Sample Size & Power Analysis Software FAQ’s

A 1. Yes. We have made every effort to make nQuery Advisor Section 508 Compliant. Click here to view the nQuery Advisor Section 508 Accessibility Statement.

Q 2. What is the correct citation to use for nQuery Advisor 7.0?

A 2. The correct citation for nQuery Advisor 7.0 is as follows :Janet D. Elashoff (2007) nQuery Advisor Version 7.0. Statistical Solutions, Cork, Ireland

Q 3. How Does nQuery Advisor Compare with Siz from Cytel?

A 3. In the comparison sheet that Cytel have used to compare nQuery Advisor and SiZ they have used three headings; Design, Simulation and Analysis. nQuery Advisor does not perform any statistical analysis, it is purely used for calculating power and sample size.
Also, nQuery Advisor uses closed form solutions for calculating sample size as much as possible, thereby eliminating the need for nQuery Advisor to perform simulations. Certain tables require simulation (for example with survival analysis) and where appropriate nQuery Advisor provides this.

In the SiZ document, there are several entries that refer to Non-inferiority/Superiority/Equivalence. In nQuery Advisor there are tables for equivalence for difference of means, ratio of means, difference of proportions. nQuery Advisor labels these tables as Equivalence but by setting the limits appropriately they can all be used for Non-Inferiority and Superiority also. There are examples of this in the nQuery Advisor manual. So it is a little misleading to say nQuery Advisor does not include these tables. The calculations can be done, they are just all under the heading of Equivalence.

nQuery Advisor also contains tables for calculating sample size for Survival Analysis, Agreement and Regression, none of which are present in SiZ. Here is a full list of the tables that nQuery Advisor offers:

http://www.statistical-solutions-software.com/nquery-advisor-sample-size-software/sample-size-tables/

Finally, the SiZ document lists an incorrect price for nQuery Advisor. nQuery Advisor is priced as follows: Academic Licenses US$795 / Euro 650 and Commercial Licenses US$995 / Euro 895. These prices are for perpetual licenses and no annual fees are payable.

A 4. It sounds like the symbol fonts that nQuery Advisor® uses (the Dixon fonts) have not installed properly on your machine. To fix this problem:
Download the following self-extracting file to your computer.
Use Explorer to view the file (dixonfonts.exe) and double click on it.
It will extract the following four files, Dixon.ttf, Dixonbd.ttf, Dixonbs.fon and Dixons.fon.
The seven instructions below will guide you through the installation process.

Go to your START button and select SETTINGS and CONTROL PANEL.
From the CONTROL PANEL select FONTS.
In the FONTS Window, select FILE and Install New Font…
Select the folder where you saved the font files as the location to install from. (NOTE If asked to overwrite or replace, click OK)
The four font files should appear in the List of Fonts field.
Click on SELECT ALL, and then click OK.
Restart the computer.

Q 5. When I am using nQuery Advisor I keep getting an error message about a .dll file. What can I do to fix this?

A 5. This error can be fixed by making sure you have all the most up to date Windows .dll files on your system. To do this, open up a command prompt, (go to Start -> Run… and type ‘cmd’ in the box and press enter). Go to the main root drive by type ‘c:’ and press enter. Now type ‘sfc /scannow’ and press enter. This runs a Windows program that will scan and repair/update any .dll files that are corrupt or out of date. In order to complete fully you may be asked to insert your Windows CD.

A 6. These TOST tests for demonstrating equivalence are based on the standard two-sample t-test in which the two groups are assumed to have equal variances. When sample sizes are large enough so that the normal approximation is reasonable, the sample size required for unequal n’s can be computed from the sample size obtained for equal n’s. So if “n per group” for the desired power is computed as n, then the total sample size required for the equal n situation is Nequal =2n. Then the formula for the total sample size in the unequal sample size situation is given below.

Penalty for unequal n’s

So, for r = 2, Nunequal = 1.125 Nequal.

Q 7. What is the difference between tables MTE0 and MTE1tg?

A 7. MTE0 is for one-sided non-inferiority testing while MTE1tg is for two one-sided t-test (TOST) equivalence testing. For the situation in which it is assumed that the test and standard really do have equal means and the upper and lower equivalence limits are symmetric about zero, the two tables will give the same required sample size if an adjustment to the requested power is made in MTE0.

MTE0 Two group t-test of equivalence in means (equal n’s)

If the desired power for the TOST situation in MTE1tg is 100(1-β), the sample size will be the same as that obtained using MTE0 and specifying a power of 100(1-β/2). That is, 90% power for each of the one-sided tests is necessary to produce 80% power for the combination of the two tests.

MTE1tg Two-group t-tests (TOST) of equivalence in means (equal n’s)

A 8. nQuery Advisor’s® Distribution Function tables can be used to make these sample size computations, using the definitions of the effect sizes shown in the tables, plus the information given in the manual and help about computational methods. We illustrate the correspondence between the results for tables MGT0 and MGT1 with those using the distribution functions for the data from Example 3 in Chapter 12 of the nQuery 5 manual.

MGT0

To compute the sample size required to get 90% power for the test of the overall null hypothesis that all 4 groups have the same mean, use table MGT0. Note that the actual power with 14 per group is 91.456.

One-way analysis of variance (equal n’s)

MGT0-1: Aid for col 1 — One-way analysis of variance (equal n’s)

To make the same computations using the Distribution functions tables, we compute the effect size, í¢Ë†” ²as shown in the table.

The mean of the four means is (5 + 10.5 + 13.5 + 12)/4 = 10.25. The term V = (5.25² + 0.25² + 3.25² + 1.75²)/4 = 10.3125, and í¢Ë†” ²= 10.3125/(6²) = .2865.

As explained in the Help: Power, sample size, or effect size are computed using central and non-central F. For MGT0, the numerator and denominator degrees of freedom are (G-1) and (N-G), and the non-centrality parameter is N times the effect size (for equal n’s N=nG).

So, for 14 per group, the non-centrality parameter is 14 x 4 x .2865 = 16.044.

Go to the Assistants menu, select Distribution Functions, and select F(Central) and fill in the table with numerator degrees of freedom G-1 = 4-1 =3, and denominator degrees of freedom, 14 x 4 – 4 = 52. Since we want the cutpoint for a 5% level test, enter .95 for the probability, to find the F-value of 2.7826 (see bottom of nQuery screen for more digits than displayed in table.)

F (Central) Distribution

Then, go to the Assistants menu, select Distribution Functions, and select F(Non-central) and fill in the table with numerator degrees of freedom G-1 = 4 – 1 = 3, and denominator degrees of freedom, 14 x 4 – 4 = 52. Then enter the non-centrality parameter, 16.044, and the cutpoint F-value of 2.7826. nQuery gives a probability of 0.085 that the observed F will be below the cutpoint, this is the probability of a beta error, so the power is 1- .085 or .915 (91.5%), which agrees with the MGT0 table.
F (Non-central) Distribution

MGT1

To compute the power obtained with 14 per group for a one degree of freedom contrast between groups 2 and 3, use table MGT1. Note that the actual power with 14 per group is only 25.47.

Single one-way between means contrast (equal n’s)

MGT1-1: Aid for col 1 — Contrast in one-way analysis of variance (equal n’s)

For MGT1, power, sample size, or effect size are computed using central and non-central F. For MGT1, the numerator and denominator degrees of freedom are 1 and (N-G), and the non-centrality parameter is the number of cases in group 1 times the square of the effect size. So, the non-centrality parameter is 14 x (0.354)_ = 1.75. Using the distribution function tables as before, we have

F (Central) Distribution

and

F (Non-central) Distribution

which gives a power of 100(1 – .745) = 25.5. You can also, solve for the non-centrality parameter in the table for non-central F and then compute the necessary n for the effect size of interest using the fact that the non-centrality parameter is the number of cases in group 1 times the square of the effect size.

A 9. For STT0, STT1, and STT2, the required number of events, E, can be obtained approximately from the equation

where z_α/2 and z_β and are the upper α/2 and upper β points, respectively, of the standard normal distribution and h is the hazard ratio. See Collett, D. (1994) Modelling Survival Data in Medical Research, Chapman & Hall, Formula 9.2
This is a modified version of the more general formula given lower in page 255 as

Where π₁ and π₂ are the proportions of individuals to be assigned to Groups 1 and 2. As you see, when equal numbers are allocated, 1/(π₁ π₂) is 4. For a 2 to 1 ratio this term equals 9/2 or 4.5. So take the number of events given by STT0, STT1, or STT2 and multiply by 4.5/4 = 1.125.

Note that this formula for the number of events is based on the assumption that the hazard ratio is constant throughout so it is not really applicable in the STT3 context in which the hazard ratio may vary from time period to time period.

A 10. There are two possible reasons why your answers do not match. First of all Table 5.4.1 of Chow & Liu gives the TOTAL sample size whereas table MTE1 of nQuery Advisor® gives “n per group”. So just multiply the answer that nQuery Advisor® gives you by 2 to get the total sample size.

Secondly, when using MTE1 for a crossover design, (which is what is presented in Chow & Liu) you must divide the CV they use by sqrt(2), so for a CV of 20% you should type in .2/sqrt(2) = 0.14142. The table entry in nQuery Advisor® is labelled “Common standard deviation” – when the study design is a crossover design, enter the standard deviation of the differences divided by 2 (sd/2), or 1/sqrt(2) times the root mean square error (RMSE) from the crossover ANOVA (se/sqrt(2)).

However, for crossover designs, table MTE1 was replaced by table MTE1co in Version 5. When using MTE1co, you can enter the squareroot of the crossover ANOVA MSE directly.

A 11. There are two reasons why the discrepancy is so large. One is that the CT and CCT are both approximations, and CCT is known to be conservative. The other, which you will find if you explore various sample sizes, is that power is related to the square root of sample size and is very insensitive to changes in n in this region. Using FET, sample sizes from 3414 to 3506 give greater than 90% power and less than 91% power and 3686 gives 92% power.

For most examples the power using Fisher’s exact test methods is in between that from CT and CCT.

Q 12. Do tables AOT2 and AOC2 use the updated formula for Lin’s Concordance Coefficient?

A 12. The formula for V(K) used in the side-tables for nQuery Advisor’s® tables AOT2 and AOC2 was published in Lin, L.I. (1992). Assay validation using the concordance correlation coefficient. Biometrics 48:599-604. An nQuery Advisor® user told us that Lin had issued a correction to this formula. The corrected formula for V(K) appears in Lin, L., Hedayat, A.S., Sinha, B., Yang, M. (2002). Statistical methods in assessing agreement: models, issues, and tools. Journal of the Amer. Stat. Assn 97:257-270.

Q 13. Does nQuery Advisor run on Windows 7?

A 13. nQuery Advisor 7.0 has been designed to run on Windows 7.

nQuery Advisor

Support Info

DOWNLOAD TRIAL VERSION

BUY NOW

Class	Price	Add
Commercial	$ 995.00
Academic	$ 795.00
5 User Comm	$ 4,000.00
5 User Acad	$ 3,000.00