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P(-1.96 < z < 1.96) = .95

P(-1.96 < z < 1.96) = .95. Shape of t Distributions for n = 3 and n = 12. Degrees of Freedom ( dof ). Example : If 10 students take a quiz with a mean of 80, we can freely assign values to the first 9 students, but the 10 th student will have a determined score.

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P(-1.96 < z < 1.96) = .95

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  1. P(-1.96 < z < 1.96) = .95

  2. Shape of tDistributions for n = 3 and n = 12

  3. Degrees of Freedom (dof) Example: If 10 students take a quiz with a mean of 80, we can freely assign values to the first 9 students, but the 10th student will have a determined score. We know the sum of all 10 has to be 800; so when we add up the first 9 that sum will establish what 10th score has to be. Because the first 9 could be freely assigned we say that there are 9 degrees of freedom. For applications in this section the Degrees of Freedom is simply the sample size minus one : dof = n-1

  4. As the number of degrees of freedom increases, the distribution becomes more symmetric. Chi-Square Distribution Chi-Square Distribution for df= 10 and df = 20 Properties of the Chi-Square Distribution The chi-square distribution is not symmetric, unlike the normal and tdistributions.

  5. A symmetric distribution is convenient, we only have to find value for a z-score (or t-score) because we know the other side will be the simply the positive or negative.

  6. Confidence Interval for σ • Since the distribution is not symmetric we have to look up 2 separate scores for a confidence interval. • The area we find will be to the right of the score. • will be the larger value. • will be the smaller value.

  7. Confidence Interval for μ with known σ: • Find sample size for mean: • Confidence Interval for μ with unknown σ: • Confidence Interval for p: • Find sample size for proportion: • Confidence Interval for σ :

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