5 ms·

Good point! Sadly, I don't have access to the Harvard study (https://pubsonline.informs.org/doi/10.1287/mksc.2019.1200 https://pubsonline.informs.org/doi/10.128

by giu 5y ago

Good point! Sadly, I don't have access to the Harvard study (https://pubsonline.informs.org/doi/10.1287/mksc.2019.1200 https://pubsonline.informs.org/doi/10.1287/mksc.2019.1200), so maybe someone with access to it could check it, but the linked article might be misleading in some aspects (depending on the results of the study shown in the paper). From the study's abstract: > A preregistered field experiment indicated that diners were 21.1% more likely to buy a bowl of chicken noodle soup when a sign revealing its ingredients also included the cafeteria’s costs to make it. From the linked article's sub-title: > Sales of a chicken noodle soup increased 21.1% when people were shown the costs of making it. The study's abstract mentions that they were more likely to buy a bowl of chicken; it's not mentioned that they actually bought it.

tuukkah 5y ago
"21.1% more likely" is how articles phrase a 21.1% increase in observed frequency. "Sales increased 21.1%" is equivalent as long as the unit price remained the same.
- vntok 5y ago
  The article is very confusing on this front. > People said they were 14.2% more likely to buy this chocolate bar when they were shown the version with a cost breakdown Surely that cannot be correct. Possibly 14.2% of the people who were asked said that they would be "somewhat" more likely to buy X with more data stuck to its label (does any data improve sales? Did they A/B the label by adding random info?). This is very different from them acting upon it though.
  - tuukkah 5y ago
    Actually, the figure 14.2% does not even appear in the research article. The experiment was between two groups of random Mechanical Turk workers: one group was not shown the cost breakdown while the other group was. The workers answered how likely they were to buy the product on a scale from 1 to 7. If you calculate the average increase from one group to the other, sure enough the number you get is 14.2% but it is not a probability.
tuukkah 5y ago
Here's open access to the full study: http://ssrn.com/abstract=2498174 http://ssrn.com/abstract=2498174

by giu 5y ago

tuukkah 5y ago

"21.1% more likely" is how articles phrase a 21.1% increase in observed frequency. "Sales increased 21.1%" is equivalent as long as the unit price remained the same.

vntok 5y ago
The article is very confusing on this front. > People said they were 14.2% more likely to buy this chocolate bar when they were shown the version with a cost breakdown Surely that cannot be correct. Possibly 14.2% of the people who were asked said that they would be "somewhat" more likely to buy X with more data stuck to its label (does any data improve sales? Did they A/B the label by adding random info?). This is very different from them acting upon it though.
- tuukkah 5y ago
  Actually, the figure 14.2% does not even appear in the research article. The experiment was between two groups of random Mechanical Turk workers: one group was not shown the cost breakdown while the other group was. The workers answered how likely they were to buy the product on a scale from 1 to 7. If you calculate the average increase from one group to the other, sure enough the number you get is 14.2% but it is not a probability.

tuukkah 5y ago

Here's open access to the full study: http://ssrn.com/abstract=2498174 http://ssrn.com/abstract=2498174