| Tag | Value |
|---|---|
| file | Inferential_Statistics_vufsw-onesidedhypothesis-1167-en_vufsw-onesidedhypothesis-1167-en |
| name | vufsw-onesidedhypothesis-1167-en |
| section | inferential statistics/nhst/hypothesis/one sided hypothesis |
| type | schoice |
| solution | FALSE, FALSE, FALSE, TRUE |
| Type | calculation |
| Program | calculator |
| Language | English |
| Level | statistical thinking |
A researcher wants to test whether the proportion of Dutch students who have ever used cannabis is higher than 20%. The alternative hypothesis of the researcher is therefore as follows: Ha: π > 0.20. The observed proportion in the mentioned sample (n = 150) is 0.24.
The researcher uses a significance level of 5% (alpha = 0.05).
Is the null hypothesis H0 rejected in this study?
The true answer is: No, because there is not enough empirical evidence to assume that the proportion of Dutch pupils who ever used cannabis is higher than 0.20.
The alternative hypothesis is that the mean is higher than 0.2 and we found a proportion of 0.24. However,
Compute the standard error (se) of the sampling distribution. se =
sqrt[ P * ( 1 - P ) / n ]
where P is the hypothesized value of population proportion in the null
hypothesis, and n is the sample size.
For this example: σ = sqrt[ P * ( 1 - P ) / n ] = sqrt (0.2)(0.8)/150=0.033
The test statistic is a z-score (z) defined by the following equation: z
= (p - P) / se
where P is the hypothesized value of population proportion in the null
hypothesis, p is the sample proportion, and se is the standard deviation
of the sampling distribution.
z = (0.24-0.20)/0.033=1.22
The critical value (one tailed) is 1.64.
Because 1.22 is smaller than 1.64 there is not enough evidence.
Language
Nederlands
Levels of Difficulty
Medium
M&T BIS
Default value
M&T Hypothesis testing: proportions
Default value