First, consider the sample variance or the sample standard deviation, and then examine the confidence interval.
The variance and the standard deviation of your sample data provide an estimate of the population variance and the population standard deviation. Because the standard deviation and the variance are based on sample data and not on the entire population, it is unlikely that the sample standard deviation and the sample variance equal the population standard deviation and the population variance. To better estimate the population standard deviation and the population variance, use the confidence interval.
The confidence interval provides a range of likely values for the population standard deviation or the population variance. For example, a 95% confidence level indicates that if you take 100 random samples from the population, you could expect approximately 95 of the samples to produce intervals that contain the population standard deviation or the population variance. The confidence interval helps you assess the practical significance of your results. Use your specialized knowledge to determine whether the confidence interval includes values that have practical significance for your situation. If the interval is too wide to be useful, consider increasing your sample size. For more information, go to Ways to get a more precise confidence interval.
When you enter a column of data, Minitab only calculates a confidence interval for the standard deviation.
Minitab cannot calculate the Bonett method with summarized data.
N | StDev | Variance | 95% CI for σ using Bonett | 95% CI for σ using Chi-Square |
---|---|---|---|---|
50 | 0.871 | 0.759 | (0.704, 1.121) | (0.728, 1.085) |
In these results, the estimate of the population standard deviation for the length of beams is 0.871, and the estimate of the population variance is 0.759. Because the data did not pass a normality test, use the Bonett method. You can be 95% confident that the population standard deviation is between 0.704 and 1.121.
When you have summarized data, Minitab cannot calculate a p-value for the Bonett method.
Null hypothesis | H₀: σ = 1 |
---|---|
Alternative hypothesis | H₁: σ ≠ 1 |
Method | Test Statistic | DF | P-Value |
---|---|---|---|
Bonett | — | — | 0.275 |
Chi-Square | 37.17 | 49 | 0.215 |
In these results, the null hypothesis states that standard deviation of beam lengths equals 1. Because the data did not pass a normality test, use the p-value for the Bonett method. Because the p-value of 0.275 is greater than the significance level of 0.05, you fail to reject the null hypothesis and cannot conclude that the standard deviation is different from 1.