Tag | Value |
---|---|
file | Inferential_Statistics_vufsw-testing-1125-en_vufsw-testing-1125-en |
name | vufsw-testing-1125-en |
section | inferential statistics/confidence intervals/testing |
type | schoice |
solution | TRUE, FALSE, FALSE, FALSE |
Type | performing analysis |
Program | calculator |
Language | English |
Level | statistical literacy |
A researcher draws a sample of 50 Dutch citizens and examines whether the confidence of the Dutch in politics is higher than the European average, of which we have a reliable population value based on previous studies.
The sample mean is 6.38 (on a scale of 0-10) and the standard deviation is 1.55. The European average is 6.0. Test, with alpha = 0.05, whether the confidence of the Dutch in politics is higher than the European average. What is the conclusion?
You can not reject the Ha hypothesis, you can only reject the
H0 hypothesis.
In order to undertake hypothesis testing you need to express your
research hypothesis as a null and alternative hypothesis. The null
hypothesis and alternative hypothesis are statements regarding the
differences or effects that occur in the population. You will use your
sample to test which statement (i.e., the null hypothesis or alternative
hypothesis) is most likely (although technically, you test the evidence
against the null hypothesis). You can reject or not reject the
H0 hypothesis.
The H0 hypothesis in this example states that the mean confidence of the Dutch in politics is 6.38. The Ha hypothesis is that the mean is higher than 6.38.
se=1.55/(sqrt 50)=0.22
t= (6.38-6.00)/0.22=1.73
The t critical value (see table) for one tailed test (df=50) is 1.68. So, there is enough evidence to reject H0
If you want to learn more about hypothesis testing watch this clip.
If you want to learn more about the statistical test about the mean watch this clip.
Language
Nederlands
Levels of Difficulty
Difficult
M&T Hypothesis testing: proportions
Default value
M&T Multivariate Analyse
Default value
M&T Multivariate Analyse CW
Default value