Statistics 104: Section 7

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1 Statistics 104: Section 7 Section Overview Reminders Comments on Midterm Common Mistakes on Problem Set 6 Statistical Week in Review Comments on Midterm Overall, the midterms were good with one notable exception: problem 3, part b (the cats & dogs question). The most common issue appeared to be conceptualizing the problem, not doing the calculations. We will go through the problem in its entirety today. Common Mistakes on Problem Set 6 The meaning of a 95% confidence interval is: In repeated sampling for a sample of size n, the 95% confidence interval will include the population mean µ in 95% of cases. It does not mean that there is a 95% probability that µ is in the interval, nor is it enough to say We have 95% confidence that the population mean µ is in this interval. P-values: A p-value is the probability of a result at least as extreme as what was observed occuring given that the null hypothesis is true. Both parts of this definition are important when giving answers on hypothesis testing questions. Statistical Week in Review t distribution and t-tests Up until now, we have worked with somewhat contrived situations where we assumed that we knew the population standard deviation (σ) of our random variable of interest. It s really more reasonable to assume that we will need to estimate not only the population mean but also the population standard deviation from data that are collected. The population mean, µ, is estimated by x, and the population standard deviation, σ, is estimated by s. Recall how we calculate s: s = n i=1 (X i x) 2 n 1 1

2 Where before we calculated a z-statistic of the form z = x µ σ/ n, we would now calculate a t-statistic, which has a very similar form: t = x µ s/ n, which follows a t distribution with n 1 degrees of freedom. The t distribution, like the normal distribution, is symmetric and bell-shaped. So, where before we were conducting z-tests to test hypotheses, we will now typically use t-tests, as we enter more realistic conditions of having to estimate the standard deviation. t-tests are among the most common basic statistical tests. They can take many different forms, depending on the structure of your question and the data that you have to answer your question. Different forms of the t-test are as follows: 1 sample t test You have one sample X 1, X 2... X n from a normal distribution with unknown mean and unknown variance (µ, σ 2 ). You want to test your data against the null H 0 : µ = µ 0. Your alternative hypothesis will determine whether you are conducting a one-tailed or a two-tailed test. 1. Calculate x = 2. Calculate s 2 = P n i=1 X i n P n i=1 (X i x) 2 n 1 3. Calculate the test statistic t = x µ 0 s 2 /n with degrees of freedom n 1 4. Look up the p-value on the t table with degrees of freedom n 1, or locate the t-value corresponding to your significance level α, and assess whether the t-statistic that you ve calculated is large enough to reject the null hypothesis. 2 sample t test You have two samples X 1, X 2,... X n1 and Y 1, Y 2,... Y n2, both of which you assume are normally distributed with unknown mean and unknown variance (µ X, σx 2 ) and (µ Y, σy 2 ). You wish to test against the null hypothesis H 0 : µ X = µ Y. First, there are several key things to calculate: 1. Sample means: x = P n1 i=1 x P n2 i j=1 n 1 and ȳ = y j n 2 2. Sample standard deviations: s 2 x = P n1 i=1 (x i x) 2 P n2 n 1 1 and s 2 j=1 y = (y j ȳ) 2 n The standard error of the sample mean for X: S.E. X= s 2 X n 1 2

3 4. The standard error of the sample mean for Y: S.E. Ȳ = s 2 Y n 2 Now what do you do? The way that you proceed from here depends upon the assumptions that you make about the equality of the variances in the two samples. Option 1: Assume different variances Here, you assume that σx 2 σ2 Y since your sample variances are unequal, by inspection: s2 X s2 Y. 1. Calculate the standard error for the difference between X and Ȳ : S.E s X = 2 Ȳ x n 1 + s2 y n 2 2. Use degrees of freedom k = min (n 1 1, n 2 1) or ask the computer for a t-test using unpooled variance. 3. Look up the test statistic for t = freedom, k. X Ȳ S.E. X Ȳ on the t table for the determined degrees of Option 2: Assume similar variances of both distributions If you can pool and get a better overall estimate of the standard deviation of the distribution, this is preferable: 1. s 2 p, the pooled variance is essentially a weighted average of the two sample variances: s 2 p = (n 1 1)s 2 x + (n 2 1)s 2 y n 1 + n 2 2 = n1 i=1 (x i x) 2 + n 2 j=1 (y j ȳ) 2 n n Then, your standard error for your estimate S.E X Ȳ = 3. Degrees of freedom is now n 1 + n 2 2. ( ) s 2 p n1 n2 4. Look up test statistic for t = X Ȳ S.E X Ȳ on t table with k = n 1 + n 2 2 degrees of freedom. Paired t test What if you had paired data (X 1, Y 1 ), (X 2, Y 2 ),... (X n, Y n )? 1. Calculate D 1 = X 1 Y 1, D 2 = X 2 Y 2,... D i = X i Y i,..., D n = X n Y n 2. Calculate mean: d = P di n 3. Calculate sd: s 2 d = P n i=1(d i d) 2 n 1 4. Perform 1 sample t test for n 1 degrees of freedom against null hypothesis H 0 : µ d = 0 3

4 Example: Lessons from Statistical Consulting A Harvard student studying the effects of autism on certain cognitive abilities, needs some statistical advice. For her honors thesis, she has taken a random sample of autistic children (assume she selected these students from as good a pool as she could find), and then connected each to a nonautistic peer a child their own age, training, ethnicity, parental income level, upbringing (if at all possible), whom she has selected from a random pool. Since the autistic children were said to be chosen from a gold standard random sample, her pairs of autistic/non-autistic children can be treated as random. However, she found only 6 autistic/non-autistic pairs to work with. She assessed the cognitive ability of each of these students with a test of verbal reasoning. Autistic Non-Autistic Scores Student Student Pair Pair Pair Pair Pair Pair She was interested in the hypothesis that autistic and non-autistic children differed in their verbal reasoning abilities. Here is an example of what not to do: a 2 sample t test (with pooled variance) Two Sample t-test t = , df = 10, p-value = percent confidence interval: (-2.33, 0.33) mean of x mean of y This has very little power, degrees of freedom is small, t value is small, and it did not take advantage of the structure of the data. What we should do: Controlling for matched pairs with a PAIRED t test We didn t take the pairing into account. We should have expressed our hypothesis in terms of the difference within pairs: Y i1 Y i2 = D i is the difference between a fitted pair of autistic and non-autistic students and is distributed D i N ( µ D, σ 2). The hypotheses for this test are: H 0 : µ D = 0 H A : µ D 0 To perform this test, we calculate d = 1 and sd D = test is n 1 = 5, and our t-statistic has a simple form of Pn i=1 (d i d) 2 n 1 d sd D / n.. The degrees of freedom for this 4

5 Paired t-test t = , df = 5, p-value = percent confidence interval: (-1.66, -0.34) mean of the differences -1 As you can see, the confidence interval for the mean difference does not include 0, a good sign. Based on the p-value associated with our t-test, we reject the null hypothesis and conclude that autistic and non-autistic children do differ in their verbal reasoning abilities, as assessed by the test administered. Now consider the following: The Sign Test What we really should have done was put some of our critical thought into the data. For instance, what would you have thought if I gave you this data set on coin flips: Coin Flips Tails Heads Throw Throw Throw Throw Throw Would you say that the coin is somewhat biased? Well, we know that if the chance really were p heads =.5 or under the null hypothesis: H 0 : p heads = p tails =.5 H A : p heads p tails.5 Under the null hypothesis the probability of getting 5 heads in a row is 1 = We would 2 5 reject the null hypothesis at an α =.05 rejection level and say that we re dealing with a biased coin. Consider: This is exactly the same as the dataset comparing autistic and non-autistic children! Assuming nothing particular about our pairs of children, we had each pair take a test, and we observed which one scored better (the last one was a tie). Every single time, the non-autistic child performed better than or even with the autistic child. If there really weren t a difference, if we d secretly just labeled an even pool of paired children as Autistic or Not Autistic, we would have a 1 32 chance of getting a dataset this extreme. This answer is remarkable in that it makes none of the assumptions the t-tests made, still shows a significant result, and is highly explainable to non-mathematicians. It is also robust to outliers in a way that t-tests can t be. This is an example of the sign test. 5

6 Practice Problems An investigator wants to test whether Gatorade brand sports drink does a better or worse job than water of keeping runners hydrated while exercising. After an appropriate warm-up, each athlete was clocked for a 1-mile run before being hydrated. Participants were given specific amounts of the appropriate fluid at specific time interval during 1 more hour of running. After the first mile plus a full hour, each runner was then clocked for another mile run. Percentage increase in the amount of time it took an athlete from the first mile to the last mile was measured for each athlete after the 1 hour of exercise. 10 observations were taken for both Gatorade and water, and the results are shown here: Obs Mean St. Dev. Gatorade Obs Mean St. Dev. Water a) What are your hypotheses? What test will you be performing? How much type I error will you allow? What is your test statistic? How many degrees of freedom do you have? What is the associated p-value? What is your conclusion? b) Actually the observations were taken from only 10 runners, and each runner was asked to perform under both conditions, Gatorade and water, a few days apart (randomized to see which liquid was used first). These results are now shown here: Obs Mean St. Dev. Diff What are your hypotheses? What test will you be performing now? What is your test statistic? How many degrees of freedom do you have? What is the associated p-value? What is your conclusion? Is this different from part A? Why or why not? c) Perform a sign test on the data above. 6