χ^{2} = (n - 1)s^{2}/σ_{0}^{2},
where σ_{0}^{2} is the hypothesized population variance.
The degrees of freedom, df = n - 1.
df = n - 1 = 30 - 1 = 29
Critical value of chi-square = 42.56 at alpha = .05.
Compute chi-square:
chi-square = (30 - 1)6.7/4.55 = 42.7
Decision: Reject H_{0}
Conclusion: The sample of 30 observations comes from a distribution with a variance greater than 4.55.
sdtesti 30 . 2.59 2.13 One-sample test of variance ------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 30 . .4728671 2.59 . . ------------------------------------------------------------------------------ Ho: sd(x) = 2.13 chi2(29) = 42.878 Ha: sd(x) < 2.13 Ha: sd(x) ~= 2.13 Ha: sd(x) > 2.13 P < chi2 = 0.9533 2*(P > chi2) = 0.0934 P > chi2 = 0.0467 use http://www.philender.com/courses/data/hsb2, clear sdtest math=10.2 One-sample test of variance ------------------------------------------------------------------------------ Variable | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- math | 200 52.645 .6624493 9.368448 51.33868 53.95132 ------------------------------------------------------------------------------ Ho: sd(math) = 10.2 chi2(199) = 167.876 Ha: sd(math) < 10.2 Ha: sd(math) ~= 10.2 Ha: sd(math) > 10.2 P < chi2 = 0.0531 2*(P < chi2) = 0.1061 P > chi2 = 0.9469
with degrees of freedom = n_{1} - 1 and n_{2} - 1.
Compute the F-test: F = s^{2}_{larger}/s^{2}_{smaller} = 42.6/16.8 = 2.54.
Compute degrees of freedom: df = 15 and 20.
The critical value of F at alpha = 0.05 is 2.20.
Decision: Reject H_{0}, the variances are significantly different.
display sqrt(16.8) 4.0987803 display sqrt(42.6) 6.5268675 sdtesti 21 84.3 4.1 16 83.7 6.53 Variance ratio test ------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 21 84.3 .8946933 4.1 82.4337 86.1663 y | 16 83.7 1.6325 6.53 80.22041 87.17959 ---------+-------------------------------------------------------------------- combined | 37 84.04054 .8573489 5.21505 82.30176 85.77932 ------------------------------------------------------------------------------ Ho: sd(x) = sd(y) F(20,15) observed = F_obs = 0.394 F(20,15) lower tail = F_L = F_obs = 0.394 F(20,15) upper tail = F_U = 1/F_obs = 2.537 Ha: sd(x) < sd(y) Ha: sd(x) ~= sd(y) Ha: sd(x) > sd(y) P < F_obs = 0.0267 P < F_L + P > F_U = 0.0622 P > F_obs = 0.9733
use http://www.gseis.ucla.edu/courses/data/hsb2 sdtest write, by(female) Variance ratio test ------------------------------------------------------------------------------ Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- male | 91 50.12088 1.080274 10.30516 47.97473 52.26703 female | 109 54.99083 .7790686 8.133715 53.44658 56.53507 ---------+-------------------------------------------------------------------- combined | 200 52.775 .6702372 9.478586 51.45332 54.09668 ------------------------------------------------------------------------------ Ho: sd(male) = sd(female) F(90,108) observed = F_obs = 1.605 F(90,108) lower tail = F_L = 1/F_obs = 0.623 F(90,108) upper tail = F_U = F_obs = 1.605 Ha: sd(1) < sd(2) Ha: sd(1) ~= sd(2) Ha: sd(1) > sd(2) P < F_obs = 0.9906 P < F_L + P > F_U = 0.0199 P > F_obs = 0.0094 histogram write, by(female) normal bin(10)
with degrees of freedom = k (number of groups) and n - 1 (number in each group).
You need to use the Table of the F-max Distribution.
Compute the F-test: F = s^{2}_{largest}/s^{2}_{smallest} = 2.214/0.857 = 2.58.
Calculate degrees of freedom: number of groups = 4, number in each group = 8,
df = 4 & 7.
Critical value of F-max at alpha = 0.05 is 8.44.
Decision: Fail to reject H_{0}, variances are not significantly different, there is homogeneity of variance.
with degrees of freedom = 1 and n - 2.
Compute F: F = r^{2}/((1-r^{2})/(n-2)) = .1024/((1-.1024)/(62-2)) = 6.84.
Compute degrees of freedom: df = 1 and n - 2 = 60.
Critical value of F at alpha = 0.05 is 4.00.
Decision: Reject H_{0}, correlation is statistically significant.