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Stat5 the t test
Stat5 the t test
Stat5 the t test
Stat5 the t test
Stat5 the t test
Stat5 the t test
Stat5 the t test
Stat5 the t test
Stat5 the t test
Stat5 the t test
Stat5 the t test
Stat5 the t test
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Stat5 the t test

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  • 1. The t test 1
  • 2. Applications To compare the mean of a sample with population mean. To compare the mean of one sample with the mean of another independent sample. To compare between the values (readings) of one sample but in 2 occasions. 2
  • 3. 1.Sample mean and population mean The general steps of testing hypothesis must be followed. Ho: Sample mean=Population mean. Degrees of freedom = n - 1 − X −µ t = SE 3
  • 4. ExampleThe following data represents hemoglobin values in gm/dl for 10 patients: 10.5 9 6.5 8 11 7 7.5 8.5 9.5 12Is the mean value for patients significantly differ from the mean value of general population(12 gm/dl) . Evaluate the role of chance. 4
  • 5. Solution Mention all steps of testing hypothesis. 8.95 − 12 t= = −5.352 1.80201 10Then compare with tabulated value, for 9 df, and 5% level of significance. It is = 2.262The calculated value>tabulated value.Reject Ho and conclude that there is a statistically significant difference between the mean of sample and population mean, and this difference is unlikely due to chance. 5
  • 6. 6
  • 7. 2.Two independent samplesThe following data represents weight in Kg for 10 males and 12 females.Males: 80 75 95 55 60 70 75 72 80 65Females: 60 70 50 85 45 60 80 65 70 62 77 82 7
  • 8. 2.Two independent samples, cont. Is there a statistically significant difference between the mean weight of males and females. Let alpha = 0.01 To solve it follow the steps and use this equation. − − X1 − X 2 t= 2 2 (n1 − 1) S1 + (n 2 − 1) S 2 1 1 ( + ) n1 + n 2 − 2 n1 n 2 8
  • 9. Results Mean1=72.7 Mean2=67.17 Variance1=128.46 Variance2=157.787 Df = n1+n2-2=20 t = 1.074 The tabulated t, 2 sides, for alpha 0.01 is 2.845 Then accept Ho and conclude that there is no significant difference between the 2 means. This difference may be due to chance. P>0.01 9
  • 10. 3.One sample in two occasions Mention steps of testing hypothesis. The df here = n – 1. − d t = sd n (∑ d ) 2 ∑ d2 − n sd = n −1 10
  • 11. Example: Blood pressure of 8patients, before & after treatmentBP before BP after d d2180 140 40 1600200 145 55 3025230 150 80 6400240 155 85 7225170 120 50 2500190 130 60 3600200 140 60 3600165 130 35 1225Mean d=465/8=58.125 ∑d=465 ∑d2=2917511
  • 12. Results and conclusion t=9.387 Tabulated t (df7), with level of significance 0.05, two tails, = 2.36 We reject Ho and conclude that there is significant difference between BP readings before and after treatment. P<0.05. 12

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