Chapter 12 Tests of Hypotheses Means 12.1 Tests of Hypotheses 12.2 Significance of Tests 12.3 Tests concerning Means 12.4 Tests concerning Means(unknown variance) 12.5 Differences between Means 12.6 Differences between Means(unknown variances) 12.7 Paired Data

12.1 Tests of Hypotheses In a law case, there are 2 possibilities for the truth—innocent or guilty Evidence is gathered to decide whether to convict the defendant. The defendant is considered innocent unless “proven” to be guilty “beyond a reasonable doubt.” Just because a defendant is not found to be guilty doesn’t prove the defendant is innocent. If there is not much evidence one way or the other the defendant is not found to be guilty.

For Statistical Hypothesis Testing H 0 =Null hypothesis (innocent) Held on to unless there is sufficient evidence to the contrary H A =Alternative hypothesis (guilty) We reject H 0 in favor of H A if there is enough evidence favoring H A

12.1 Tests of Hypotheses Distribution(s) or population(s): Parameter(s) such as mean and variance Assertion or conjecture about the population(s) – statistical hypotheses 1. About parameter(s): means or variances 2. About the type of populations: normal, binomial, or …

Example 12.1 Is a coin balanced? This is the same as to ask if p=0.5 Is the average lifetime of a light bulb equal to 1000 hours? The assertion is μ=1000

Null Hypotheses and Alternatives We call the above two assertions Null Hypotheses Notation: H 0 : p=0.5 and H 0 : μ=1000 If we reject the above null hypotheses, the appropriate conclusions we arrive are called alternative hypotheses H A : p  0.5 H A : μ  1000

Null Hypothesis vs Alternative H 0 : p=0.5 vs H A : p  0.5 H 0 :μ=1000 vs H A : μ  1000 It is possible for you to specify other alternatives H A : p>0.5 or H A : p<0.5 H A : μ>1000 or H A : μ<1000

12.2 Significance of Tests A company claims its light bulbs last on average 1000 hours. We are going to test that claim. We might take the null and alternative hypotheses to be H 0 :μ=1000 vs H A : μ  1000 or may be H 0 :μ=1000 vs H A : μ<1000

Mistakes or errors: Law case—convict an innocent defendant; or fail to convict a guilty defendant. The law system is set up so that the chance of convicting an innocent person is small. Innocent until “proven guilty” beyond a reasonable doubt.

Two Types of Errors in statistical testing Type I error -- reject H 0 when it is true (convict innocent person) Type II error -- accept H 0 when it is not true (find guilty person innocent)

Statistical hypotheses are set up to Control type I error  =P(type I error) =P(reject H 0 when H 0 true) (a small number) Minimize type II error  =P(type II error) =P(accept H 0 when H 0 false)

Control Types of Errors In practice,  is set at some small values, usually 0.05 If you want to control  at some small values, you need to figure out how large a sample size (n) is required to have a small  also. 1-  is called the power of the test 1-  =Power=P(reject H 0 when H 0 false)

Example 12.2 X=breaking strength of a fish line, normal distributed with σ =0.10. Claim: mean is  =10 H 0 :  =10 vs H A :  10 A random sample of size n=10 is taken, and sample mean is calculated Accept H 0 if Type I error? Type II error when  =10.10?

Solution Type I error=P(reject H 0 when  =10)

Solution Type II error=P(accept H 0 when H 0 false) Power= =0.9429

12.3 & 12.4 Tests concerning Means A company claims its light bulbs last an average 1,000 hours 5 steps to set up a statistical hypothesis test

5 steps: step 1 1. Set up H 0 and H A H 0 :  =1,000 vs H A :  <1,000 This is a one-sided alternative. Other possibilities include H 0 :  =1,000 vs H A :  1,000 (Two sided alternative) Note: we could write H 0 :  ≥1,000, but in this book H 0 is always written with an equal (=) sign.

5 steps: step 2 and 3 2. Specify  =P(type I error): level of significance.  =0.05 usually. This corresponds to 95% confidence. 3. Decide on sample size, n, and specify when to reject H 0 based on some statistic so that  =P(Reject H 0 when H 0 is true)

Step 3 continued Suppose we use n=10 bulbs. Find the sample mean, and compare to We need to set a probability to  =0.05, so we want a statistic we can compare to a table of probabilities. If we know , then set z has a standard normal distribution if  =1,000 and then we can use the normal table.

Step 3 continued Reject H 0 :  =1,000 in favor of H A :  <1,000 if the sample mean is too far below This will give us a negative value of z. How far below 0 does z have to be for us to reject H 0 ? The rejection region is set up so that the probability of rejecting H 0 is only a=5% if H 0 is true. So we reject H 0 if

Step 3 continued: if  is unknown If  is unknown, the usual situation, and the population is normal, we use a t distribution. Calculate sample deviation s: Rejection region:

5 steps: step 4 4. Collect the data and compute the statistics: z or t Suppose, s=30, n=10 then

5 steps: step 5 5. Decide whether to reject H 0 t=-3.16< is in the rejection region Reject H 0 :  =1,000 in favor of H A :  <1,000 at  =0.05 level.

5 steps summary 1. hypothesis statement 2. Specify level of significance  3. determine the rejection region 4. compute the test statistic from data 5. conclusion

Relationship Between Hypotheses Testing and Confidence Intervals For two tailed test: To accept null hypothesis at level  H 0 :  =  0 is equivalent to showing  0 is in the (1-  ) Confidence Interval for .

Example 12.3 Normal population.  unknown H 0 :  =750 vs H A :  750 Define Reject H 0 if the sample mean is too far from 750 in either direction Rejection region:

Example Let’s take  =0.05, n=20 (df=19) Data turned out to be Get t: t =2.093 |t|<2.093 Conclusion: accept H 0

Example 12.3 (continued)  =50 is known,  =0.05, n=20 H 0 :  =750 vs H A :  750 Reject region |z|>z 0.025=1.96 Calculate z |z|<1.96. Accept H 0 Question: if  =0.10, what is the conclusion?

More cases H 0 :  =1000 vs H A :  >1000 Define t or z statistics (  unknown or known) Rejection regions: t>t  or z>z 

Rejection Regions: Alternative Hypotheses  >  0  <  0   0 Rejection Regions z>z  t>t  z<-z  t<-t  z>z  /2 or z<-z  / t>t  /2 or t<-t  /2

P-value In practice more commonly one performs the test by computing a p-value. The book describes revised steps 3, 4, 5 as 3.’ We specify the test statistic. 4.’ Using the data we compute the test statistic and find its p-value. 5.’ If p-value< , reject H 0.

P-value The book’s definition of p-value: A p-value is the lowest level of  at which we could reject H 0. A more usual way to think about p-values: If H 0 is true, what is the probability of observing data with this much or more evidence against H 0.

Example H 0 :  =120 vs H A :  <120  =0.05 From the data z=-1.78 Evidence against H 0 is sample mean less than 120, meaning z<0. P-value<0.05  Reject H 0 :  =120 in favor of H A :  <120 The reject/accept H 0 decision is the same as comparing z to , but the p-value gives more information—How inconsistent are the data with H 0 ?

If z is less than , then the p- value is less than Comparing the p-value to 0.05 is the same as comparing the z value to For t tests we can find the exact p- value without a calculator or software.

P-value for 2-sided test H 0 :  =120 vs H A :  ≠120  =0.1 From the data z=1.32 Evidence against H 0 is z values away from 0 in either direction P-value=2*0.0934= P-value> , Do not reject H 0. 2-sided p-value=2*(1-sided p-value)

Exercise Given that n=25,  =100, and sample mean is 1050, 1. Test the hypotheses H 0 :  =1000 vs H A :  <1000 at level  = Test the hypotheses H 0 :  =1000 vs H A :  ≠1000 at level  =0.05.

Solution More evidence against H 0 is smaller values of z Evidence against H 0 is z values away from 0 in either direction

A word of caution NOTE: Accepting H 0 does not prove H 0 is true. There are many other possible values in the confidence interval. OPINION: In most situations it would be more useful to report confidence intervals rather than results of hypothesis tests.