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T-test and ANOVA Exercises

Assignment

Nursing

Pages 6 (1506 words)

PART 1 1. The number of employed women were 436 and the number unemployed were 524. 2. The total sample size is 436+524=960 3. The mean CES-D score in unemployed women was( 20.8965 +12.46) and that for employed was (15.8239 +10.13) 4. Leven’s test is an inferential statistic used to assess the equality of variances (squared deviations of the difference of raw scores and mean score for a variable) which is calculated for two or more groups…

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## Introduction

Thus, the null hypothesis of equal variances is rejected and it is concluded that there is a difference between the variances in the population.It is called heteroscadasticity.Some of the procedures typically assuming homoscedasticity are the anova , t-tests.Levene's test is often used before a comparison of means. When Levene's test shows significance, one should switch to generalized tests, free from homoscedasticity assumptions. Since the p value is less than 0.05 ( 0.000) in this case hence homoscadasticity or equal variances cannot be assumed 5. The computed t statistic is 6.95 which is more than the assumed or critical t statistic of 6.82 for assuming equal variance. The degree of freedom is hence retained as 957.51 and the p values as stated is <0.05 which is 0.000 6. ...

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