure. A test of H 0 : μ = 34 versus H a : μ ≠ 34 will be based on a large sample of tires so that the z test statistic will have approximately a standard normal distribution. Determine the value of z and the corresponding P-value in each of the following cases: a) n = 50, x ̅ = 34.43, s = 1.06 ° = ± − ? 0 ² ³´ µ = 34.43 34 1.06 ³ 50 µ = 2.868 P-value = 2P (Z > 2.868) = 2(1 – P(Z ≤ 2.87)) = 2(1 – 0.9979) = 2(0.0021) = 0.0042 b) n = 50, x ̅ = 33.57, s = 1.06 ° = ± − ? 0 ² ³´ µ = 33.57 34 1.06 ³ 50 µ = 2.868 P-value = 2P (Z < -2.87) = 2(0.0021) = 0.0042
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2 c) n = 32, x ̅ = 33.25, s = 1.89 ° = ± − ? 0 ² ³´ µ = 33.25 34 1.89 ³ 32 µ = 2.24 P-value = 2P (Z < -2.24) = 2(0.0125) = 0.0250 d) n = 36, x ̅ = 34.66, s = 2.53 ° = ± − ? 0 ² ³´ µ = 34.66 34 2.53 ³ 36 µ = 1.565 P-value = 2P (Z > 1.565) = 2(1 – P(Z ≤ 1.57)) = 2(1 – 0.9418) = 2(0.0582) = 0.1164 (1.5 pt.) 8.16 (use the 7 steps, p.355). To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 35 specimens are buried in soil for an extended period. The maximum penetration (in mils) is then measured for each specimen, yielding a sample mean penetration of 52.7 and a sample standard deviation of 4.8. The conduits were manufactured with the specification that the true average penetration be at most 50 mils. Does the sample data indicate that specifications have not been met? 1. μ = the true average maximum penetration distance for corrosion of a certain type of steel conduit. 2. H 0 : μ = 50 H a : μ > 50 3. α = 0.05 4. ° = ± − ? 0 ² ³´ µ = 52.7 50 4.8 ³ 35 µ = 3.328 5. P = P(Z > 3.33) = 1 – P(Z ≤ 3.33) = 1 – 0.9996 = 0.0004 6. Reject H 0 because 0.0004 > 0.05 7. The data does give strong support (P-value = 0.0004) to the claim that the true average maximum penetration distance for corrosion of a certain type of steel conduit is greater than 50 mils. (1.5 pts.) 8.18 (these do NOT use the 7 steps, p.367). The paint used to make lines on
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Homework7key - Homework 7 (15.2 pts) due Oct. 28 (1.2 pts.)...

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