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Hypothesis testing I

Two-tailed tests

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Tutor: Bilal

Summary

Two-tailed tests

In a nutshell

The procedure of conducting two-tailed hypothesis tests is very similar to that of one-tailed tests. The main difference is that the relevant probability is compared to half of the significance level.



How to conduct a two-tailed test

To conduct a two-tailed test, follow this procedure.


procedure

1.

Define the test statistic and model it with the appropriate binomial distribution.

2.

Write down the null and alternative hypotheses.

3.

Assume the null hypothesis to be true.

4.

Calculate the relevant cumulative probability and compare it with half of the significance level.

  • If the probability is less than half of the significance level, reject the null hypothesis.
  • If the probability is greater than half of the significance level, do not reject the null hypothesis.

5.

Write a conclusion based on the context of the question.


Deciding the relevant probability

Because the test is two-tailed, the alternative hypothesis does not indicate the inequality for the relevant probability. Instead, use the mean:


The mean of a binomial distribution with parameters nnn and ppp is npnpnp.


If the observed value is less than the mean, then the relevant probability will have a ≤\leq≤ sign.​


Example 1

A casino is suspected of using weighted dice. A particular die is tested for whether or not it is biased towards or against landing on 111. The die is rolled 303030 times and it lands on 111​ only once. Test, at the 10%10\%10%​ significance level, whether or not the die is biased.


Define the test statistic and write down the hypotheses:

Let XXX be the number of times the die lands on 111.
​X∼B(30,p)X\sim B(30,p)X∼B(30,p)​​


​H0:p=16H_0:p=\dfrac{1}{6}H0​:p=61​​​


​H1:p≠16H_1:p\neq\dfrac{1}{6}H1​:p=61​​​


Assume the null hypothesis to be true:

​p=16⇒X∼B(30,16)p=\dfrac{1}{6}\Rightarrow X\sim B(30,\dfrac16)p=61​⇒X∼B(30,61​)​​


Calculate the relevant probability:

The mean of this binomial distribution is np=30×16=5np=30\times\dfrac16=5np=30×61​=5​. Because 1<51<51<5, the relevant probability is P(X≤1)P(X\leq 1)P(X≤1).


​P(X≤1)=0.0295P(X\le1)=0.0295P(X≤1)=0.0295​​


Compare this with half of the significance level:

​0.0295<0.1÷2=0.050.0295 < 0.1\div2=0.050.0295<0.1÷2=0.05​, so reject the null hypothesis.


Conclude based on the context of the question:

There is sufficient evidence to suggest that the die is biased at the 10% level of significance.


Note: It is also possible to find the critical values of the test and see whether or not the observed value lies in the critical region, but the procedure above is quicker.


​

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FAQs - Frequently Asked Questions

How do you decide whether or not to reject the null hypothesis?

If the relevant probability is less than half of the significance value, then reject the null hypothesis.

How do you deduce the inequality for the relevant probability?

Compare the observed value to the mean of the distribution. If X ~ B(n,p), then the mean of X is np.

What is a two-tailed test?

A two-tailed test is a hypothesis test with an alternative hypothesis of the form H_1: p ≠ _.

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