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Hypothesis testing applied to means. Characteristics of the Sampling Distribution of the mean. The sampling distribution of means will have the same mean as the population : 0 = :. Characteristics of the Sampling Distribution of the mean.

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## Hypothesis testing applied to means

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**Characteristics of the Sampling Distribution of the mean**The sampling distribution of means will have the same mean as the population :0 = :**Characteristics of the Sampling Distribution of the mean**The sampling distribution of means has a smaller variance. s 2 s 2 s = s = x x N N s = standard deviation of the mean = standard error x This is because the means of samples are less likely to be extreme compared to individual scores.**Characteristics of the Sampling Distribution of the mean**The shape of the sampling distribution approximates a normal curve if either: the population of individual cases is normally distributed the sample size being considered is 30 or more**X**- m 115 - 100 z = = = 1 . 0 s 15 Probability is approximately .16**Example**Jill now has to choose 25 intelligent people. 0 = 106 IQ test: := 100 F = 15. Hypotheses: H1: :1:2 H0: :1 = :2 = 100**Sampling Distribution**:1 = :2 = 100 s 15 15 s = = = = 3 x N 25 5 X - m X - m 106 - 100 6 z = z = = = = 2 . 00 s s 3 3 x x Look up area in the tail of the curve = .0228 (one tailed) or .0456 (two tailed) If signficance level (") = .05, reject the Null Hypothesis**t Distribution**• What happens if we do not have the population standard deviation? • We can use the sample standard deviation and an estimate. • Problem • Cannot use z and the normal distribution to estimate probability. • Sample variance tends to underestimate the population variance • Have to use a slightly different distribution - Student=s t Distribution X - m z = s x X - m t = s x**Degrees of freedom**df = n - 1 Different t distribution for each degree of freedom**Sample: N = 56 0 = 104.3 S = 12.58**Norms: : = 100 Hypotheses: H1: :1:2 H0: :1 = :2 = 100**X**- m X - m = t = s s x N 104 . 13 - 100 4 . 125 = = 12 . 58 1 . 682 56 t = 2 . 45 df = 56 - 1 = 55 tcrit or t.025 = " 2.009 tobs > tcrit Reject H0**Percentage Points of the t Distribution**See Howell page 247**Matched-Sample t-test**Use related samples In SPSS called the >Paired-Samples t-test' Data: Howell p 193**m**= m - m = 0 H0: 1 2 D D - m D - 0 D - 0 t = = = s s s D D D N**D**- 0 - 4 . 29 - 0 - 4 . 29 = = = - 1 . 49 s 16 . 04 2 . 88 D 31 N df = N - 1 (N is number of pairs of observations) df = 31 - 1 = 30 t.025(30) = " 2.042 tobs < tcrit Fail to reject H0**Two Independent Samples**Distribution of Differences between Means**2**2 s s 1 2 Variance of the Distributions of the Means & N N 1 2 s s Standard Error of the Distributions of the Means 1 2 & N N 1 2 2 2 s s Variance of the Distribution of Mean Difference 2 1 2 s + X - X = N N 1 2 1 2 2 2 s s Standard Error of the Distribution of Mean Difference 1 2 s = + X - X N N 1 2 1 2 m - m = 0 Mean of the Distribution of Mean Differences 1 2**t - test for two independent samples**X - m t = s x ( X - X ) - ( m - m ) 1 2 1 2 t = s X - X 1 2 ( X - X ) - ( m - m ) 1 2 1 2 = 2 2 s s 1 2 + N N 1 2 ( X - X ) 1 2 = 2 2 s s 1 2 + N N 1 2**Pooled Variance estimate**2 2 ( N - 1 ) s + ( N - 1 ) s 2 1 1 2 2 s = p N + N - 2 1 2 ( X - X ) ( X - X ) 1 2 = 1 2 t = 2 2 s s 2 2 s s p p + 1 2 + N N N N 1 2 1 2 ( X - X ) 1 2 t = æ ö 1 1 2 s + ç ÷ p N N è ø 1 2 df = (N1 - 1) + (N2 - 1) = N1 + N2 -2 Degrees of Freedom:

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