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Psych 5500/6500. The t Test for a Single Group Mean (Part 1): Two-tail Tests & Confidence Intervals. Fall, 2008. Back to Our Example.
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Psych 5500/6500 The t Test for a Single Group Mean (Part 1): Two-tail Tests & Confidence Intervals Fall, 2008
Back to Our Example We are testing a theory which predicts that Elbonians should have a different mean IQ than people in the USA. In general H0 is the hypothesis of ‘no difference’, while HA is the hypothesis that reflects what the theory predicts. H0: μElbonia = 100 (same as USA) HA: μElbonia 100 (different than the USA)
Challenge The sample mean of the Elbonians is 106, the challenge is to determine whether this is because the population mean of Elbonians is different than 100, or if the Elbonians have a population mean of 100 and we obtained a sample mean six above that just due to chance (i.e. our sample had random bias).
Approach • Determine the probability of obtaining our data if H0 were true. • If that probability is low enough (less than or equal to our significance level of .05) then reject H0.
Test Statistic The statistic that is going to make or break H0 is the sample mean, this is our ‘test statistic’. We need to see what values the test statistic could take on if H0 were true.
Sampling Distribution This is the population of values the sample mean could be if H0 is true.
Normality of the Sampling Distribution We will be estimating the standard deviation of the population so the SDM will be modeled using the t distribution, rather than the normal distribution. For the probabilities in the t table to be accurate we need to either sample our scores from a normal population or make sure that our sample N is 30 or greater. As we will be using a small N in this example (to make number crunching easier) let’s assume that the IQ scores of Elbonians are normally distributed.
Mean of the SDM We are looking at the sampling distribution of the mean assuming H0 is true. If H0 is true, then the mean IQ of Elbonians is 100, we will use that to determine the mean of the SDM.
Standard Error To find the standard deviation of the SDM (i.e. the ‘standard error’) we need to know the standard deviation of the population. We don’t, so we will have to estimate it from our sample of 6 scores.
Rejection Regions Now, we are interested in those values of the sample mean that have a 5% chance or less of happening if H0 were true. If we get a sample mean in that area of the curve we will decide to ‘reject H0’. If we get a sample mean close to what H0 predicts (i.e. close to 100) then we will ‘not reject H0’.
tcritical values The next step is to find out how many standard deviations above and below the mean we have to go to cut off the 5% most unlikely means (2.5% on each tail). d.f.= N -1 = 6 -1= 5 Looking in the t table we see that leads to a t value of 2.571 (the lower tail would be -2.571). As this establishes where we make our decision about H0, these are called the ‘tcritical values’.
tobtained value OK, our criteria are set, there is only a 5% chance that the sample mean will fall 2.571 or more standard deviations away from the mean if H0 is true, if the sample mean does fall that far away from the mean then we will reject H0. Now, the sample mean we obtained was 106, we need to change that into a standard score to see where if falls on our curve. The standard score of the sample mean is called the ‘tobtained value’.
tobtained value (cont.) Remember that a standard score is always some point on a curve minus the mean of the curve divided by the standard deviation of the curve. So, our sample mean of 106 falls almost four standard deviations above the mean on the curve representing what H0 predicts.
SDM (so far) Our decision is “to reject H0”
Equivalent Approach We could also have found what values of the sample mean fall at the rejection regions. From the graph above it is clear that our sample mean of 106 would lead to a decision to reject H0.
Stating the Decision Remember: H0: μElbonia = 100 (same as USA) HA: μElbonia 100 (different than the USA) Decision: ‘We reject H0” Since ‘rejecting H0’ implies ‘accepting HA’, we can go on to say... We can conclude that the mean IQ of the population of Elbonians does not equal 100’.
Possible Error If our decision is to ‘reject H0’, then if we are wrong that would be a type 1 error. If the null hypothesis happens to actually be true (μElbonia = 100) then there is a .05 chance that we would obtain a sample mean that would lead us to erroneously reject H0 (i.e. α=.05)
What if... Let’s take a look an an example where everything is exactly the same except the data lead to ‘not rejecting’. I’ll simply subtract 4 from each of the scores to give us a sample mean closer to 100, without changing d.f. or our estimate of the standard deviation. Y=102, 99, 104, 96, 105, 106 Mean=102
tobtained value Remember that a standard score is always some point on a curve minus the mean of the curve divided by the standard deviation of the curve. So, our sample mean of 102 falls a little more than one standard deviation above the mean of the curve representing what H0 predicts.
SDM Our decision is “do not reject H0”
Equivalent Approach It is clear that a sample mean of 102 would lead to not rejecting H0
Stating the Decision Remember: H0: μElbonia = 100 (same as USA) HA: μElbonia 100 (different than the USA) Decision: Do not reject H0. Since ‘not rejecting H0’ implies ‘not accepting HA’, we can say... • “We can not conclude that the mean IQ of the population of Elbonians differs from 100”,or clearer still, • “We cannot determine whether or not the mean IQ of Elbonians differs from 100” (I like this wording as it best conveys the ambiguity of the result). However, we can’t say... • “We can conclude that the mean IQ of Elbonians equals 100.”
Huh? Under most circumstances, when the mean falls in the ‘do not reject H0’ area, you can say you ‘fail to reject H0’ but you can’t say that you ‘accept H0’ or that you’ve ‘proven H0 is correct’. The reason for this will be developed in the lecture on ‘power’. For now, think of it this way: You set out to try to find a difference between the actual population mean and the value proposed by H0. If you find a difference, then that result is interpretable (reject H0). If you don’t find a difference, is that because the difference doesn’t exist or because you didn’t search hard enough to find it?
Confidence Intervals We can now return to confidence intervals, as we now have the tool we need to generate them (i.e. the t table). The 95% confidence interval of the mean can be generated using the following formula, where td.f.,α=.05,2-tail stands for the tc value for the two-tailed alpha of .05 with the appropriate degrees of freedom. For our sample that had a mean of 106...
99% Confidence Interval For other intervals use a different value for alpha. For the 99% confidence interval use the two-tailed tc for alpha=.01
Using Confidence Intervals Confidence intervals can be used to make decisions about a priori hypotheses. Any hypothesis that proposed (a priori) a μ outside that range can be rejected at the .05 significance level.
Using Confidence Intervals The 95% confidence interval for our data was: 101.96 μY 110.04 Note that H0 said that μ =100, thus H0 can be rejected using the confidence interval. There is absolutely no difference between testing H0 using a two-tailed test and doing the same thing using confidence intervals. Confidence intervals are useful in other ways, however, and we will take a look at those towards the end of the semester when we examine alternatives to null hypothesis testing.
