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solution5 - Stat 350 (Spring 2011) Section 7.2 7. Homework...

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Stat 350 (Spring 2011) Homework 5 Solution Section 7.2 7. (a) The entry in the 3.0 row and .09 column of the z table is .9990. Similarly, the entry for -3.09 is .0010. Therefore, the area under the z curve between -3.09 and +3.09 is .9990 - .0010 = .9980. The confidence level is then 99.8%. (b) Following the example in part (a), the z‐table entries corresponding to z = ‐2.81 and z = +2.81 are .9975 and .0025, respectively. Therefore the area between these two z values is .9975 ‐ .0025 = .9950. The confidence level is then 99.5%. (c) The z‐table entries corresponding to z = ‐1.4 and z = +1.44 are .9251 and .0749, respectively. Therefore, the area between these two z values is .9251 ‐ .0749 = .8502. The confidence level is then 85.02%. (d) The coefficient of s/ is not written, but is understood to be 1.00. The z‐table entries corresponding to z = ‐1.001 and z = +1.00are .8413 and .1587, respectively. Therefore, the area between these two z values is .8413 ‐ .1587 = .6826. The confidence level is then 68.26%. 8. (a) 98% of the standard normal curve area must be captured. This requires that 1% of the area is to be captured in each tail of the distribution. So, P(Z < -z critical value) = .01 and P(Z > z critical value) = .01. Thus, the z critical value = 2.33. (b) 85% of the standard normal curve area must be captured. This requires that 7.5% of the area is to be captured in each tail of the distribution. So, P(Z < -z critical value) = .075 and P(Z > z critical value) = .075. Thus, the z critical value = 1.44. (b) 75% of the standard normal curve area must be captured. This requires that 12.5% of the area is to be captured in each tail of the distribution. So, P(Z < -z critical value) = .125 and P(Z > z critical value) = .125. Thus, the z critical value = 1.15.
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solution5 - Stat 350 (Spring 2011) Section 7.2 7. Homework...

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