|Explanations & examples:
Calculate the 95 % confidence interval and the 95 % prediction interval of one or more mean values. Or choose a different percentual level than 95 and get the intervals of your choice. Enter the info about the mean that you have; either
1) the mean, the standard deviation and the sample size (N)
2) the mean, the standard error and the sample size (N), or
3) the mean, the 95 % confidence interval and the sample size (N).
In the section "Description of Means" you will have both the approximated 95 % confidence interval and the exact 95 % confidence interval of the mean value that you have chosen with the drop down selector to the left. The approximated interval is calculated using the z-distribution and the exact interval using the t-distribution. Furthermore, you can test whether the mean value could be equal to a certain value using both a z-test and a t-test. This is done by entering the value into the field "H0" and press enter. If the p-value in the test of your choice (either z or t) is below 0.05 then the null hypothesis H0 can be rejected on a five percent significance level. The null hypothesis stated that the mean value is equal to the value entered into the input field. You can also test whether the mean value could be either greater or smaller than the input value by changing the "=" sign to either ">" or "<" in the selector option.
In the next section "Description of Standard Deviations" you can get the 95 % confidence interval of the standard deviation (SD) of the mean values that you have chosen in the drop-down selector menus. You can also get a different level confidence interval than 95 by choosing another percentage with the percentage button. You can test whether the standard deviation of a mean could be equal to a certain value by entering the value into the H0 input field of the. The null hypothesis H0 will then be; the SD is equal to the entered value (i.e. there is no statistically significant difference between them). If the p-value in the chi-squared test is below 0.05 then the null hypothesis is rejected on a five percent significance level and the alternative hypothesis H1 is accepted instead; namely that the SD value is different from the entered value.
In the section "Test" it is tested whether two selected mean values could be equal, using both a z-test and a t-test. You can choose which two mean value you want to test using the drop-down selector boxes (provided you have entered at least two mean values in the section "Means" above). In both tests the null hypothesis is the same, namely that there is no statistically significant difference between the two means (in other words that their MD, "mean difference", could be equal to 0). When using a t-test, however, it is necessary to first determine which version of the t-test should be used;
1) the one where the two variances (SD2) of the mean values are assumed the same, or
2) the one where they are assumed different.
To determine if the two SD2 values could be the same (and thus which version t-test to use), an f-test is being performed on the two SD2 values. If the p-value in the f-test is below 0.05 we reject the null hypothesis that the two SD2 could be equal. In this case the t-test version "assuming diffferent variances" should be used. Otherwise the t-test version "assuming same variance" is used, in this case the common variance (SD2) value for both mean values is being calculated. If the p-value of the corresponding t-test is below 0.05 then the null hypothesis (that the two mean values are equal) is being rejected on a five percent significance level. If p > 0.05 then the alternative hypothesis H1 is being accepted as true, namely that the two mean values in question are equal (there is no statistically significant difference between them). If the p-value is below 0.05 then the two 95 % confidence intervals of the mean values will not overlap each other. If p > 0.05 then one of the mean values is included in the confidence interval of the other mean (or they are both included in each other's intervals). If there is a small overlap between the intervals (and none of the means is included in the other mean's interval), then we can't say anything beforehand about the p-value but will have to perform the test. For more info on the calculations and the formulas used in the tests, please see the page medical statistics formulas.
In the section "Weighted Average of all Means" the weighted average (weighted estimate) of all the mean values entered above is being calculated; taking into consideration the different sample sizes N in each group. The weighted average is the common mean value for all groups combined into one. It is only meaningful to use and interpret the weighted average in this way if the groups being combined do not have significantly different mean values from each other (p-value above 0.05). If the groups involved have different mean values (p < 0.05) then it isn't reasonable to combine the means into one using the weighted average.
Example:Out of 656 babies born in a country, 333 were boys and 323 were girls. The mean weight of the boys was 3.670 kilograms with standard deviation 0.452 kilograms. The mean weight of the girls was 3.538 kg with SD = 0.463 kg. We want to investigate whether boys and girls could have the same mean weight:
The null hypothesis H0 in this case is; the mean weight of baby boys is the same as the mean weight of baby girls in the country (alternatively: there is no statistically significant difference between the weight of newborn boys and girls in the country on average). First it is noticed that there is no overlap between the 95 % confidence intervals of the mean values, namely [3.6215 : 3.7185] and [3.4875 : 3.5885] respectively. Since the intervals do not even overlap each other, the two mean values could never be equal to each other on a 5 % significance level and we can therefore already now reject the null hypothesis without even doing the test; since we know already that the p-value will be below 0.05 in both the z-test and the t-test. If you want to do the test anyway; it is noticed that (in this case) the p-value is exactly the same in both the z-test as well as both versions of the t-test, coincidently, namely p = 0.0002. In any case it's less than 0.05 wherefore H0 is rejected on a five percent significance level. In this example we would use the t-test version "assuming same variance" since the p-value in the f-test is 0.6640 which is more than 0.05. So we can't reject that the two variances are the same.
Having rejected the null hypothesis in this case we conclude that there is, in fact, a significant difference between the mean weight of newborn boys and the mean weight of newborn girls. This result is then valid for the entire population (all the newborns in the country), tested on this sample of 656 babies taken from the population.
No. of groups:
What is known?
What is known?