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> I guess that is fair, especially with inflation nowadays. The accumulated inflation in the US since June 2019 is c. 17%.[1] This means US$ 55 in June 2019 is

> I guess that is fair, especially with inflation nowadays. The accumulated inflation in the US since June 2019 is c. 17%.[1] This means US$ 55 in June 2019 is "worth" c. US$ 64.35 today. So it seems that you get a 4GB Raspi 5 for c. 7.25% less today. The "US Real Average Hourly Earnings" have increase by ca. 22% in the same period, from US$ 27.75 to US$ 33.82.[2] So an average person needs to work c. 12% less to buy a 4GB Raspi 5 today. However, I think the issue is more complicated: There is inherent deflation in electronics, which is included in the inflation rate. you can observe it when looking at the current price of a 4GB Raspi 4 at Amazone, which is c. US$ 67. So if the introductory price for a Rapsi 5 is really going to be US$ 60, you get something better for more than 10% less now. [1] https://www.usinflationcalculator.com/inflation/current-inflation-rates/ https://www.usinflationcalculator.com/inflation/current-infl... [2] https://www.bls.gov/news.release/archives/realer_07112019.htm https://www.bls.gov/news.release/archives/realer_07112019.ht... and https://www.bls.gov/news.release/empsit.t19.htm https://www.bls.gov/news.release/empsit.t19.htm

wredue 3y ago
Please stop using “average” for wage comparisons. It’s an utterly useless metric for showing if wages are moving or not. If you must use average, then you should remove the top 5% from the average.
- sapiogram 3y ago
  > If you must use average, then you should remove the top 5% from the average. 5% is too arbitrary, just use the median.
  - adolph 3y ago
    Given outliers in a skewed dataset, wouldn't median be a worse description than mean without outliers? https://openstax.org/books/statistics/pages/2-6-skewness-and-the-mean-median-and-mode https://openstax.org/books/statistics/pages/2-6-skewness-and...
    - ghufran_syed 3y ago
      I think the median works better in this situation because the huge population means that both the outliers and the skewness have much less effect than they do on the mean. The problem with using the mean “without outliers” is that you have to make arbitrary decisions about what data to exclude as an outlier, unlike with the median.
    - sapiogram 3y ago
      The answer is always "it depends", which is exactly why I prefer the median in most cases. Once you choose to use the median, there are no more choices/degrees of freedom - it's just the median. On the other hand, "mean without outliers" requires you to make a subsequent value judgement on what exactly is an "outlier".
      - adolph 3y ago
        > "mean without outliers" requires you to make a subsequent value judgement Do you think that comparison of outliers to interquartile range is not a relatively objective method of determining outliers? The interquartile range is a number that indicates the spread of the middle half, or the middle 50 percent of the data. It is the difference between the third quartile (Q3) and the first quartile (Q1) . . . The IQR can help to determine potential outliers. A value is suspected to be a potential outlier if it is less than 1.5 × IQR below the first quartile or more than 1.5 × IQR above the third quartile. Potential outliers always require further investigation. https://openstax.org/books/statistics/pages/2-3-measures-of-the-location-of-the-data https://openstax.org/books/statistics/pages/2-3-measures-of-...
  - wredue 3y ago
    5% is not arbitrary. It is removing massively skewing outliers who also see a massive growth outlier compared to everyone else. Looking at wage movement as slices of where one falls gives a clearer picture. Additionally, the reason it doesn’t make sense to remove low wage outliers while removing high wage outliers for average is because the median wage is closer to low wage outliers than it is to high wage outliers. For example, if you take the median and then +- for your dataset (if median is 48,000, then use 0 thru 96,000), you’ll be removing more than just the top 5%, and yet, this also gives a far far far better picture of the dire economic position and what is happening with real wage movement.
    - deleted 3y ago
      [deleted]
    - squeaky-clean 3y ago
      It's arbitrary because why not 4% or 6%. 5 just happens to be a pleasant number but there's nothing special about it in an economic sense.
    - Gormo 3y ago
      Why would you cut off a fixed percentage rather than use a standard deviation range?
      - ralphm 3y ago
        Because wages are non-Gaussian, standard deviation doesn't mean anything in this context.
        Gormo 3y ago
        Interesting. Most income graphs I've seen appear to be normally distributed -- could you explain what would cause them not to be, or provide any examples that demonstrate non-Gaussian distributions? But if wages are non-Gaussian, how would trimming the top and bottom 5% off of your sample be any better than using standard deviation ranges to control for outliers? The assumption that outliers are to be found on the top and bottom of your range is one that seems to apply to Gaussian distributions, and doesn't necessarily hold for others, regardless of whether you are using fixed percentage values or standard deviation thresholds. For example, in a bimodal distribution, outliers might be found in the center.
  - permo-w 3y ago
    median is a type of average. the parent commenter means the mean
    - ryanisnan 3y ago
      Is this correct? I know no definition where median can be interpreted as an average.
      - llbeansandrice 3y ago
        It's frequently used in when talking about incomes specifically. [0] > For example, the average personal income is often given as the median—the number below which are 50% of personal incomes and above which are 50% of personal incomes—because the mean would be higher by including personal incomes from a few billionaires. For this reason, it is recommended to avoid using the word "average" when discussing measures of central tendency and specifically specify which type of measure of average is being used. [0] https://en.wikipedia.org/wiki/Average https://en.wikipedia.org/wiki/Average
      - KolenCh 3y ago
        The median minimizes the L1 norm and the mean minimizes the L2 norm. (Of the error made between the estimator and the data roughly speaking.) In this sense yes, median is some kind of average. But then it is insensitive to the outliners, which is the reason it is suggested above.
      - mlyle 3y ago
        e.g. M-w.com: > 1 a > : a single value (such as a mean, mode, or median) that summarizes or represents the general significance of a set of unequal values Average isn't well defined-- if you have to guess, people probably intend "arithmetic mean" when they say it. But if you're using it in the sense of a single number that represents the typical value of a dataset, you may want the median, mode, or midrange, or even a geometric or harmonic mean depending upon circumstance. So it's best to use that term. And it's best to be charitable to not jump on people saying "average" when the median might be the best measure of central tendency for a task.
- thereisnojesus 3y ago
  no
- reissbaker 3y ago
  "Real Average Hourly Earnings" is already an adjusted metric to make sure that it's not skewed to the top earners (hence why it's called "Real Average" and not just "Average"). In the linked post you'll note the following explanation of what "Real Hourly Average Earnings" means: "Data relate[d] to production employees in mining and logging and manufacturing, construction employees in construction, and nonsupervisory employees in the service-providing industries," which apparently covers roughly 80% of private sector jobs in the US.
  - mediaman 3y ago
    You are right that this metric excludes things like capital gains income, and covers just hourly earnings. “Real”, however, refers to it being inflation adjusted, not an exclusion of high earners (who are excluded anyway because of the hourly earnings metric).
Dah00n 3y ago
Amazon is not a good place to get prices from for a comparison. All the enshittification has ruined the prices. As it forces prices up outside Amazon too even outside prices are not useful. To do a proper comparison that doesn't include changes forced by Amazon IMO you need to look at non-US non-Amazon non-Amazon-sellers prices and work with those. Quite annoying really.
- naikrovek 3y ago
  I've seen the word "enshittification" three times in three days now, and it is already rubbing my last nerve raw. we make fun of business jargon and buzzword fads, and we do it just as much as anyone else, and it sickens me.
  - afavour 3y ago
    Why does it sicken you? We make fun of business jargon because it’s an attempt to imbue authority where it isn’t due. “Enshittification” is… not that.
  - whyenot 3y ago
    If you were on HN a decade ago, you would have seen "pivot" all over the place. Times change, but we are still human (mostly) and it seems to be part of who we are to mold language to fit our needs and circumstances. Take "bipping," a word I had never seen before a month ago, but that is now all over the place. (At least in some communities in the Bay Area; I guess all the car break-ins helped it's adoption).
    - ShadowBanThis01 3y ago
      Never heard of that, and when I do I will gleefully mock whoever is using it.
      - Operyl 3y ago
        It’s almost as if the English language adapts over time. I’m sure in the 1580s when the word “Gleeful” was seemingly introduced people felt weird about it too, heh.
grrowl 3y ago
Worth mentioning Raspberry Pis are designed, and in most cases built and assembled, in the UK.
runeks 3y ago
> The accumulated inflation in the US since June 2019 is c. 17%.[1] This means US$ 55 in June 2019 is "worth" c. US$ 64.35 today. So it seems that you get a 4GB Raspi 5 for c. 7.25% less today. Included in that 17% figure is the price increase for e.g. corn, rent and steaks. Why should the change in price of either of these things be relevant to the price of a Raspberry Pi?
- handsclean 3y ago
  How else would you define how much a dollar is?
  - runeks 3y ago
    That my point: you can’t define how much a dollar is worth in dollars because it’s a circular definition. You can, of course, calculate the change in price over time of a specific basket of goods (rent, steaks, corn). But there’s no reason this is connected in any way to the change in price of an entirely different basket of goods (Raspberry Pi’s). In other words, if you try to define how much the value of a dollar has changed over time, you’ll get a different answer depending on the basket of goods you choose. So the change is clearly not inherent to the dollar.

wredue 3y ago

Please stop using “average” for wage comparisons. It’s an utterly useless metric for showing if wages are moving or not. If you must use average, then you should remove the top 5% from the average.

sapiogram 3y ago
> If you must use average, then you should remove the top 5% from the average. 5% is too arbitrary, just use the median.
- adolph 3y ago
  Given outliers in a skewed dataset, wouldn't median be a worse description than mean without outliers? https://openstax.org/books/statistics/pages/2-6-skewness-and-the-mean-median-and-mode https://openstax.org/books/statistics/pages/2-6-skewness-and...
  - ghufran_syed 3y ago
    I think the median works better in this situation because the huge population means that both the outliers and the skewness have much less effect than they do on the mean. The problem with using the mean “without outliers” is that you have to make arbitrary decisions about what data to exclude as an outlier, unlike with the median.
  - sapiogram 3y ago
    The answer is always "it depends", which is exactly why I prefer the median in most cases. Once you choose to use the median, there are no more choices/degrees of freedom - it's just the median. On the other hand, "mean without outliers" requires you to make a subsequent value judgement on what exactly is an "outlier".
    - adolph 3y ago
      > "mean without outliers" requires you to make a subsequent value judgement Do you think that comparison of outliers to interquartile range is not a relatively objective method of determining outliers? The interquartile range is a number that indicates the spread of the middle half, or the middle 50 percent of the data. It is the difference between the third quartile (Q3) and the first quartile (Q1) . . . The IQR can help to determine potential outliers. A value is suspected to be a potential outlier if it is less than 1.5 × IQR below the first quartile or more than 1.5 × IQR above the third quartile. Potential outliers always require further investigation. https://openstax.org/books/statistics/pages/2-3-measures-of-the-location-of-the-data https://openstax.org/books/statistics/pages/2-3-measures-of-...
- wredue 3y ago
  5% is not arbitrary. It is removing massively skewing outliers who also see a massive growth outlier compared to everyone else. Looking at wage movement as slices of where one falls gives a clearer picture. Additionally, the reason it doesn’t make sense to remove low wage outliers while removing high wage outliers for average is because the median wage is closer to low wage outliers than it is to high wage outliers. For example, if you take the median and then +- for your dataset (if median is 48,000, then use 0 thru 96,000), you’ll be removing more than just the top 5%, and yet, this also gives a far far far better picture of the dire economic position and what is happening with real wage movement.
  - deleted 3y ago
    [deleted]
  - squeaky-clean 3y ago
    It's arbitrary because why not 4% or 6%. 5 just happens to be a pleasant number but there's nothing special about it in an economic sense.
  - Gormo 3y ago
    Why would you cut off a fixed percentage rather than use a standard deviation range?
    - ralphm 3y ago
      Because wages are non-Gaussian, standard deviation doesn't mean anything in this context.
      - Gormo 3y ago
        Interesting. Most income graphs I've seen appear to be normally distributed -- could you explain what would cause them not to be, or provide any examples that demonstrate non-Gaussian distributions? But if wages are non-Gaussian, how would trimming the top and bottom 5% off of your sample be any better than using standard deviation ranges to control for outliers? The assumption that outliers are to be found on the top and bottom of your range is one that seems to apply to Gaussian distributions, and doesn't necessarily hold for others, regardless of whether you are using fixed percentage values or standard deviation thresholds. For example, in a bimodal distribution, outliers might be found in the center.
- permo-w 3y ago
  median is a type of average. the parent commenter means the mean
  - ryanisnan 3y ago
    Is this correct? I know no definition where median can be interpreted as an average.
    - llbeansandrice 3y ago
      It's frequently used in when talking about incomes specifically. [0] > For example, the average personal income is often given as the median—the number below which are 50% of personal incomes and above which are 50% of personal incomes—because the mean would be higher by including personal incomes from a few billionaires. For this reason, it is recommended to avoid using the word "average" when discussing measures of central tendency and specifically specify which type of measure of average is being used. [0] https://en.wikipedia.org/wiki/Average https://en.wikipedia.org/wiki/Average
    - KolenCh 3y ago
      The median minimizes the L1 norm and the mean minimizes the L2 norm. (Of the error made between the estimator and the data roughly speaking.) In this sense yes, median is some kind of average. But then it is insensitive to the outliners, which is the reason it is suggested above.
    - mlyle 3y ago
      e.g. M-w.com: > 1 a > : a single value (such as a mean, mode, or median) that summarizes or represents the general significance of a set of unequal values Average isn't well defined-- if you have to guess, people probably intend "arithmetic mean" when they say it. But if you're using it in the sense of a single number that represents the typical value of a dataset, you may want the median, mode, or midrange, or even a geometric or harmonic mean depending upon circumstance. So it's best to use that term. And it's best to be charitable to not jump on people saying "average" when the median might be the best measure of central tendency for a task.
thereisnojesus 3y ago
no
reissbaker 3y ago
"Real Average Hourly Earnings" is already an adjusted metric to make sure that it's not skewed to the top earners (hence why it's called "Real Average" and not just "Average"). In the linked post you'll note the following explanation of what "Real Hourly Average Earnings" means: "Data relate[d] to production employees in mining and logging and manufacturing, construction employees in construction, and nonsupervisory employees in the service-providing industries," which apparently covers roughly 80% of private sector jobs in the US.
- mediaman 3y ago
  You are right that this metric excludes things like capital gains income, and covers just hourly earnings. “Real”, however, refers to it being inflation adjusted, not an exclusion of high earners (who are excluded anyway because of the hourly earnings metric).

Dah00n 3y ago

Amazon is not a good place to get prices from for a comparison. All the enshittification has ruined the prices. As it forces prices up outside Amazon too even outside prices are not useful. To do a proper comparison that doesn't include changes forced by Amazon IMO you need to look at non-US non-Amazon non-Amazon-sellers prices and work with those. Quite annoying really.

naikrovek 3y ago
I've seen the word "enshittification" three times in three days now, and it is already rubbing my last nerve raw. we make fun of business jargon and buzzword fads, and we do it just as much as anyone else, and it sickens me.
- afavour 3y ago
  Why does it sicken you? We make fun of business jargon because it’s an attempt to imbue authority where it isn’t due. “Enshittification” is… not that.
- whyenot 3y ago
  If you were on HN a decade ago, you would have seen "pivot" all over the place. Times change, but we are still human (mostly) and it seems to be part of who we are to mold language to fit our needs and circumstances. Take "bipping," a word I had never seen before a month ago, but that is now all over the place. (At least in some communities in the Bay Area; I guess all the car break-ins helped it's adoption).
  - ShadowBanThis01 3y ago
    Never heard of that, and when I do I will gleefully mock whoever is using it.
    - Operyl 3y ago
      It’s almost as if the English language adapts over time. I’m sure in the 1580s when the word “Gleeful” was seemingly introduced people felt weird about it too, heh.

grrowl 3y ago

Worth mentioning Raspberry Pis are designed, and in most cases built and assembled, in the UK.

runeks 3y ago

> The accumulated inflation in the US since June 2019 is c. 17%.[1] This means US$ 55 in June 2019 is "worth" c. US$ 64.35 today. So it seems that you get a 4GB Raspi 5 for c. 7.25% less today. Included in that 17% figure is the price increase for e.g. corn, rent and steaks. Why should the change in price of either of these things be relevant to the price of a Raspberry Pi?

handsclean 3y ago
How else would you define how much a dollar is?
- runeks 3y ago
  That my point: you can’t define how much a dollar is worth in dollars because it’s a circular definition. You can, of course, calculate the change in price over time of a specific basket of goods (rent, steaks, corn). But there’s no reason this is connected in any way to the change in price of an entirely different basket of goods (Raspberry Pi’s). In other words, if you try to define how much the value of a dollar has changed over time, you’ll get a different answer depending on the basket of goods you choose. So the change is clearly not inherent to the dollar.