For the variable PHE we reject the hypothesis of equal variances therefore we

For the variable phe we reject the hypothesis of

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For the variable PHE, we reject the hypothesis of equal variances; therefore, we look to the t test results in the bottom row, which are based on formulas 4.14 and 4.15. The null hypothesis of equal means is also rejected, now with higher significance since p = 0.002. Note that the means of the two groups are more than three times the standard error apart. ± Figure 4.10. a) Window of STATISTICA Power Analysis module used for the specifications of Example 4.10; b) Results window for the previous specifications. Example 4.10 Q: Compute the power for the ASP variable (aspartame content) of the previous Example 4.9, for a one-sided test at 5% level, assuming that as an alternative hypothesis white wines have more aspartame content than red wines. Determine what is the minimum distance between the population means that guarantees a power above 90% under the same conditions as the studied samples. A: The one-sided test for this RS situation (see section 4.2) is formalised as: H 0 : µ 1 µ 2 ; H 1 : µ 1 > µ 2 . (White wines have more aspartame than red wines.) The observed level of significance is half of the value shown in Table 4.6, i.e., p = 0.011; therefore, the null hypothesis is rejected at the 5% level. When the data analyst investigated the ASP variable, he wanted to draw conclusions with protection against a Type II Error, i.e., he wanted a low probability of wrongly not detecting the alternative hypothesis when true. Figure 4.10a shows the
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4.4 Inference on Two Populations 137 STATISTICA specification window needed for the power computation. Note the specification of the one-sided hypothesis. Figure 4.10b shows that the power is very high when the alternative hypothesis is formalised with population means having the same values as the sample means; i.e., in this case the probability of erroneously deciding H 0 is negligible. Note the computed value of the standardised effect ( µ 1 µ 2 )/s = 2.27, which is very large (see section 4.2). Figure 4.11 shows the power curve depending on the standardised effect, from where we see that in order to have at least 90% power we need E s = 0.75, i.e., we are guaranteed to detect aspartame differences of about 2 mg/l apart (precisely, 0.75 × 2.64 = 1.98). ± 0.0 0.5 1.0 1.5 2.0 2.5 .3 .4 .5 .6 .7 .8 .9 1.0 Power Standardized Effect (Es) Power vs. Es (N1 = 30, N2 = 37, Alpha = 0.05) Figure 4.11. Power curve, obtained with STATISTICA, for the wine data Example 4.10. Commands 4.3. SPSS, STATISTICA, MATLAB and R commands used to perform the two independent samples t test. SPSS Analyze; Compare Means; Independent Samples T Test STATISTICA Statistics; Basic Statistics and Tables; t-test, independent, by groups MATLAB [h,sig,ci] = ttest2(x,y,alpha,tail] R t.test(formula, var.equal = FALSE) The MATLAB function ttest2 works in the same way as the function ttest described in 4.3.1, with x and y representing two independent sample vectors. The function ttest2 assumes that the variances of the samples are equal.
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138 4 Parametric Tests of Hypotheses The R function t.test , already mentioned in Commands 4.1, can also be used to perform the two-sample t test. This function has several arguments the most important of which are mentioned above. Let us illustrate its use with Example 4.9.
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