I am analyzing the accuracy of different clinicians when it comes to providing presumptive diagnoses. I am comparing dentists and physicians for accuracy of referrals. My data is as follows: 33 cases the MD diagnosed correct, 14 cases the MD was incorrect. 115 cases the DDS diagnosed correctly. 76 cases the DDS was incorrect.

Is there a statistical difference between these two groups? How would I calculate that using Excel?

    Correct Incorrect   
MD   33         14  
DDS  115        76  
  • $\begingroup$ When you added the [statistical] tag, did you notice it said, "This tag is deprecated. DO NOT USE IT"? $\endgroup$ – gung - Reinstate Monica Mar 2 '17 at 15:45
  • $\begingroup$ Nope. But I am new to this so I wasn't paying much attention. $\endgroup$ – friese Mar 2 '17 at 20:44

I think what you are looking for is a hypotheses contrast about the difference in proportions.

Let $p_{MD}$, $p_{DDS}$ the (population) proportion of correctly diagnosed cases for the two clinician types, and $\hat{p}_{MD}=\frac{33}{33+14}$, $\hat{p}_{DDS}=\frac{115}{115+76}$.

You test the null hypothesis $H_0:\,\,\,p_{DDS}-p_{MD}=0$ against the alternative $H_1:\,\,\,p_{DDS}-p_{MD}\neq 0$.

The total proportion of correct diagnoses for both samples is $\hat{p}=\frac{33+115}{33+14+115+76}$.

The sample sizes are $n_{DDS}=115+76$, $n_{MD}=33+14$.

You build the statistic


This follows a standard normal distribution, so you have to pick a threshold of that distribution to compare to. For instance, for an $\alpha=0.05$ significance (or $95\%$ confidence in your response), you pick $z_{\frac{\alpha}{2}}=1.96$ and $-z_{\frac{\alpha}{2}}=-1.96$.

If the $Z$ statistic you computed is greater than $z_{\frac{\alpha}{2}}$ or smaller than $-z_{\frac{\alpha}{2}}$, then you can state the proportions are different at the $95\%$ level. Otherwise, they are not significantly different.


You can use a chi-squared test of independence to answer the question "Is diagnosis accuracy different in DDSs vs MDs?"

In excel, you'll want to enter your data more or less as you have presented them here, in a 2x2 cell arrangement. Then you have to make a second similar 2x2 table of the expected counts for each cell. Each cell should be $row total * column total / grand total$

For example, in your case, the expected counts for the Correct MD cell would be

$(33 + 14) * (33 + 115)/(33 + 14 + 115 + 76) = 22.23$

Fill that in for all four cells in the expected counts table. Then you can use the function CHISQ.TEST() to get the p value associated with the chi-squared test of independence --- a significant p value (generally, p < .05) means the two factors are NOT independent, i.e. the diagnostic accuracy depends on which degree they have.


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