Computational Statistics & Data Analysis 3 (1986) 219-226 North-Holland

219

An exact test for the mean of a normal distribution with a known coeff ic ient of variation

Ramesh GUPTA Department of Mathematics, University of Maine, Orono, ME 04469, USA

Ram T R I P A T H I Department of Mathematics, University of Texas at San Antonio, San Antonio, TX 78285, USA

Joel M I C H A L E K and Thomas W H I T E Data Sciences Division, USAF School of Aerospace Medicine, .San Antonio, TX 78235-5000, USA

Received 24 December 1984 Revised 3 September 1985

Abstract: Tables are presented for the critical value, power and sample size required to attain a given power at a given alternative for the likelihood ratio statistic for testing H 0 : 0 = 0 o versus H~: 0 = 01 > 00, where 0 is the mean of a normal distribution having variance a202 with the coefficient of variation, a, assumed known.

Keywords: Coefficient of variation, Normal distribution, Exact test.

1. Introduction

In the physical, biological and medical sciences there are situations where the standard deviation, o, of an observation is proportional to the mean, 0, and the ratio a = o /0 , termed the coefficient of variation, is known while 0 and o may be unknown; see Khan [6] for several examples of this kind of data. While these occurrences have been recognized for many years, statistical theory directed at inference on the mean when the coefficient of variation is known has been weak and has only recently attracted the attention of mathematical statisticians. See, for example, Hinkley [5], Gleser and Healy [4], Khatri and Ratani [7], Amemiya [1], and Gerig and Sen [3].

0167-9473/86/$3.50 © 1986, Elsevier Science Publishers B.V. (North-Holland)

220 R. Gupta et aL / Test for the mean of a normal distribution

The purpose of this article is to complete work begun by Khan [6] regarding the problem of testing the mean of a normal distribution with a known coefficient of variation. He considered the problem of testing H 0 : 0 = 0 0 > 0 versus Hi: 0 = 0 a > 00 based on a random sample Xl, x2 , . . . , xn from a normal, N(0, a202), population with known coefficient of variation a > 0. Khan showed that applica- tion of the Neyman-Pearson lemma gives the following test:

Reject Ho if T= ~ (x~-~n) 2 ( x n - k ) 2 ,= aO ° + n afro > c

where ~, is the sample mean, k = 0o01/(0o + 01) and c is a suitable positive constant. The critical value, c, for this test is determined by

C

1 - a = f o [~(d+ ¢7-

where a = P ( T > clO=Oo),

x ) - * ( d - c ~ - x ) ] f ~ _ l ( X ) d x ,

d=v~Oo/[a(Oo + 01)],

(1)

q~(x) is the standard normal distribution function and fro(X) denotes the prob- ability density function of a X 2 distribution. Khan remarked that (1) appeared difficult to solve for c and also left open the problem of finding the power function and of finding n for a given a and power.

In this paper we show how to compute, for any a > O, (a) the critical value for the test T, (b) the power of T at 0 ! for a given a and n, (c) the sample size needed to attain a given power at a given alternative 01 and (d) the p-value associated with an observed value of T.

2. Formula development

Let F(c) denote the left hand side minus the right hand side of (1). The critical value c is then uniquely determined by the equation F(c) -- 0. This root-finding problem is made numerically tractable using the change of variable v/c - x = v/c sin 0 under the integral sign in F, giving

1 r ( c ) = l - a - F [ ( n - 1)/2] erfc - - -

)] -erfc ~ - + ~-/2 sin 0

-(fC-/2 cos O) "-2 e x p [ - (fc-/2

d V/C-/2 sin 0) d-

- - cos 0)2](V~/2 sin O)dO, (2)

where d = ~/n/[a(1 + r)l and r = 01/00. This change of variable converts the interval of integration to (0, 4 /2) for all c and n >~ 2 and spreads the very steep rise and fall of the integrand in (1) over a larger part of the interval. The conversion of the difference in the normal distibution in (1) to a difference in complementary error functions allows high accuracy calculation in the tail areas

R. Gupta et al. / Test for the mean of a normal distribution 221

for large values of their arguments when appropriate erfc subroutines are used, e.g., Cody [2]. The difference of the two values of the complementary error function does not cause a loss of accuracy in the evaluation fo the integral. Examination of the integrand over the range of integration (0, ,rr/2) for n, d and c at the corners of Table 1 shows that when this difference is below about 10 -5, the product of the remaining factors in the integrand is below about 10-38

The power of the test T at the alternative 01 can easily be derived as

¢.(01)=1- f0 e n c-y g._l(y)dy,

r02--02, 2 where g.-a(Y) is the pdf of [ 1 / o ]Xn-1, which reduces to

-q~( 1/~-(01 - k) a01 010°c~-y) ]g , ,_ l (y )dy , (3)

with k = OoOa/(O o + 01). The same change of variable, ~c - y = ~ sin 0, brings the expression (3) into a numericaly tractable form, which is also algebraically similar to (2), given by

1 f0~'/2[ ( r d 1 ~-/2 sin 0 ) fin(01)= 1 - F [ ( n - 1)/2] erfc ~ r

sin01] - e r f c ~ - + r

• ( 1 ~ - / 2 cos 0) ".2 e x p [ - ( 1 ~ - / 2 cos 0)2](1~- , /2 sin 0 ) d 0 .

(4) Formulas (2) and (4) can be used to numerically determine, for any value of a > 0, the critical value for the test T, its power at 01 for a given a and n, the sample size needed to attain a given power, and the p-value associated with an observed value of T. These calculations are described in Section 3, wherein examples are presented.

3. Numerical evaluations

3.1. Critical value

Figures 1 and 2 show plots for a = 0.01 and 0.05 of n versus c for various values of at.

The tabular equivalents of Figures 1 and 2 are Table 1 for a = 0.01 and Table 2 for a = 0.05.

222 R. Gupta et al. / Test for the mean of a normal distribution

10 3

9

8

7

6

5 -

4 -

3 -

2 LLI

.-J

.-J

o°: 10~

~- 8 o: 7

6

0 2 5 .

2 0 .

1 5 .

1 0 .

7.

=

4 .

3 .

2.

1 °

10 t . ~ I / 1 I I I I I I I I 10 ° 2 3 4 S 6 7 8 910 t 2

SRMPLE SIZE

/

I I I I I I I I

3 4 5 6 7 O 9 1 0 =

Fig. 1. Critical value vs. sample size with d as parameter, a--- 0.01.

Table 1 Critical values for a = 0.01

n d = 0 . 5 d = l d = 2 d = 3 d = 4 d = 5 d = 7 d = 1 0 d = 1 5 d = 2 0 d = 2 5

2 10.3 i2.9 20.2 29.7 41.3 54.9 88.1 153.0 301.0 500.0 748.0 5 15.8 17.9 24.5 33.7 45.0 58.5 91.6 156.0 305.0 503.0 751.0

10 23.8 25.4 31.4 40.1 51.2 64.5 97.3 162.0 310.0 508.0 756.0 15 31.1 32.6 38.1 46.5 57.3 70.5 103.0 167.0 315.0 513.0 761.0 20 38.0 39.4 44.7 52.8 63.5 76.4 109.0, 173.0 321.0 519.0 767.0 25 44.8 46.1 51.1 59.0 69.5 82.3 115.0 179.0 326.0 524.0 772.0 40 64.0 65.3 69.9 77.4 87.5 99.9 132.0 195.0 342.0 540.0 788.0 70 101.0 102.0 106.0 113.0 123.0 135.0 165.0 228.0 374.0 571.0 819.0

100 136.0 137.0 141.0 148.0 157.0 169.0 199.0 261.0 406.0 603.0 850.0

R. Gupta et aL / Test for the mean of a normal distribution 223

i0 ~ 9 8 7 6

2 --z

._J

_J

~- B

7 rJ

B

B

D

- 25.

20.

15.

I0.

7.

5.

4.

3,

/

J J

.

I0 t L_

10 ° I , ~ / ~ / ' 4 5 B 7 BglO t 2 3 4 5 B 7 BglO z

BFIMPLE SIZE /

Fig. 2. Critical value vs. sample size with d as parameter, a = 0.05.

Table 2 Critical values for a = 0.05

n d = 0 . 5 d = l d = 2 d = 3 d = 4 d = 5 d = 7 d = 1 0 d = 1 5 d = 2 0 d = 2 5

2 6.72 8.65 14.6 22.8 33.1 45.3 75.8 137.0 278.0 470.0 711:0 5 11.6 13.2 18.6 26.5 36.6 48.8 79.2 140.0 281.0 473.0 714.0

10 18.8 20.1 25.1 32.6 42.5 54.5 84.7 145.0 287.0 478.0 719.0 15 25.4 26.6 31.3 38.7 48.3 60.2 90.2 151.0 292.0 483.0 724.0 20 31.8 33.0 37.5 44.6 54.1 65.9 95.8 156.0 297.0 488.0 730.0 25 38.0 39.2 43.5 50.5 59.9 71.6 101.0 162.0 302.0 493.0 735.0 40 56.1 57.1 61.3 67.9 77.0 88.4 118.0 178.0 318.0 509.0 750.0 70 90.9 91.8 95.7 102.0 111.0 122.0 151.0 210.0 350.0 540.0 781.0

100 125.0 126.0 129.0 135.0 144.0 155.0 183.0 242.0 381.0 571.0 812.0

224 R. Gupta et al. / Test for the mean of a normal distribution

Table 3 Critical value/power for n, r, a, and a = 0.01

n a = l

r = l . 2 r = l . 4 r = l . 6 r = l . 8 r = 2 . 0

5 17.9 0.083 17.5 0.246 17.2 0.439 16.9 0.607 16.7 0.732 10 27.7 0.135 27.0 0.420 26.5 0.689 26.0 0.853 25.7 0.934 15 36.5 0.186 35.6 0.566 34.9 0.838 34.3 0.950 33.9 0.985 20 44.1 0.237 43.8 0.683 42.8 0.920 42.2 0.984 41.6 0.997 25 53.0 0.288 51.7 0.773 50.7 0.961 49.8 0.995 49.1 0.999 50 91.3 0.523 89.0 0.965 87.1 0.999 85.6 1.00 84.4 1.00

100 163.0 0.820 159.0 1.00 155.0 1.00 153.0 1.00 151.0 1.00 a = 2

5 15.8 0.068 15.7 0.194 15.6 0.350 15.6 0.501 15.5 0.627 10 24.4 0.106 24.2 0.330 24.1 0.572 24.0 0.754 23.9 0.866 15 32.1 0.144 31.9 0.452 31.7 0.731 31.6 0.887 31.4 0.956 20 39.5 0.182 39.2 0.560 39.0 0.837 38.8 0.951 38.6 0.986 25 46.6 0.220 46.2 0.652 45.9 0.903 45.7 0.979 45.5 0.996 50 80.1 0.405 79.4 0.908 79.0 0.995 78.6 1.00 78.3 1.00

100 143.0 0.696 142.0 0.996 141.0 1.00 140.0 1.00 140.0 1.00 a = 3

5 15.4 0.065 15.4 0.182 15.3 0.331 15.3 0.478 15.3 0.603 10 23.8 0.101 23.7 0.310 23.6 0.547 23.6 0.730 23.5 0.848 15 31.3 0.136 31.1 0.430 31.1 0.705 31.0 0.870 31.0 0.946 20 38.5 0.171 38.3 0.532 38.2 0.814 38.1 0.940 38.0 0.982 25 45.5 0.206 45.2 0.624 45.1 0.886 45.0 0.973 44.9 0.994 50 77.9 0.381 77.6 0.890 77.4 0.992 77.2 1.00 77.1 1.00

100 139.0 0.667 138.0 0.994 138.0 1.00 138.0 1.00 138.0 1.00

3.2. Power

Power is determined directly from (4) after specification of r, c, d and n or, alternatively, r, c, a and n. Table 3, for a = 0.01, and Table 4, for a -- 0.05, give the power and the corresponding critical value for several values of n, r and a.

3.3. Minimum sample size

The sample size needed to attain a given power, /3, at the alternative 01, given a, a and 00 is determined by solving the equation

a ( n ) - 1 + /3=0 where G(n) denotes the integral on the fight hand side of (4). Tables 5 and 6 give minimum sample size requirements to attain a power of 0.9, for a = 0.01 and 0.05, at various values of a and r.

3.4. p-value

The p-value is computed from equation (2) with c replaced by the observed value of test statistic.

R. Gupta et al. / Test for the mean of a normal distribution

Table 4

Crit ical v a l u e / p o w e r for n, r, a and a = 0.05

225

n a = l

r --- 1.2 r --- 1.4 r = 1.6 r = 1.8 r = 2.0

5 13.2 0.213 12.9 0.432 12.7 0.622 12.4 0.757 12.3 0.845

10 21.9 0.303 21.4 0.625 20.9 0.832 20.6 0.931 20.3 0.972

15 30.0 0.381 29.2 0.754 28.6 0.928 28.1 0.981 27.7 0.995

20 37.7 0.450 36.7 0.841 35.9 0.969 35.3 0.995 34.8 0.999

25 45.2 0.512 44.0 0.899 43.1 0.987 42.4 0.999 41.8 1.00

50 81.1 0.740 79.0 0.990 77.3 1.00 76.0 1.00 74.9 1.00

100 150.0 0.934 146.0 1.00 142.0 1.00 140.0 1.00 138.0 1.00

a = 2

5 11.6 0.185 11.6 0.366 11.5 0.537 11.4 0.671 11.4 0.769

10 19.2 0.257 19.1 0.535 19.0 0.748 18.9 0.872 18.8 0.937

15 26.3 0.320 26.1 0.660 25.9 0.865 25.8 0.952 25.7 0.984

20 33.0 0.377 32.8 0.754 32.6 0.929 32.4 0.982 32.3 0.996

25 39.6 0.430 39.3 0.823 39.0 0.963 38.8 0.994 38.7 0.999

50 71.0 0.640 70.4 0.969 70.0 0.999 69.7 1.00 69.4 1.00

100 130.0 0.867 129.0 0.999 129.0 1.00 128.0 1.00 128.0 1.00

a = 3

5 11.3 0.179 11.3 0.353 11.3 0.519 11.2 0.652 11.2 0.751

15 18.7 0.248 18.7 0.516 18.6 0.728 18.6 0.857 18.5 0.927

20 32.1 0.363 32.0 0.733 31.9 0.917 31.9 0.978 31.8 0.994

25 38.5 0.413 38.4 0.804 38.3 0.955 38.2 0.992 38.1 0.998

50 69.1 0.618 68.8 0.961 68.6 0.998 68.5 1.00 68.3 1.00

100 127.0 0.849 127.0 0.999 126.0 1.00 126.0 1.00 126.0 1.00

Table 5

Minimum sample size for power = 0.9, a = 0.01

a r = 1.2 r = 1.4 r = 1.6 r = 1.8 r = 2.0

1.0 127 37 19 12 9

1.5 156 45 23 15 11

2.0 169 49 25 16 12

2.5 176 51 26 17 12

3.0 181 52 27 17 13

Table 6

Minimum sample size for power = 0.9, a = 0.05

a r = 1.2 r = 1.4 r = 1.6 r = 1.8 r = 2.0

1.0 86 26 14 9 7

1.5 105 31 16 11 8

2.0 114 34 18 12 9 2.5 119 35 19 12 9

3.0 122 36 19 12 9

226 R. Gupta et al. / Test for the mean of a normal distribution

Acknowledgement

This paper was written while Ramesh Gupta was a University Resident Research Associate at the USAF School of Aerospace Medicine.

References

[1] T. Amemiya, Regression analysis when the variance of the dependent variable is proportional to the square of its expectation, J. Amer. Statist. Assoc. 68 (1973) 928-934.

[2] W.J. Cody, Rational Chebyshev approximations for the error function, Math. Comput. 23 (107) (1969) 631-637.

[3] T.M. Gerig and A.R. Sen, MLE in two normal samples with equal but unknown population coefficients of variation, J. Amer. Statist. Assoc. 75 (1980) 704-708.

[4] L.J. Gleser and J.D. Healy, Estimating the mean of a normal distribution with a known coefficient of variation, J. Amer. Statist. Assoc. 71 (1976) 977-981.

[5] D.V. Hinkley, Conditional inference about a normal mean with known coefficient of variation, Biometrika 64 (1977) 105-108.

[6] R. Khan, A note on testing the means of normal distributions with known coefficients of variation, Comm. Statist.-Theory Methods A 7(9) (1978) 867-876.

[7] C.G. Khatri and R.T. Ratani, On estimation of the mean parameter of a truncated normal distribution with a known coefficient of variation, Comm. Statist.- Theory Methods A 8(3) (1979) 237-244.