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User1865345
User1865345
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  1. With each variable having a probability of false positive of 0.05,$0.05, $ the expected value is 0.05 * number of variables = 0.05 * 100 = 5$0.05 \times\text{ number of variables} = 0.05 \times 100 = 5.$

  2. For "at least" situations, it is generally easier to find the "other cases" and subtract from one. The probability of zero null rejections is the probability of a null NOT rejected raised to the power of the number of experiments = 0.95 ^ 3 = 0.8574. The$= 0.95 ^ 3 = 0.8574. $ The odds of at least 1$1$ is thus 1 - 0.8574 = 0.1426$1 - 0.8574 = 0.1426$ (probably no rejections).

  3. 1 - 0.95^100 = 0.9941$1 - 0.95^{100} = 0.9941$ (almost surely one or more will be rejected).

  4. 1 - 0.95^n >= 0.5$1 - 0.95^n \geq 0.5.$ This was actually just solved about an hour ago at Probability and statistical significance

  1. With each variable having a probability of false positive of 0.05, the expected value is 0.05 * number of variables = 0.05 * 100 = 5

  2. For "at least" situations, it is generally easier to find the "other cases" and subtract from one. The probability of zero null rejections is the probability of a null NOT rejected raised to the power of the number of experiments = 0.95 ^ 3 = 0.8574. The odds of at least 1 is thus 1 - 0.8574 = 0.1426 (probably no rejections)

  3. 1 - 0.95^100 = 0.9941 (almost surely one or more will be rejected)

  4. 1 - 0.95^n >= 0.5 This was actually just solved about an hour ago at Probability and statistical significance

  1. With each variable having a probability of false positive of $0.05, $ the expected value is $0.05 \times\text{ number of variables} = 0.05 \times 100 = 5.$

  2. For "at least" situations, it is generally easier to find the "other cases" and subtract from one. The probability of zero null rejections is the probability of a null NOT rejected raised to the power of the number of experiments $= 0.95 ^ 3 = 0.8574. $ The odds of at least $1$ is thus $1 - 0.8574 = 0.1426$ (probably no rejections).

  3. $1 - 0.95^{100} = 0.9941$ (almost surely one or more will be rejected).

  4. $1 - 0.95^n \geq 0.5.$ This was actually just solved about an hour ago at Probability and statistical significance

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Community Bot
Community Bot
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  1. With each variable having a probability of false positive of 0.05, the expected value is 0.05 * number of variables = 0.05 * 100 = 5

  2. For "at least" situations, it is generally easier to find the "other cases" and subtract from one. The probability of zero null rejections is the probability of a null NOT rejected raised to the power of the number of experiments = 0.95 ^ 3 = 0.8574. The odds of at least 1 is thus 1 - 0.8574 = 0.1426 (probably no rejections)

  3. 1 - 0.95^100 = 0.9941 (almost surely one or more will be rejected)

  4. 1 - 0.95^n >= 0.5 This was actually just solved about an hour ago at Probability and statistical significanceProbability and statistical significance

  1. With each variable having a probability of false positive of 0.05, the expected value is 0.05 * number of variables = 0.05 * 100 = 5

  2. For "at least" situations, it is generally easier to find the "other cases" and subtract from one. The probability of zero null rejections is the probability of a null NOT rejected raised to the power of the number of experiments = 0.95 ^ 3 = 0.8574. The odds of at least 1 is thus 1 - 0.8574 = 0.1426 (probably no rejections)

  3. 1 - 0.95^100 = 0.9941 (almost surely one or more will be rejected)

  4. 1 - 0.95^n >= 0.5 This was actually just solved about an hour ago at Probability and statistical significance

  1. With each variable having a probability of false positive of 0.05, the expected value is 0.05 * number of variables = 0.05 * 100 = 5

  2. For "at least" situations, it is generally easier to find the "other cases" and subtract from one. The probability of zero null rejections is the probability of a null NOT rejected raised to the power of the number of experiments = 0.95 ^ 3 = 0.8574. The odds of at least 1 is thus 1 - 0.8574 = 0.1426 (probably no rejections)

  3. 1 - 0.95^100 = 0.9941 (almost surely one or more will be rejected)

  4. 1 - 0.95^n >= 0.5 This was actually just solved about an hour ago at Probability and statistical significance

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MikeP
MikeP
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  1. With each variable having a probability of false positive of 0.05, the expected value is 0.05 * number of variables = 0.05 * 100 = 5

  2. For "at least" situations, it is generally easier to find the "other cases" and subtract from one. The probability of zero null rejections is the probability of a null NOT rejected raised to the power of the number of experiments = 0.95 ^ 3 = 0.8574. The odds of at least 1 is thus 1 - 0.8574 = 0.1426 (probably no rejections)

  3. 1 - 0.95^100 = 0.9941 (almost surely one or more will be rejected)

  4. 1 - 0.95^n >= 0.5 This was actually just solved about an hour ago at Probability and statistical significance