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Part 1: Chapters 7 to 9. 95% within 2 standard deviations 68% within 1 standard deviations 99.7% within 3 standard deviations.

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Presentation on theme: "Part 1: Chapters 7 to 9. 95% within 2 standard deviations 68% within 1 standard deviations 99.7% within 3 standard deviations."— Presentation transcript:

1 Part 1: Chapters 7 to 9

2 95% within 2 standard deviations 68% within 1 standard deviations 99.7% within 3 standard deviations

3 Mean is zero Standard deviation is 1 0 The scale (Z) is the Number of Standard Deviations From the mean

4 We estimate the population parameters with a point estimate that is a statistic from the sample. Because different samples will produce different point estimates, we provide a range of values called a Confidence Interval. While the point estimate is a statistic from the sample, we add and subtract a margin of error (E) to create this range (or interval) of values.

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8 Null Hypothesis is the status quo Alternate Hypothesis is an Inequality or ≠ Which determines whether it’s a left, right or 2-tailed test.

9 Means: If you know σ, the population standard deviation, you can use the standard normal distribution (z). Otherwise, use student’s t-distribution With degree of freedom (n-1). Proportions : use the standard normal distribution. (Assumption: np>5 and nq>5)

10 Calculate the sample test statistic and find the p-value. If σ is known, use standard normal distribution On calculator: normalcdf(low,high) = p-value If z 0 normalcdf(z,5)

11 Since σ is known We calculate z And use the Standard Normal Distribution. Indicates a right-tailed test

12 Since σ is known We calculate z And use the Standard Normal Distribution. 2.56 – 2.0 = 0.56 Z = 0.56 / 0.221 = 2.534

13 2.56 – 2.0 0.56 Z = 2.534 normalcdf(low, high) normalcdf(2.534, 5) =0.0056 =p-value

14 If the p-value < α (significance: usually.01 or.05), then we can reject the null hypothesis. Otherwise, we “fail to reject”. p-value =0.0056 Use α=1%=0.01 0.0056 < 0.01 so, we reject the null hypothesis

15 If we reject the null hypothesis, we accept the alternate. Your interpretation should use language from the problem. The evidence indicates that the mean concentration of lead exceeds the reported mean of 2.0 ug/l.

16 Sample standard deviation

17 Stat 4: ClrList 2nd 1 ClrList L1 First clear the list you plan to use (L1)

18 Stat Edit Edit the list Type each number and hit enter 9.1 9.5 10.2 9.8 When all numbers are entered:

19 Stat Edit Edit the list Type each number and hit enter 9.1 9.5 10.2 9.8 When all numbers are entered: Stat Calc

20 When all numbers are entered: Stat Calc 1: 1-Var Stats 10.49166667 125.9 1341.19 1.358112549 1.30029377

21 Indicates a right-tailed test

22 Calculate the sample test statistic and find the p-value. If σ is unknown, use student’s t-distribution t =1.248

23 Calculate the sample test statistic and find the p-value. Degrees of Freedom = n-1 12 -1 = 11 Use the student’s t-distribution t =1.248 P-value will be > 0.10

24 You get a range of p-values (from t-distribution). If the larger p-value ≤ α we reject. Otherwise, we “fail to reject”. P-value will be > 0.10

25 For a Right-Tailed Test, 99% confidence is equivalent to 1% significance d.f = n-1 12 -1 = 11

26 Interval: ( 0.49 – 1.067, 0.49 + 1.067) ( -0.577, 1.557)


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