Hypothesis Testing

A hypothesis test involves two hypotheses: Null hypothesis and alternative hypothesis.

 The null hypothesis denoted by  is the hypothesis to be tested.

The alternative hypothesis, or      is the research hypothesis. 

Problem: Should the null hypothesis be rejected in favor of the alternative hypothesis or should we fail to reject the null hypothesis?

Step 1. Choose the hypothesis.  Null hypothesis: should always specify a single value for the parameter. ______________

Step 2. Alternative hypothesis.  The choice of the alternative hypothesis depends upon what you are trying to test.  Three choices are possible:  _____________ , ____________, ___________

Identify the above hypotheses as a two tailed test, a left tailed test or a right tailed test.

Example 1.  An orange grower claims that his oranges have an average weight of 8.6 ounces. A random sample of 36 oranges from his orange grove produced a  mean of 7.8 ounces with a standard deviation of 2.1 ounces.  Does the data present sufficient evidence at the 5% level of significance to conclude that the average weight of oranges from this grove is less than 8.6 ounces? State the null and alternative hypothesis.

Terms

 Test statistic:

Rejection region

Non rejection region

Critical values

Type I and Type II errors:

Type I error:

Type II error:

Significance level:  The probability of making a type I error, that is, of rejecting a true null hypothesis , is called the significance level .

Possible conclusions for a hypothesis test.

· If the null hypothesis is rejected, we conclude that the alternative hypothesis is true.

· If the null hypothesis is not rejected, we conclude that the data do not provide sufficient evidence to support the alternative hypothesis. We fail to reject the null hypothesis. We never say “accept the null hypothesis”.

Graphical display of the decision criterion :

Procedure for the one-sample z-test for a population mean.

ASSUMPTIONS:  1. Normal population or large sample.

2.  is known .

1) Specify the null and alternative hypothesis.

2) Decide on the significance level, .

3) Determine the critical values ( based on one tailed, two tailed test) and specify the criterion for rejection of the null hypothesis.

4) Compute the value of the test statistic

5) Reject or fail to reject the null hypothesis based on the value of the test statistic and the criterion specified in step 3.

6) state the conclusion in words using non technical terms.

Example 1:  In tests of a computer component, it is found that the mean time between failures is 937 hours. A modification is made which is supposed to increase reliability by increasing the time between failures.  Tests on a sample of 36 modified components product a mean time between failures of 983 hours , with a standard deviation of 52 hours. At the 0.01 level of significance, test the claim that the modified components have a longer mean time between failures.  Interpret your results. ( we can assume that the standard deviation of the sample is a good approximation for the population standard deviation) .

Solution: Assumptions: normal population or large sample. Here we have a large sample ( 36) so it makes no difference if population is normal or not. We do not know population standard deviation but , in this case, we can use the sample standard deviation to estimate the population standard deviation.

1. Decide on the null and alternative hypothesis.

: = 937

Because the alternative hypothesis specifies that the mean is greater than 937, this is a one tailed test with the rejection region located in the right tail.

2. Significance level is 0.01, which is the same as the probability of a type I error. The number 0.01 is also the probability of rejecting the null hypothesis when in fact, it is true.

3. We will reject the null hypothesis in favor of the alternative hypothesis if the calculated value of z exceeds = 2.33. Our rejection region is located in the right tail of the normal curve and to the right of z = 2.33.

4. Compute the value of the test statistic: . So , z = = 5.30, which greatly exceeds the critical z value of 2.33.

5. Since =5.30 falls in the rejection region, our decision is to reject the null hypothesis in favor of the alternative hypothesis and state that the mean time between failures is greater than 937 hours. The probability that reject the null hypothesis when in fact it is true is less than 0.01.