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Text: Sullivan, Michael I. Fundamentals of Statistics, 5th Edition.

Scenario: The mean weight gain during pregnancy is 30 pounds, with a standard deviation of 12.9 pounds. Weight gain during pregnancy is skewed right. An obstetrician obtains a random sample of 35 low-income patients and determines their mean weight gain during pregnancy was 36.2 pounds. Does this result suggest anything unusual?

I understand that the sampling distribution of $\bar X$ for a sample of size 35 is approximately normal with mean $\mu_{\bar X} = 30$ and standard error $\sigma_{\bar X} = \frac{12.9}{\sqrt{35}}$.

So to determine whether having a sample of size 35 with a sample mean of 36.2 is unusual I want to find $P(\bar X \geq 36.2)$. I can do this by standardizing and using a z-score table, so I understand that $P(\bar X \geq 36.2) \approx 0.0023$

Interpretation: The text says the following -- We can conclude one of two things based on this result:

  1. The mean weight gain for low-income patients is 30 pounds, and we happened to select women who, on average, gained more weight.
  2. The mean weight gain for low-income patients is more than 30 pounds.

We are inclined to accept the second explanation over the first since our sample was obtained randomly.

I am having trouble seeing why that is the interpretation we should make in this scenario. For a sample of size 35, it is unusual to see a sample mean of 36.2 or more. How does this translate to the mean weight gain being more than 30 pounds?

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2 Answers 2

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If the population we sampled from (low income) would indeed have an expected weight gain of 30 pounds, seeing 36.2 as an average would be very unlikely. Thus, it is more plausible that the population we sampled from has a higher average weight gain.

The idea is that maybe the unconditional population mean is 30, but that the weight gain is negatively correlated with income. Thus, the conditional expectation of weight gain conditional on being poor might be higher. This could explain the high average in our sample.

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  • $\begingroup$ So obtaining an unusual sample should lead to us thinking that maybe the population characteristics are misspecified? $\endgroup$
    – user235125
    Commented Mar 28, 2019 at 23:24
  • $\begingroup$ If the sample average is unusually large, then the population mean is probably higher than we initially thought. If the sample average is unusually small, then the population mean is probably lower than we initially thought. Is this the idea you are trying to give me? $\endgroup$
    – user235125
    Commented Mar 28, 2019 at 23:27
  • $\begingroup$ yes, I'm pretty sure that's what your book is trying to teach you. In the end of course once in a while such an unusual result will appear just due to chance, although the population mean was correctly specified. However, in this case the other explanation (that poor women are different) is not unlikely at all, so this is the most plausible explanation. $\endgroup$
    – user1587692
    Commented Mar 28, 2019 at 23:37
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The distribution being skewed right made me wonder if the idea of the question was to use something different from the normal distribution.

If we apply Cantelli's inequality to this case we get: $$ Pr(\bar{X} - \mu_X \geq \lambda) \leq \frac{\sigma_{\bar{X}}^2}{\sigma_{\bar{X}}^2 + \lambda^2} \ . $$ For $\lambda = 36.2-30 = 6.2$ and $\sigma_{\bar{X}} = \frac{12.9}{\sqrt{35}} = 2.18$, we get $Pr \leq 0.1101$.

Hence, in this case we could conclude that nothing significantly unusual happened.

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  • $\begingroup$ Don't you actually get that $Pr \leq 0.1101$??? $\endgroup$
    – beta1_equals_beta2
    Commented Mar 29, 2019 at 2:19
  • $\begingroup$ @beta1_equals_beta2 Yes, of course. I'll correct it. Thank you. $\endgroup$
    – Ertxiem - reinstate Monica
    Commented Mar 29, 2019 at 2:20
  • $\begingroup$ The fact that we're dealing with a mean of a large number of values makes a difference. The worst case over all distributions is way too wide for this situation (you don't come close to attaining it with averages). There are other bounds on sums that can be used in various situations - e.g. bounded, subGaussian, subexponential ... $\endgroup$
    – Glen_b
    Commented Mar 29, 2019 at 5:33
  • $\begingroup$ So you're using Cantelli's Inequality to show that for a sample of size 35, the probability of the sample mean exceeding 36.2 can be at most 0.1101. You say we can conclude that nothing significantly unusual happened, so why is it that the author says that the population mean is higher than 30? If obtaining a sample of size 35 that gives a sample mean of 36.2 is not significantly unusual, then why does the author conclude that the population mean is more than 30? $\endgroup$
    – user235125
    Commented Mar 29, 2019 at 22:11
  • $\begingroup$ The author said that there are two possibilities of interpretation and that they "are inclined" to follow the second one. Therefore, they should be doing an assumption regarding the distribution while knowing that it's skewed right. From what you have written, I can't tell for sure what was that assumption. (Although I could say that changing the assumption on the distribution could lead us to a different conclusion.) $\endgroup$
    – Ertxiem - reinstate Monica
    Commented Mar 30, 2019 at 2:02

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