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EXAM REVIEW T TESTS.doc

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Department
Psychology
Course Code
PSYB07H3
Professor
Douglas Bors

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AFTER MIDTERM EXAM NOTES T-TEST ONE-SAMPLE T-TEST -if μ is known and σ is unknown, we must estimate σ with s i.e. if you2 don’t know the population variance use the sample variance to estimate -because we use s, we can no longer declare the answer to be a z but rather it is a t WHY? -the sample variance (s ) is an unbiased estimator of the population 2 2 2 variance (σ ) but s is not a normal distribution  s tends to be positively skewed (if z is used) we more often underestimate 2 population variance (σ ) than overestimate esp. with a small n (underestimating is more of a bigger mistake) -THEREFORE the z table cannot be used as the t is likely to be larger 2 than z because s is in the denominator  z table can be only used in this situation if we have an infinity observations but this is never -t table has critical values instead of t-values Differences in formulas: x − μ x − μ x − μ vs. (substitute s for σ ) 2 x − μ x − μ x − μ z = σ = σ = 2 t = S = S = 2 x σ x S N N N N - To treat t as a Z would give us too many significant results Therefore, for the student’s t distribution, we switch to the t-table when 2 we use s -unlike the z, the distribution is a function of the degrees of freedom with n = infinity and t = z  z = 1.96 but for t value is dependent of degrees of freedom  with an infinity number of df, it approaches 1.96 -for one-sample cases, degrees of freedom: df = n-1  for sample but if given μ, no need for n-1 just n -1 df is lost because we used (sample mean) to calculate s 2 Σ(x-) = 0, all x can vary save for 1 Example 1: the effect of statistic tutorials Last 100 years: μ = 76.0 (no tutorials) This year’s: = 79.3 (tutorials) N = 20 s = 6.4 1. State the null and the alternative hypothesis H 0 μ = 76 H : 1 ≠ 76 2. Plug information into formula x − μ SampleMean -Pop'n mean t = = S x standard error x − μ = s3. Check up on t-table -not area (p) above or below value of 1 -fives t values that cut off at critical areas i.e. 0.05 -t also defined for each degree of freedom THEREFORE, at α = 0.05, T0.0519) is ± 2.093 (critical value) 79.3−76 3.3 = 6.4 = 1.43 = 2.31 20 2.31>2.093  reject null hypothesis  evidence that this year’s group is different from previous years’ (but there is still a 5% chance of making a Type I error) Factors Affecting the Magnitude of t and Decision 1. Difference between and μ ( - μ)  the larger the numerator, the larger th2 t-value  effect size or observed difference 2. Size of s  as n increases, denominator decreases, t increases 3. Size of n  as n increases, denominator decreases, t increases 4. α level 5. one- or two-tailed test (always do two tailed test) Confidence Limits on Mean -point estimate  a specific value taken as estimator of a parameter -interval estimates  a range of values estimated to include parameter -confidence limits  a range of values that has a specific (p) of bracketing the parameter (i.e. 95%)  End Points = confidence limits -How large or small μ could be without rejecting H if we0ran a t-test on the obtained sample mean Confidence Limits (C.I.) Using Example 1 data, x −μ x −μ t = = S x S N -we know , s and N and we know the critical value for t at α = 0.05: = t (19)= 2.093 .05 -solve for μ by rearranging the equation: 79.3−μ 79.3−μ ±2.093 = = 6.4 1.43 20 μ = ±2.093(1.43)+79.3 μ = ±2.993+79.3Using +2.993 and –2.993 More on Confidence Limits (from Internet) μ upp= −2.993+79.3 = 76.31 lower C.I .95 76.31≤ μ ≤ 82.29 Confidence limits for the mean are an interval estimate for the mean. Interval estimates are often desirable because the estimate of the mean varies from sample to sample. Instead of a single estimate for the mean, a confidence interval generates a lower and upper limit for the mean. The interval estimate gives an indication of how much uncertainty there is in our estimate of the true mean. The narrower the interval, the more precise is our estimate. Confidence limits are expressed in terms of a confidence coefficient. Although the choice of confidence coefficient is somewhat arbitrary, in practice 90%, 95%, and 99% intervals are often used, with 95% being the most commonly used. As a technical note, a 95% confidence interval does not mean that there is a 95% probability that the interval contains the true mean. The interval computed from a given sample either contains the true mean or it does not. Instead, the level of confidence is associated with the method of calculating the interval. The confidence coefficient is simply the proportion of samples of a given size that may be expected to contain the true mean. That is, for a 95% confidence interval, if many samples are collected and the confidence interval computed, in the long run about 95% of these intervals would contain the true mean. The width of the interval is controlled by two factors: 1. As N increases, the interval gets narrower from the term. That is, one way to obtain more precise estimates for the mean is to increase the sample size. 2. The larger the sample standard deviation, the larger the confidence interval. This simply means that noisy data, i.e., data with a large standard deviation, are going to generate wider intervals than data with a smaller standard deviation. To test whether the population mean has a specific value, , against the two-sided alternative that it does not have a value , the confidence interval is converted to hypothesis-test form. The test is a one-sample t- test. TWO RELATED SAMPLE T TEST - design in which the same subject is observed under more than one condition (therefore also known as repeated measures, matched samples) -each subject will have 2 measures x and x 1hat wil2 be correlated  this must be taken into account Example 1: Promoting social skills in adolescents  before and after intervention 1. State the hypotheses DIFFERENCE SCORES: Set of scores representing the difference between the subject’s performance on two occasion (x and x ) 1 2 x1 x2 Difference( D ) 18 12 6 The difference scores are the new data and the standard deviation and mean of 5 4 1 19 17 2 the difference scores are the ones that 13 11 2 are used therefore we are testing the 12 8
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