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Online Test Scenario:

A random sample of $500$ is taken from a large population, which is known to be equally divided between males and females, and values for the quantity of interest are recorded. On examination of the results, it is found that the sample taken includes $200$ females for which the mean is $10.2$ with standard deviation of $0.6$ and $300$ males for which the mean is $14.8$ and standard deviation $2.4$.

Question:

Which one of the following statements in NOT correct?

  1. Taking the mean value of the data for the $500$ sampled would over-estimate the true population mean

  2. The most accurate estimate of the mean of the quantity of interest would have been obtained by sampling equal numbers of males and females

  3. Given the sample that was taken, the best estimate of the population mean is the average of the means of the males and females i.e. $12.5$

  4. The estimate from the sample of the mean for females is likely to be more accurate than that for males

My Attempt:

My guess is that option $4)$ is incorrect because your told in the beginning the population is known to be divided equally?

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  • $\begingroup$ I agree with you that option 4 is incorrect, but because it should say "The estimate .. for females is less likely to be more accurate than that of males" because of the smaller sample size for females. $\endgroup$
    – RayDansh
    Jul 4, 2019 at 16:53
  • $\begingroup$ The distribution for females is much more homogeneous, so I think it’s arguably likely to be more accurate, despite the smaller sample. I think #3 is false, because averaging over the sample over-represents the males and their higher average quantity of interest. (That said, this is a pretty unusual random sample! The chance of a 200/300 split if the population has equally many males and females is extremely unlikely.) $\endgroup$
    – Steve Kass
    Jul 4, 2019 at 16:58
  • $\begingroup$ @SteveKass #3 says averaging the means, not over the sample. They did $\frac{10.2+14.8}2=12.5$ $\endgroup$
    – RayDansh
    Jul 4, 2019 at 16:59
  • $\begingroup$ @RayDansh so would you say $#3$ is correct? $\endgroup$
    – UniStuffz
    Jul 4, 2019 at 17:10
  • $\begingroup$ Yes, it makes sense, because the population is said to contain an equal proportion of males and females. $\endgroup$
    – RayDansh
    Jul 4, 2019 at 17:11

2 Answers 2

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Statement #4 is likely true.

If the sample is indeed random, then the $200$ females constitute a random sample of the females in the population. Similarly, the $300$ males constitute a random sample of the males in the population.

A rough estimate of the how far a sample mean is from a population mean is the sample’s standard error. Here, the standard error for females is ${0.6\over\sqrt{200}}\approx0.04$ and for males is ${2.4\over\sqrt{300}}\approx 0.14$. It’s reasonable to assume then that the expected inaccuracy of the female sample mean (as an estimate of the female population mean) is about ${0.04\over0.14}\approx 0.29$ of the expected inaccuracy of the male sample mean as an estimate of the male population mean.

Statement #3 is likely true.

The best estimate we have from the sample is that the males population average is $14.8$ and the female population average is $10.2$. Based on these best estimates, and using the given that the population is half female and half male, one would expect the population average across males and females to be $10.2+14.8\over2$. (This is different from the sample average, which would be a weighted average of the two means.)

Statement #1 is likely true. The particular sample happens to have more than the expected number of males, and males appear to have a higher value of the quantity of interest, so the sample mean is likely an over-estimate.

Statement #2 is likely false. Assuming, as appears likely from the sample, that males’ values of the quantity of interest are more spread out, a better estimate would be found by sampling more males. Consider this extreme scenario: A quantity of interest is constant and equal to $1$ for females in a population, but it is spread out between $0$ and $2$ for males, with an unknown mean. To estimate the sample mean, it would not be useful to sample more females than needed to suspect that the female average was constant.

[While not part of the question, it’s worth noting that a random sample with the given male-female split is very unusual. Within the distribution of all random samples of size 500 from a population with equally many males as females, the $z$-score of a sample with $200$ females (instead of the expected $250$) is about $4.47$. Only $0.08$% of samples, or one in about $1250$, would have such an unbalanced distribution of males to females, if the population were indeed equally divided. This would make me question the premises of the question!]

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  • $\begingroup$ Why is statement #3 false though? It averages the means of the samples, not over the entire sample. $\endgroup$
    – RayDansh
    Jul 4, 2019 at 17:47
  • $\begingroup$ I'm confused... didn't you say earlier that the standard errors meant that #4 was true? $\endgroup$
    – RayDansh
    Jul 4, 2019 at 17:52
  • $\begingroup$ Sorry, I messed up the statement numbers in my first version of the answer. Fixed now, I hope! $\endgroup$
    – Steve Kass
    Jul 4, 2019 at 17:54
  • $\begingroup$ Did you not say above that #2 is false? $\endgroup$
    – UniStuffz
    Jul 4, 2019 at 17:54
  • $\begingroup$ I think my answer is clear and correct now. $\endgroup$
    – Steve Kass
    Jul 4, 2019 at 18:01
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Let $M$ be the mean of the males. Let $F$ be the mean of the females.

$1$) is correct because taking the mean value of the sample would mean the $M$ is weighted more than $F$ ($60$%-$40$%).

$2$) is kinda correct, though really as long as you take the average of $M$ and $F$, not of the entire sample group, you don't need to get equal samples of males and females.

$3$) is correct because it reiterates the idea of averaging $M$ and $F$, and not taking the mean of the entire sample itself.

$4$) [Credit to SteveKass] is correct because the larger standard deviation associated with the sample of males offsets the larger sample size, so there is less accuracy in $M$.

So in conclusion, $2$ is iffy, because since we're averaging $M$ and $F$, sample size isn't as important.

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  • $\begingroup$ Could you explain why 2 is iffy in comparison to the rest? i haven't covered "mean of the quantity of interest" so i'm googling it now. $\endgroup$
    – UniStuffz
    Jul 4, 2019 at 17:28
  • $\begingroup$ "Mean of the quantity of interest" is just saying population mean; "quantity of interest" was the fancy word the question used for the thing measured. $\endgroup$
    – RayDansh
    Jul 4, 2019 at 17:29
  • $\begingroup$ I would argue that I brought up more than a good point... A $p$% confidence interval for the males here will be much wider than for the females, due to the differences in sample sizes and standard deviations, both of which matter directly. While the population distribution for the quantity of interest is unknown, for most real-life distributions the estimate for males will be less accurate because of the 4-times greater standard deviation. Put another way, the standard error for males is about $0.13$ and for females is about $0.04$. This is roughly an estimate of the accuracy of prediction. $\endgroup$
    – Steve Kass
    Jul 4, 2019 at 17:29
  • $\begingroup$ @RayDansh: Think about a real example. Suppose a quantity of interest is very homogeneous for females (say it is between $0.99$ and $1.01$ for $99$% of females, but not for males (say it is between $0.8$ and $1.2$ for the middle $90$% of males). A sample average for females regardless of sample size will be within $0.02$ of the true female population mean. Many sample averages for males, even for large samples, will be further than $0.02$ away from the male population mean. $\endgroup$
    – Steve Kass
    Jul 4, 2019 at 17:36
  • $\begingroup$ @SteveKass So I see what you're saying, would you agree then that statement 2 is incorrect? $\endgroup$
    – RayDansh
    Jul 4, 2019 at 17:37

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