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Lecture 9

Lecture 9 - The t Statistic (note: a & d = answers to last multiple choice qs)

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Alison Luby

PSY201 Lecture 9; Nov. 17, 2011 ○ Rmbr s = SS/(n-1) ○  s = √(SS/df) Introduction to the t Statistic The t Statistic: An Alternative to z • sMused more in terms of sample variance, since • When hypothesis testing: provides accurate & unbiased estimate of the ○ Use a sample mean M to approximate the pop variance σ2 pop mean µ. ○ Standard error, σM, measures how well M approximates µ ○  Estimated sample error = =  σM= σ/√n ○ Rmbr must know sample mean before ○ Then compare M to µ by computing the computing sample variance, so there is a relevant z-score test statistic restriction on sample variability  z = (M - µ)/ M ○ Only n - 1 scores are independent & free to  Use to determine whether the obtained vary dif is greater than expected by chance ○  n - 1 = degrees of freedom (df) for sample by looking at unit normal table (for a variance normal distribution) • Like z-score, t statistic allows researchers to us• t statistic (like the z-score) forms a ratio: sample data to test hypotheses about an ○ Numerator: Obtained dif btwn M unknown µ hypothesized µ • Particular advantage of t statistic: ○ Denominator: Estimated standard error ○ Doesn’t require any knowledge of the pop (measures how much dif expected by σwhich we typically don’t know chance) ○ So can be used to test hypotheses about a completely unknown pop, ie:  Both µ & σ unknown • A large value for t (a large ratio) indicates  Only available info about pop comes obtained difference btwn data & hypothesis > from sample expected if treatment has no effect (ie. is just • All required for a hypothesis test with t is a sample & a reasonable hypothesis about µ error) ○ W large sample, df large, so estimation is • When we don’t haveσcan estimate it using the very good sample variability  So t statistic will be very similar to z- ○ Much like estimating µ using M score • Can use t statistic when determining whether ○ W small samples, df small, so t statistic treatment causes a change in µ provides relatively poor estimate of z • Sample is obtained from the pop  treatment is • Can think of the t statistic as "estimated zscore." administered to sample ○ As usual, if resulting sample M is significantly different from original µ, can conclude that ; treatment has a significant effect • Like before, hypothesis test attempts to decide • Just like we can make a distribution of z-scores, btwn: we can make a t distr ○ H o Is it reasonable that the discrepancy ○ Complete set of t values computed for every possible random sample for a specific btwn M & µ is simply due to sampling error and not result of treatment effect? sample size n, or specific degrees of freedom ○ H 1 Is the discrepancy btwn M & µ more than df expected by sampling error alone? • As df approaches infinity, t distr  ie. Is M significantly different from µ approximates normal distr st • How well it approximates normal distr • Critical 1 step for t statistic hypothesis test: ○ Calculate exactly how much dif btwn M & µ is depends on df reasonable to expect ○ ie. There’s a “family” of t distrs so there’s a ○ But since σ is unknown, impossible to dif sampling distr of t for each possible df • Larger n  larger n – 1  better the t distr compute standard error of M (σ =Mσ/√n) as approximates the normal distr done w z-scores ○  t statistic requires to use sample • W small values for df  t distr flatter & more data’s variance, s to compute spread out than a normal distr estimated standard error, s M •  s M s/√n ○ ie. obtained dif btwn data & hypothesis (ratio in numerator) is significantly larger than dif expected if no treatment effect (standard error in the denominator) • When M >> µ (compared to estimated standard error), obtain large value for t, so can reject H 0 • When mean dif small compared to s , faMl to reject H0 • Example 1 ○ z distr (red) & t-distr (green) (n-1 = 4 = df) ○ t peak lower than z, sides higher than z • Researcher wanted to know whether people who (more density away from mean) practice yoga have dif (includes greater and lesser) depression level from general pop; uses a depression inventory that has mean score of 25 for pop • 14 indivs who regularly practice yoga complete the inventory • Scores: 15, 19, 24, 27, 18, 22, 23, 25, 17, 16, 20, 21, 17, 18 • Do the yoga students show a level of depression significantly dif from general pop? • • n = 14; µ = 25; M = 20.14; α = 0.05 ○ z distr (red) & t-distr (purple) (n-1= 40 = df) • Hypotheses: • As you can see, both distrs are unimodal & ○ H o µyoga= 25 symmetrical about the mean (= 0) ○ H 1 µyoga≠ 25 • However, have a dif shaped distrs when • Since we don’t know pop variance, have to use t small n statistic, so need to find df • And t has larger and more variable 2 ○ df = n – 1 = 13 variance, s • Now, determine our critical t value from the ○ ie. estimated standard error varies btwn table samples ○ t = 2.16  s = √(s /n) critical M • Then, determine the test statistic & reject H of >  As n & df ↑, variability in t distr ↓, tcritical becomes more normal ○ SS = 5852 – 79524/14 = 171.71 • To evaluate t statistic from hypothesis test: ○ Must select an α level ○ Estimated sample error = s = M = ○ Then find the appropriate value of df ○ Determine the critical t statistic from the t 0.9713 distr table, Appendix B, page 729  SS/13 = 13.21  #s in table are values of t that separate  13.21/14 = 0.94 ○ tobtained(20.14-25)/0.9713 = -5.003 the tail from the main body of the distr  Please look at this table to see that as df
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