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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.

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. Calculate 95 % confidence intervals

If this is possible, I would be incredibly grateful if anyone could please check the answers to the questions stated below by providing advice. Many thanks in advance for your help.

Quantile Table

enter image description here

Question 1

I would like to establish if the variance of the length of survivors is compatible with the calculated confidence interval for non-survivors?

Sample unit = total length of sparrows

  • Calculate the 95 % CI for the variance of 15.07 mm2 for total length of 28 sparrows that did not survive

0.975 and 0.025 Chi-Square (χ2) distribution for non-survivors (n=28) inferred from the quantile table:

  1. χ2(28)(0.025) = 15.31
  2. χ2(28)(0.975) = 44.46

Calculate Confidence Intervals:

  • σ² (non-survivors) = 15.07 mm2

Confidence interval = (n-1) * σ² / χ2

  1. 0.975 CI = (28-1)*(15.07)/15.31 = 26.58
  2. 0.05 CI = (28-1)*(15.07)/44.46 = 9.15

The variance for survivors (n=21) is 11.05 mm2

The variance of the length of survivors (i.e 11.05 mm2) is compatible with the confidence interval range for non-survivors (0.975 = 26.58, 0.025 = 9.15) Therefore, from these calculations we are 95 % confident that the true population variance for survivors (n=21) does fall within the confidence intervals for non-survivors (n=28).

Alice Hobbs
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  • Question 1: I think you meant $(n-1)\sigma^2/\chi^2$ not $(n-1)\mu/\chi^2$. Also, check your degrees of freedom. –  Dec 16 '16 at 09:55
  • Thank you Blue-Footed Booby, an oversight. Much appreciated :) Otherwise is the methodology accurate? – Alice Hobbs Dec 16 '16 at 11:21
  • Comparing the variance of length strikes me as unduly awkward way to go about this. Presumably these are some teacher's questions, but just plotting the data would be immensely more useful. Then you could check whether lengths are normal (instead of assuming that). – Nick Cox Dec 16 '16 at 11:31
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    The 0.975 confidence limit (not interval) would usually be paired with the 0.025 limit (not the 0.05 limit). That pair of limits would then be the 0.95 or 95% confidence interval. – Nick Cox Dec 16 '16 at 11:32
  • I started editing this but then got bored. I'll leave you to find instances where variance of length is stated to be in mm (mostly you got that right). – Nick Cox Dec 16 '16 at 11:35
  • Thank you Nick Cox, I have amended the error for CI (0.025). Unfortunately we were not given the data, only the information provided. Your advice is much appreciated. – Alice Hobbs Dec 16 '16 at 11:57
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    I doubt that it is intentional but this is an echo of mediocre statistical practice from say the 1950s.... – Nick Cox Dec 16 '16 at 12:27
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    I think this post is too broad. Break up your questions into separate posts. Then you will have a chance to get some answers instead of just comments. – Michael R. Chernick Dec 16 '16 at 14:25
  • It would be more likely for you to get useful answers if you split this up as @MichaelChernick suggested (a) someone may be able to help with only one part (b) it is rather forbidding at the moment and volunteer helpers are going to be put off reading through so much. – mdewey Dec 21 '16 at 13:31
  • How did you arrive at your statement that $\sigma^2 = 3.882$? I think there is an error there. – mdewey Dec 22 '16 at 17:26
  • I can now see my mistake, I was using the standard deviation (σ = 3.882) rather than the variance for non-survivors – Alice Hobbs Dec 22 '16 at 17:55

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