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Information about the data

For this study, I would like to statistically test how the total length of sparrows affects their survival. Consider here one of the characteristics of the sparrows recorded was total length, which is assumed to be normally distributed.

If this is possible, I would be incredibly grateful if anyone could please help me understand if my assumptions are correct. Many thanks in advance for your help.

Information regarding the data:

  1. In a group of 21 surviving sparrows, the variance of the total length was 11.05 mm2.

  2. In a group of 28 sparrows that subsequently died, the variance of the total length was 15.07 mm2.

  3. The µ of the total length of sparrows that survived was 157.4 mm2

  4. The mean µ of the total length of sparrows that subsequently died was 158.4 mm2

Problems to solve:

I am unsure if I have understood the concepts correctly underpinning how to: -

  1. Read quantile tables properly for Chi-square and F distributions
  2. Correct inferences from hypothesis testing

Question

Test the hypothesis the variance of the total length is the same in the two groups, sparrows that did not survive and surviving sparrows.

Quantile Table

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Definitions:

  • Variance Length of Survivors (VLS)
  • Variance Length of Non-Survivors (VLNS)

Hypothesis testing:

  1. H0: VLS = VLNS
  2. H1: VLS ≠ VLNS (two-sided test)
  3. Significance level, p = 0.05

F statistics – testing differences between the mean of two populations (i.e. survivors and non-survivors)

Chi-Squared and F distributions inferred from the quantile table (above):

F distribution for VLS and VLNS:-

F (21, 28) (0.025) = 0.4300

∴ Ratio of the variances for VLNS/VLS = 15.07/11.05 = 1.36

The null hypothesis has been accepted showing the variance of the total lengths of VLS and VLNS are not same when compared to the non-significant F-statistic distribution 0.4943. The ratio of VLNS/VLS is far smaller than the combined F-distribution which tested differences between the mean of both populations. The null hypothesis has been accepted

Alice Hobbs
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    The units of mean length are mm not mm$^2$. This is a repost of a previous question http://stats.stackexchange.com/questions/251926/help-quantile-tables-and-confidence-intervals, and you did get advice to slim that down. But it's better now for you to delete that post, not to abandon it. – Nick Cox Dec 22 '16 at 12:26
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    Yes that's true. Users who visited the original question proclaimed that it was far too broad and daunting to read; therefore, I slimmed the original post down into three separate questions on Cross Validated. Overall, I felt this was good advice. Most users who assessed the original question provided advice on question 1, but not 2 or 3. I will definitely ask concise questions from now on. – Alice Hobbs Dec 22 '16 at 12:55
  • To compare variances you divide one by the other and then use F. I do not see where you did that. Which one you divide by which and which value of F you use depends on whether you have a directional hypothesis. Does that help at all? – mdewey Dec 22 '16 at 13:42
  • Fine, but the advice to delete the original thread still stands. – Nick Cox Dec 22 '16 at 13:51
  • No, you divide the variances. Whatever you are trying to achieve by looking up values of $\chi^2$ is irrelevant here. – mdewey Dec 22 '16 at 14:55
  • You fixed the units of mean length, but falsified those of its variance, which really will be mm$^2$ for measurements in mm. – Nick Cox Dec 22 '16 at 15:16
  • Thank you mdewey, this makes a lot more sense. I was getting far too lost after reading different sources. Additionally, thank you for all your help Nick Cox – Alice Hobbs Dec 22 '16 at 16:38

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