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Quiz 7 afternoon solutions

# Quiz 7 afternoon solutions - Level Count Std Dev 0 5...

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Quiz 7 afternoon solutions STA 6166 Fall 2005 Of interest is whether the mean plasma polyamine differs between two age groups. To test the hypothesis, five newborns and five children age 1 yo were randomly selected for measurement. The results are shown below. Test the hypothesis of interest using a Type I error rate of 0.03. Be sure to state the hypotheses clearly and identify any symbols you use in the hypotheses. Means for Oneway Anova Level Number Mean Std Error Lower 95% Upper 95% 0 5 12.5760 0.73608 10.879 14.273 1 5 10.0320 0.73608 8.335 11.729 Std Error uses a pooled estimate of error variance t Test Assuming equal variances Difference t Test DF Prob > |t| Estimate 2.54400 2.444 8 0.0403 Std Error 1.04098 Lower 95% 0.14350 Upper 95% 4.94450 UnEqual Variances Difference t Test DF Prob > |t| Estimate 2.5440 2.444 5.01519 0.0582 Std Error 1.0410 Lower 95% -0.1295 Upper 95% 5.2175 Tests that the Variances are Equal
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Unformatted text preview: Level Count Std Dev 0 5 0.786848 1 5 2.190678 Test F Ratio DFNum DFDen Prob > F O'Brien[.5] 1.6494 1 8 0.0235 Brown-Forsythe 1.5571 1 8 0.0247 Levene 1.5157 1 8 0.0253 Bartlett 3.2151 1 . 0.0730 Warning: Small sample sizes. Use Caution. H : μ newborns = μ 1YO where μ is the population mean plasma polyamine level H A : μ newborns ≠ μ 1YO Test statistic value: t = 2.444. Note: The tests of variance indicate that there is some evidence that the two populations have different variances (Levene’s test, F = 1.5157, p-value = 0.0253), hence we will use the unequal variance test. The p-value of the t-test is 0.0582. Conclusion: since the p-value > alpha, we fail to reject the null hypothesis and conclude that with significance level of 0.03, there is insufficient evidence to indicate that mean plasma polyamine levels differ between newborn and 1 year old children....
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