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Text: Section 9.1, 9.2, 9.3, 9.
Hypothesis Tests Continued
Recall that we agreed to conduct our tests with the following 6 steps:
- State clearly what your variables are (define your terms).
- State the null and alternative hypotheses.
- Decide upon a level of significance, ๐ผ.
- Compute a test statistic (๐ง, ๐ก, ๐^2 , ๐๐๐ ๐น are popular stats).
- Find the ๐-value corresponding to your test statistic (for left/right/or two tailed test).
- Form a conclusion: if ๐ < ๐ผ (improbable data) reject ๐ป 0 otherwise do not reject. We never accept, just like the courts never say that someone is innocent.
Hereโs another quote: โOnce you leave behind such class concerns as how to balance the peas on the back of a fork, all the important rules surely boil down to one: remember you are with other people; show some consideration.โ (Truss) I include it because in Step 6. You should never say something like โreject the nullโ but instead say something like โThe data are inconsistent with STA 100 students having a mean IQ of 100. We conclude that their mean IQ is higher at the 99% level of significanceโ or something like that.
Tests Concerning a Sample Proportion
We have already seen how to test a sample proportion with our coin toss example. Hereโs another example. Try to work it out yourself before looking at the answer.
A newspaper article contains the statement that, nationwide, 60% of all college seniors have a job prior to graduation. The director of a college placement office at a large university is interested in
testing this claim for her university. If a random sample of 75 recent graduates showed that 40 had a job prior to graduation, what conclusion can be drawn? Use a 0.05 level of significance.
We work through the โ6 Stepsโ
- We will use the symbol ๐๐ข๐๐๐ฃ๐๐๐ ๐๐ก๐ฆ for the proportion of students who hold a job at the particular university under discussion. Note that ๐๐๐๐ก๐๐๐๐๐ = 0.6 but we donโt know ๐๐ข๐๐๐ฃ๐๐๐ ๐๐ก๐ฆ. 2. ๐ป 0 : ๐๐ข๐๐๐ฃ๐๐๐ ๐๐ก๐ฆ = 0.6, ๐ป 1 : ๐๐ข๐๐๐ฃ๐๐๐ ๐๐ก๐ฆ < 0.6. (Note that weโve set the alternative hypothesis up to โtripโ if we get a sample proportion too small to be consistent with a local value of 60%. If you are the counselor you will probably set the test up this way so that the test really just checks to see if the university is too deficient.)
- ๐ผ = 0.05 (There is a strategy here, too. If you are the counselor you would probably take ๐ผ = 0.01 so that it is more difficult to conclude that you are doing poorly compared to the national average).
- We are dealing with proportions and decently large sample sizes, so compute a ๐ง. (If the sample size was much smaller, as it can be in medical studies, you would have to compute your statistic using the binomial random variable directly, not the large sample approximation).
๐ง = ๐ โ ๐๐ข๐๐๐ฃ๐๐๐ ๐๐ก๐ฆ ๐๐ข๐๐๐ฃ๐๐๐ ๐๐ก๐ฆ 1 โ ๐๐ข๐๐๐ฃ๐๐๐ ๐๐ก๐ฆ ๐
75 โ^ 0.
- This is a โleft tailedโ test, so we compute the probability of observing a sample statistic this low (40/75=53%) if the proportion at the university is really 60%. Using the z table: ๐ = 0.1190.
- Since ๐ > ๐ผ (that is, our data are not โtoo unlikelyโ, we fail to reject the null hypothesis. Remember we reject in the context of improbable data (reject small p). We would write: Since the data are consistent with the proportion at the university being equal to 60% we do not have sufficient evidence to conclude that ๐๐ข๐๐๐ฃ๐๐๐ ๐๐ก๐ฆ < 0.6.
valid if you can assume that the population is at least approximately normally distributed. If not, and if your sample size is small, you are in trouble. We have ๐ก = ๐ฅ โ ๐ ๐ ๐ ๐ก๐๐๐ก ๐
= 708.0097.1642^ โ^650
- Find the ๐-value corresponding to your test statistic (for left/right/or two tailed test). If you must use the table in the book this is slightly tricky. We wonโt find the t-value 2.4612 in our table exactly, so we need to โsandwichโ our value between those given in the table. I see the following along the row for ๐๐ = ๐ โ 1 = 16 : One-tail area 0.025 0. 16 2.120 2.4612 2.
Since our t-value is between 2.120 and 2.583 we can say our p-value (one tail area) is between 0.025 and 0.01. That is, it is small.
- Form a conclusion: if ๐ < ๐ผ (improbable data) reject ๐ป 0 otherwise do not reject. We have ๐ < ๐ผ = 0.05 so we reject the null hypothesis and conclude: The observed data are inconsistent with an average selling price of $650 at the 5% level of significance and so we conclude that the mean selling price for used American Stratocasters is greater than $650.
Obviously since this test is so common Excel will help you along your way. You can just type: = ๐๐ท๐ผ๐๐(2.4612,16,1) where 2.4612 is the t value you would like an area for, 16 is the degrees of freedom, and 1 means 1 tail (itโs a one tailed test). I got 0.012796222 which is between 0.01 and 0.025.
What happens when we have more than one population?
Tests Concerning Two Population Means- Independent Samples
Suppose you are now wondering, which keeps its value better: A Fender Stratocaster Guitar or a Fender Jazz Bass? Luckily I have some data available for you.
Table 1 Fender American Stratocaster Guitar: sample mean=708.00, sample standard deviation=97.1642, sample size= 770 775 700 800 721 700 629 849 752 788 661 669 860 560 510 690 602
Table 2 Fender American Jazz Bass: sample mean=748.44, sample standard deviation=170.13, sample size= 699 469 760 899 620 860 899 960 570
Now recall what we meant by independent samples when we were constructing confidence intervals: an individualโs inclusion on one group has nothing to do with their inclusion in another group. This is in contrast to Before/After, Father/Son, etc.
For instance, if you wanted to know โIs the average ring finger length greater than the average index finger length?โ you could collect your data in two obvious ways:
- Walk up to 50 people and measure their index fingers, walk up to another 50 people and measure their ring finger lengths and then compare the averages. This is the independent samples approach.
- Walk up to 50 people and measure their index fingers, measure their ring finger lengths and compare the averages. This is the paired samples approach.
In both cases we have 50 ring finger measurements and 50 index finger measurements, but the analysis is different. The first situation is for independent samples while the second is for โpaired dataโ. There are reasons why we generally prefer paired data in addition to the obvious reduction in effort to obtain data exhibited in the example above.
- Find the ๐-value corresponding to your test statistic (for left/right/or two tailed test). If you like to use the table, look up t values that โsurroundโ the value you want a p-value for. I see Two-tail area 0. 8 0.659 0. Our value is too large for the table the way it is set up. Notice that the table assumes you can figure out that the t distribution is symmetrical, so we can find areas on the positive side to get areas on the negative side. That is, just drop the sign on your p- value and use the table.
Using Excel, the command = ๐๐ท๐ผ๐๐ 0.659,8, returns the value 0.528403918. If you try to run this with a negative sign youโll get an error.
- Form a conclusion: if ๐ < ๐ผ (improbable data) reject ๐ป 0 otherwise do not reject. We never accept, just like the courts never say that someone is innocent. Since our p-value is very large our data are not very improbable at all, in fact a mean difference of 708.00 โ 748.44 = โ40.44 is fairly likely even if the population means are equal. There is so much variability, and your sample sizes are so small, that a difference as large as this may easily arise by chance.
Here is another possible presentation topic.
The following problem is taken from Devore and Peck, Introductory Statistics. The paper Anthropometric and Physical Performance Characteristics of Male Volleyball Players, (Canadian J. of Appl. Sports Sci. (1982): 182-188) reported on a comparison study of Russian and Finnish volleyball players. One of the variables studied was the height increase in body center of gravity (cm) during a vertical jumping test. Summary data appears in the accompanying table.
Nationality Sample Size Sample Mean Sample SD Finns n1 = 14 x1 = 46.0 s1 = 3. Russians n2 = 10 x2 = 49.4 s2 = 4.
The authors of the paper assumed for purposes of analysis that both height-increase population distributions (Finns and Russians) were normal with the same unknown standard deviation. Do you believe that there is a significant difference between the sample means?
And another:
The following is taken from Moore and McCabe, Introduction to the Practice of Statistics. In a double blind experiment to determine whether calcium would be effective in the lowering of human blood pressure a control group was given a placebo while another group was given a calcium supplement. The data are summarized as follows:
Treatment Sample Size Mean Decrease in BP Sample SD Calcium n1 = 10 x1 = 5.0 s1 = 8. Placebo n2 = 11 x2 = โ0.273 s2 = 5.
Do the data support the claim that calcium lowers blood pressure?
