Hypothesis Testing Is also called significance testing Tests a claim about a parameter using evidence (data in a sample The technique is introduced by considering a one-sample (Z- test and T-test) The procedure is broken into four steps Each element of the procedure must be understood
Null and Alternative Hypotheses Convert the research question to null and alternative hypotheses The null hypothesis ( H ) is a claim of “no difference in the population” (currently accepted value of a parameter). The alternative hypothesis ( H a ) claims “ H is false” Collect data and seek evidence against H as a way of rejecting H a (deduction)
SHORT EXAMPLE The problem: In the 1970s, 20–29 year old men in the U G . had a mean μ body weight of 170 pounds. Standard deviation σ was 40 pounds. We test whether mean body weight in the population now differs. Null hypothesis H 0: μ = 170 (“no difference”) The alternative hypothesis H a: μ ≠ 170 ( two-sided test )
Test Statistic Calculate the necessary sample statistics.The test statistic either Z statistic o r t statistic according to its characteristics. Z-test: This test is used if the population variance is known or if the sample size is larger than 30
continuation T-test: This test is used when the sample size is less than 30 and the population variance is known
When to Reject H and its Consequences Select a method to use for the test either P-Value method or Traditional approach (Critical value Method) . Draw a conclusion and interprete the decision Reject Ho if P-Value <α or the test statistic in the rejection region Fail to reject Ho if the P-value >α or the test statistic not in the rejection region.
ILLUSTRATIVE EXAMPLE I.Label on a large can of hilt-op coffee states that the average weight of coffee contained in all cans it produces is 3 pounds of coffee .A coffee drinker association claims average weight is less than 3 pounds of coffee, Suppose a random sample of 30 cans has an average weight of 2.95 pounds and standard deviation of s=0.18.Does data support Coffee drinker association’s claim at α=0.05?
P VALUE APPROACH SOLUTION n=30 ẋ=2.95 s=0.18 α=0.05 P-value approach Ho: μ=3 Ha:μ<3 (Left tailed test) Test statistic P value=P(ẋ≤2.95)=P(t≤-1.52)=0.06 Conclusion Since P value =0.06>α=0.05 , we fail to reject Ho μ=3 Therefore data doesn’t support the coffee drinkers association claim.
TRADITIONAL APPROACH HO: μ=3 Ha:μ<3 (Left tailed test) Test statistic Df=n-1=30-1=29 Critical value from table is -1.699 ≈ 1.7 Conclusion Since t=-1.52<to=-1.699 We fail to reject Ho μ=3 Therefore data doesn’t support the coffee drinkers association claim.
Possible Errors of Decision In decision-making, there is therefore the possibility of committing an error, which could either be an error of Type I or an error of Type II.
Assessing the Two Types of Errors From the table in the preceding slide, we have: Type I error : committed when H is rejected when in reality it is true. Type II error : committed when H is not rejected when in reality it is false. Just like in the court trial, an error of Type I is considered to be a more serious type of error (“convicting an innocent man”). Therefore, we try to minimize the probability of committing the Type I error.