DSCI 346 Yamasaki Lecture 1 Hypothesis Tests for Single Population DSCI 346 Lecture 1 (22 pages)1

Terminology: 1. Null Hypothesis (H 0 ) e.g. - nothing new - no difference -no association 2. Alternative Hypothesis (H A ) 3. Test Statistic 4. Rejection Region, Critical Value 5. Type I error = P(reject H 0 when H 0 is true) = , also called significance level 6. Type II error = P(not rejecting H 0 when H 0 is false) =  7. Power = P(rejecting H 0 when H 0 is false) = 1-  2DSCI 346 Lecture 1 (22 pages)

H 0 : x ~ N(  0,  2 ) H A : x ~ N(  A,  2 )  A >  0 Suppose we decide to reject H 0 in favor of H A if x ≥ critical value 00 AA  critical value 3DSCI 346 Lecture 1 (22 pages)

One more bit of terminology The prior example is known as a one tailed test since the alternative was in only one direction, >. There are two tailed tests where the alternative is in either direction, ≠. AA 00 AA critical value  /2 4DSCI 346 Lecture 1 (22 pages)

Typical steps to Hypothesis Testing 1. Make some observations 2. Set up null hypothesis 3. Set up alternative hypothesis 4. Set significance level (  ) 5. Determine test statistic 6. (May have a predetermined , so need to determine minimum sample size) 7. Collect data 8. Calculate test statistic 9. Compare test statistic to critical value 10. Reject or not reject H 0 5DSCI 346 Lecture 1 (22 pages)

Example: A drug manufacturer claimed that the mean potency of one of its antibiotics was 80%. A random sample of 100 capsules were tested and produced a sample mean of x = 79.7%, with a standard deviation s =.8%. Do the data present sufficient evidence to refute the manufacturer’s claim? Let  =.05. H 0 :  = 80 a) Suppose that > 80 is okay. H A :  < 80  0 = 80  A < 80 critical value,c 6DSCI 346 Lecture 1 (22 pages)

If  0 is true, then we would expect x to be close to 80. We would reject H 0 if x is too low. How low is too low? P H 0 (x ≤ c) =  =.05 if H 0 is true, x N(  0,  2 /n) P H 0 (x ≤ c)=.05 P H 0 ( )= DSCI 346 Lecture 1 (22 pages)

Reject if We don’t have , but n is large enough to pretend  = s. n= 100 x = 79.7  =.8  0 = 80 Since z < , we can reject H 0 and conclude that  is significantly < 80% 8DSCI 346 Lecture 1 (22 pages)

b) Now let’s suppose that we want to reject if too high or too low (e.g. too low is not effective, too high is toxic). H 0 :  = 80 H A :  ≠ 80  =.05 Reject H 0 if z ≤ or if z ≥ z = ≤ -1.96, so we can reject H 0 and conclude  is significantly different from 80% DSCI 346 Lecture 1 (22 pages)

Note: 1) Suppose out statistic turned out to be We would not have been able to reject our one tailed test in part a). 2) Suppose our test statistics turned out to be We would not have been able to reject our two tailed test. By doing a one tailed test we concentrate the rejection region into one tail and thus have a better chance of rejecting the null hypothesis; however we need to be sure we aren’t interested in the other tail. 10DSCI 346 Lecture 1 (22 pages)

Note: 1) Suppose out statistic turned out to be We would not have been able to reject our one tailed test in part a). 2) Suppose our test statistics turned out to be We would not have been able to reject our two tailed test. By doing a one tailed test we concentrate the rejection region into one tail and thus have a better chance of rejecting the null hypothesis; however we need to be sure we aren’t interested in the other tail. 11DSCI 346 Lecture 1 (22 pages)

Definition: P value is the probability of getting what we observed or more extreme, if the null hypothesis were indeed true. More extreme meaning further away from what would be expected. Example: Suppose we have H 0 :  =  0 H A :  ≠  0  =.05 We reject if z ≤ - z  /2 or z ≥ + z  /2 Suppose the value of our test statistic was, say, z= DSCI 346 Lecture 1 (22 pages)

If the null hypothesis were true, what is more extreme than z=2.90? z≥ P H 0 (z≥2.90) = Since we have a two tailed alternative we need to add in P H 0 (z≤ -2.90)=.0019 So P value = = = DSCI 346 Lecture 1 (22 pages)

If we had H A  >  0 then more extreme is only z ≥ 2.90 so P value is If we had H A  <  0 then more extreme is z ≤ 2.90 so P value is DSCI 346 Lecture 1 (22 pages)

Let’s assume that we are doing a two tailed test in our last example. Would we reject at  =.01? Yes How about  =.005? Yes We would be able to reject all the way down to  = So p value is the minimum significance level at which we can still reject H 0. Final note: Hypothesis tests, via p values, can give you an idea of how unlikely your results are under the null hypothesis. Confidence intervals give you a guess as to where the true parameter is. 15DSCI 346 Lecture 1 (22 pages)

Small Sample Hypothesis Testing for a single mean Before we had Now we have So other than having to use a different table, we are doing the same thing as we did with a large sample. 16DSCI 346 Lecture 1 (22 pages)

Example: In 1985 the national mean for the verbal SAT score was 431 which was lower than anticipated. There is a concern that the average is still falling. In order to investigate this, a sampling of 20 scores from current high school students was drawn with the following results: x = 419 s = 57 Do the data provide sufficient evidence to indicate the average is still falling?  = Since -.94 > we conclude there is insufficient evidence to indicate averages are still falling 17DSCI 346 Lecture 1 (22 pages)

What is the P-value of your test statistic? Since P(t 19 ≤ ) =.10 we know P value = P(t 19 ≤ -.94) >.10 And P(t 19 ≤ ) =.20 we know P value = P(t 19 ≤ -.94) <.20 So.10 < P value <.20 18DSCI 346 Lecture 1 (22 pages)

The equivalence between confidence intervals and hypothesis tests involving means. There is an equivalence between confidence intervals and hypothesis tests involving means in that any value of the parameter that is not in a (1-  )% confidence interval will be rejected in a two tailed,  significance level hypothesis test and vice versa. 19DSCI 346 Lecture 1 (22 pages)

Hypothesis Test involving a single proportion Earlier we stated that by the Central Limit Theorem, we have, when n large enough Now lets consider a one tailed hypothesis test H 0 :  =  0 H A :  >  0 So if H 0 is true then we have Not only is the mean specified, but also the variance 20DSCI 346 Lecture 1 (22 pages)

Example: A financial company has established a standard of a 2% noncompliance rate. If more than 2% of its accounts do not have appropriate documentation, then the internal controls are not effective. The company samples 600 accounts and finds 18 are lacking adequate documentation. Is there sufficient evidence to indicate the internal controls are not effective Test using  =.01. H 0 :  =.02 H A :  >.02 21DSCI 346 Lecture 1 (22 pages)

 =.01 So reject H 0 if z ≥ Since 1.75 < we have insufficient evidence to reject H 0 ; that is there is insufficient evidence to conclude that the internal controls are not effective DSCI 346 Lecture 1 (22 pages)