Introduction
Statistical hypothesis testing attempts to answer the following questions: whether the null hypothesis is true, if there any extreme results which are favorable to the alternative hypothesis and what is the probability of getting the results observed?
The current paper aims to analyze whether the assumption that was made is true with the help of the statistical methodology of hypothesis testing.
It is important to mention that the principles of the hypothesis testing do not give a logical proof of its correctness. The results obtained from hypothesis testing can be considered a statement which is credible enough and does not contradict an existing experience. If the null hypothesis describes the data observations that result from chance, the alternative hypothesis reflects the data that result from some non-random cause.
In order to test a hypothesis, the null hypothesis and alternative hypothesis should be identified. Examination of a sample given is important for an identification of the null and the alternative hypothesis. The null hypothesis generally corresponds to a default position. The alternative hypothesis represents a competitive statement.
Testing hypothesis consists of several stages:
formulating the null and the alternative hypothesis;
choosing the level of significance;
imposition of the appropriate criterion for the sample with the normal distribution;
definition of the field of the hypothesis acceptance;
computation of the factual value of the criterion and checking the condition.
In this example the following data was analyzed: a sample of 20 respondents was taken to define whether they prefer to learn the information from printed newspapers or from the web resources. The survey results showed that the majority of the sampling (18 respondents or 90%) of 20 answered that they use web resources for news and 2 respondents (10%) answered they prefer printed newspapers.
In this case, the null hypothesis (H0) is as follows: more people use printed newspapers to learn news rather than the web sites.
The alternative hypothesis (H1) is as follows: more people use the news and the web sites for news entertainment rather than printed newspapers.
Let us assume that the distribution of p is normal and the level of significance is equal α = 0.05.The significance level was taken as 0.05 because this is one of the most common levels of significance used in statistics. There are other popular levels of significance which are subjective and are taken intuitively.
1). One of the ways of testing a hypothesis is to z-value computation. The following formula should be used for z-value computation:
H0 µ = 10, where µ is mean.
H1 µ > 10.
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Z = (x - µ):(σ: √n),
where x – value of variable, µ - mean of values, σ –standard deviation, n – sample size.
Putting the data from this example, the following equation will take place:
Z = (18-10):(4:4.47) = 9.0
In order to compute standard deviation, we will use the following formula:
n
σ = √ (∑ (Xi – μ)²)/N, where
i=1
Xi – the i-th data value, μ – the sample mean, N – the size of a sample.
Putting the data given we will receive the following equation:
σ = √ ((18 – 10)² + (2 – 10)²)/2 = 4.0.
The rejection of the null hypothesis will satisfy the following criteria:
z = (x - µ):(σ: √n) > 1.83 when α = 0.05.
As 9.0>1.83 the null hypothesis has to be rejected, i.e. more respondents prefer derive news from the web sources rather than from printed newspapers. The probability of the null hypothesis acceptance is 1.83% which is much less that the level of significance α = 0.05.
2). Another way to test a hypothesis is to compute p-value. P-value means the probability of the null hypothesis acceptance. P-value identifies the circumstances under which the null hypothesis should be accepted. The rejection rule is when p-value is ≤ α, or p ≤ 0.05.
The hypothesis testing is:
H0: p > α; H1: p ≤ ± α if α = 0.05. H0 is an assumption that more people use printed newspapers for news and H1 is an assumption that more people use web for news.
The test statistic obtained is: z = 9
The p-value = P( Z < z ) = P( Z < 9 ) = 0.0001
The two-tailed p-value is less than 0.0001. This difference is extremely statistically significant by conventional criteria.
Since the p-value is less than the significance level of 0.05 the null hypothesis should be rejected. To conclude, the alternate hypothesis p ≤ 0.05 is true.
Conclusion
The current paper aimed to answer a question whether people chosen for the survey prefer to derive news from the web resources rather than from printed newspapers. For analysis of a sample of twenty respondents was taken. Aiming to test a hypothesis the statistical methods of defining z-value and p-value were applied. Hypothesis testing results showed that the null hypothesis should be rejected because the probability of the null hypothesis acceptance was insignificant. Thus, the alternative hypothesis was accepted, namely: more people prefer to derive news from web resources.
