Stats II week 5

Stats II week 5 - Standard error of mean = s √ n Standard...

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Sarah Anderson Stats week 5 The following Exercises: (10 points) Exercises 16 (done) 18 (page 348) done Exercise 22 (page 352) done Exercise 28 (page 355) 16.)Compute t = (xbar - 100) / s /√6 s being the sample standard deviation and xbar the sample mean. If t > critical t with 5 degrees of freedom, reject H0; critical t (0.05,5) = 2.571 b) Number of cases 6 To find the mean, add all of the observations and divide by 6 Mean 111.6667 Squared deviations (118-(111.6667))^2 = (6.3333)^2 = 40.1111 (105-(111.6667))^2 = (-6.6667)^2 = 44.4444 (112-(111.6667))^2 = (0.3333)^2 = 0.1111 (119-(111.6667))^2 = (7.3333)^2 = 53.7778 (105-(111.6667))^2 = (-6.6667)^2 = 44.4444 (111-(111.6667))^2 = (-0.6667)^2 = 0.4444 Add the squared deviations and divide by 5 Variance (using n-1) = 183.3333 / 5 Variance 36.6667 Standard deviation (using n-1) = sqrt(variance) = 6.0553 H0: μ = 100 HA: μ ≠ 100 Sample mean = 111.6667 Standard deviation = 6.0553
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Unformatted text preview: Standard error of mean = s / √ n Standard error of mean = 6.0553 / √ 6 SE = 6.0553/2.4495 Standard error of mean 2.4721 t = (xbar- μ ) / SE t = (111.6667-100) / 2.4721 t = 4.7194 c) test statistic t=4.7154 exceeds critical t of 2.571 Therefore, reject H0. From t-probability table, the p-value lies between 0.002 and 0.01, so less than 0.05 (significance level). This leads to the rejection of the null hypothesis. 18.) Data: 2159,2160,2167,2170,2171,2179,2180,2181,… n= 9 Σx= 19552 , Σx²= 42476338 , xbar= 2172.444444444444 , σn-1=9.382312, z= ( x - µ ) / σ: x= 2172.4-->z= 1.32; P( X> 2172.4 )= P( z > 1.32)=0.093147 p-value is 0.093147 22.) n = 120 p = 0.3 q = 0.7 Sigma(p) = sqrt(0.3*0.7/120) = .4183300133e-1 Decision Rule: If abs(0. - 0.4)/.4183300133e-1 > 1.96, Rejct H0. If 9.56 > 1.96, we Reject H0. Conclusion: π is different at .40 28.)...
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Stats II week 5 - Standard error of mean = s √ n Standard...

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