# 086 95 ci 996 431 2086 996 899 10 189 k310

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Unformatted text preview: 6 ± 4.31t ∗ We don’t have software handy, so we calculate df = min(n1 − 1, n2 − 1) = min(20, 22) = 20 K310 Statistical Techniques Inference for Distributions 20 / 26 Two-sample t conﬁdence interval example What is the 95% conﬁdence interval for the mean improvement by the students in the previous example? (¯1 − x2 ) ± t ∗ x ¯ 2 s1 s2 +2 n1 n2 = (51.48 − 41.52) ± t ∗ 11.011 17.152 + 21 23 = 9.96 ± 4.31t ∗ We don’t have software handy, so we calculate df = min(n1 − 1, n2 − 1) = min(20, 22) = 20 Look up the value for t ∗ for C = 95%: t ∗ = 2.086. 95% C.I. = 9.96 ± (4.31 × 2.086) = 9.96 ± 8.99 = (1.0, 18.9) K310 Statistical Techniques Inference for Distributions 20 / 26 Software approximation for df Software for calculating two-sample t statistics often gives a decimal value for df How does the software get this value? 2 s2 s1 +2 n1 n2 df = 1 n1 − 1 2 s1 n1 2 + 2 1 n2 − 1 2 s2 n2 2 This is called the Welch approximation We can still get critical t ∗ values from software using a decimal df value. We just can’t easily use the table. Provides a better approximation than df = min(n1 − 1, n2 − 1) (which is more conservative) Messy formula. Look it up if you need it! K310 Statistical Techniques Inference for Distributions 21 / 26 Pooled two-sample t procedure So far, our discussion has been for populations with unequal variances Things can simplify if we have equal variances Use the pooled variance 2 sp = 2 2 (n1 − 1)s1 + (n2 − 1)s2 n1 + n2 − 2 Then use the pooled variance to calculate the standard error s(¯1 −x2 ) = x¯ 2 2 sp sp + = sp n1 n2 1 1 + n1 n2 Often the assumption of equal variance isn’t a bad one, so we’ll often use this method K310 Statistical Techniques Inference for Distributions 22 / 26 Pooled two-sample t conﬁdence interval Suppose we have random sample of size n1 from normal population #1 random sample of size n2 from normal population #2 and we don’t know the means µ1 , µ2 , but σ1 = σ2 . The conﬁdence interval for µ1 − µ2 is (¯1 − x2 ) ± t ∗ sp x ¯ 1 1 + n1 n2 where t ∗ is from the t (df1 + df2 ) = t (n1 + n2 − 2) distribution with a C level of conﬁdence K310 Statistical Techniques Inference for Distributions 23 / 26 Pooled two-sample t hypothesis test To test the hypothesis H0 : µ1 = µ2 , use the test statistic t= x1 − x2 ¯ ¯ 1 1 sp + n1 n2 coming from a t (n1 + n2 − 2) distribution Alternative P -value Ha : µ1 − µ2 > 0 P (T ≥ t ) Ha : µ1 − µ2 < 0 P (T ≤ t ) Ha : µ1 − µ2 = 0 Picture 2P (T ≥ |t |) t t |t | K310 Statistical Techniques Inference for Distributions 24 / 26 Pooled two-sample t test example Q: Does increasing the amount of calcium intake reduce blood pressure? Calcium group Begin End Decrease 107 100 7 110 114 −4 123 105 18 129 112 17 112 115 −3 111 116 −5 107 106 1 112 102 10 136 125 11 102 104 −2 Placebo group Begin End Decrease 123 124 −1 109 97 12 112 113 −1 102 105 −3 98 95 3 114 119 −5 119 114 5 114 112 2 110 121 −11 117 118 −1 130 133 −3 Summary statistics: Group 1 2 K310 Statistical Techniques Treatment Calcium Placebo n 10 11 x ¯ 5.000 −0.273 s 8.743 5.901 Inference for Distributions 25 / 26 Pooled two-sample t test example Group 1 2 Treatment Calcium Placebo n 10 11 x ¯ 5.000 −0.273 1 Hypotheses: 2 Test statistic: 3 P -value; look at α = 0.10 and = 0.05: 4 s 8.743 5.901 Conclusion: K310 Statistical Techniques Inference for Distributions 26 / 26...
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## This note was uploaded on 04/16/2013 for the course MATH-M 310 taught by Professor Palanivelmanoharan during the Spring '13 term at Indiana.

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