[R-meta] sample variance estimation of an effect size (reponse ratio) using confidence limits
Viechtbauer, Wolfgang (SP)
wo||g@ng@v|echtb@uer @end|ng |rom m@@@tr|chtun|ver@|ty@n|
Thu Mar 18 10:04:08 CET 2021
Dear Diego,
Do not include n in this equation. The sample size is already incorporated into the variance that is back-calculated with the equation without n.
Best,
Wolfgang
>-----Original Message-----
>From: R-sig-meta-analysis [mailto:r-sig-meta-analysis-bounces using r-project.org] On
>Behalf Of Diego Grados Bedoya
>Sent: Thursday, 18 March, 2021 9:47
>To: James Pustejovsky
>Cc: r-sig-meta-analysis using r-project.org
>Subject: Re: [R-meta] sample variance estimation of an effect size (reponse ratio)
>using confidence limits
>
>James,
>
>Thank you for the example (I followed those steps). However, it is not
>clear to me yet if including the number of comparisons (n) of the effect
>size (if it is available) will result in a different estimate of the
>variance (based on *variance = (n^2 * (mean_ effect_size - lower_limit_
>confidence_interval)) / 1.96)*. For instance, if n would have been provided
>in your example?
>
>Diego
>
>On Wed, 17 Mar 2021 at 16:42, James Pustejovsky <jepusto using gmail.com> wrote:
>
>> The variances will be quite different. Only the one based on the log scale
>> is correct. Here's an example:
>> Percentage change = +24.6%, 95% CI [18.5%, 31.0%]
>>
>> Back-transform to log response ratio using LRR = log[(% change) / 100 + 1]:
>> LRR = 0.22, 95% CI [0.17, 0.27]
>>
>> Calculate SE of log response ratio using SE_LRR = [U - L] / (2 *
>> z_critical)
>> SE_LRR = [0.27 - 0.17] / (2 * 1.96) = 0.025
>>
>> On Wed, Mar 17, 2021 at 10:30 AM Diego Grados Bedoya <
>> diegogradosb using gmail.com> wrote:
>>
>>> James,
>>>
>>> The effect size was originally reported as a percentage (based on the log
>>> response ratio). I back-transformed it using the equation (exp(RR) - 1) *
>>> 100% and I did the same for the confidence intervals. Based on these
>>> back-transformed values I am estimating the variance.
>>>
>>> If I use both equations, the values of the variance are
>>> complicity different for each equation.
>>>
>>> Am I missing something?
>>>
>>> Thank you,
>>>
>>> Diego
>>>
>>> On Wed, 17 Mar 2021 at 16:12, James Pustejovsky <jepusto using gmail.com>
>>> wrote:
>>>
>>>> Diego,
>>>>
>>>> Is the effect size reported on the log scale (log response ratio, with
>>>> range from negative infinity to positive infinity and null value of zero)
>>>> or on the ratio scale (range from 0 to infinity, null value of 1)?
>>>> Typically, confidence intervals are calculated on the log scale. If the
>>>> effect size is reported on the ratio scale, then you can use the formula
>>>> you described but you'll first have to convert the response ratio and
>>>> confidence limits to the log scale.
>>>>
>>>> James
>>>>
>>>> On Wed, Mar 17, 2021 at 10:09 AM Diego Grados Bedoya <
>>>> diegogradosb using gmail.com> wrote:
>>>>
>>>>> Dear all,
>>>>>
>>>>> I am trying to estimate the sample error variance of an effect size
>>>>> (reported as response ratio) based on the confidence intervals (assuming
>>>>> that it follows a Gaussian distribution). Is it a valid approach to use
>>>>> the
>>>>> following equation if I do not have the number of comparisons used in
>>>>> the
>>>>> calculation of the effect size?
>>>>>
>>>>> *variance = ((mean_ effect_size - lower_limit_confidence_interval) /
>>>>> 1.96)^2*
>>>>>
>>>>> If I would have the number of comparisons (n), should I go for the
>>>>> following equation?
>>>>>
>>>>> *variance = (n^2 * (mean_ effect_size - lower_limit_
>>>>> confidence_interval))
>>>>> / 1.96*
>>>>>
>>>>> Thanks in advance,
>>>>>
>>>>> Kind regards,
>>>>>
>>>>> Diego
>>>>>
>>>>> [[alternative HTML version deleted]]
>>>>>
>>>>> _______________________________________________
>>>>> R-sig-meta-analysis mailing list
>>>>> R-sig-meta-analysis using r-project.org
>>>>> https://stat.ethz.ch/mailman/listinfo/r-sig-meta-analysis
>>>>>
>>>>
>
> [[alternative HTML version deleted]]
>
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