Statistical Inference Basic Principle Underlying all inferential statistics… samples are representative of the population from which they are drawn Hypotheses Testing Estimation –The inferential process of using sample statistics to estimate population parameters is called estimation Accuracy / Precision –Small error Unbias-ness –Expected Value of the Estimator is the population parameter

When to make Estimation ? Goal: Estimate the value of an unknown population parameter, usually the value for an unknown population mean Estimate effect after the null hypothesis is rejected Already know an effect and wish to estimate for it Population parameter is unknown and wish to gain info

Two Ways of making Estimates Point estimation –Use a single number to estimate an unknown population parameter Interval estimation –Because the estimate is associated with confidence, it is also called confidence interval –Use a range of values as an estimate –Advantage: increase confidence –Disadvantage: decrease precision

Procedure for Interval Estimation of Population Mean µ Begin by calculating the sample mean. This provides a point estimate for µ. Best bet is to predict that the sample mean is located somewhere in the centre of the distribution Determine the amount of confidence required, say 95% Estimate that our sample mean is somewhere in the middle 95% of the sample means. This section of the distribution is bounded by –z scores of -1.96, +1.96or –t scores of - t 0.025, df, + t 0.025, df Compute the limits of the 95% Confidence Interval for µ –Using z:sample mean +/- z * standard error –Using t :sample mean +/- t * standard error

Example 12.2 (page 381) Recent studies allowed psychologists to establish definite links between specific foods and specific brain functions. E.g. lecithin (found in soybeans, eggs, and liver) has been shown to increase the conc of certain brain chemicals that help regulate memory and motor coordination. This expt is designed to demo the importance of this particular food substance. Two independent samples of rats, 10 with normal diet with standard amounts of lecithin, 5 in another group are fed a special diet which contains almost no lecithin. After 6 months, each of the rats is tested on a specially designed learning problem that requires both memory and motor coordination. The purpose of expt is to demo the deficit in performance that results from lecithin deprivation. Score for each animal is number of errors it makes before it solves the learning problem. Data:No-lecithin diet: n=5, x1-bar=33, SS=140 Regular diet:n=10, x2-bar=25, SS=250

Example continued Note: S 2 2 =250/9=28 S 1 2 =140/4=35 Estimate µ1-µ2 =33-25=8 S p 2 = ( )/(9+4) =30 S (x1-bar - x2-bar) = (30/10 +30/5) **.5 =3 t.025, 13 =2.16 A 95% CI for µ1-µ2 is 8 +/- 3*2.16 = 1.52, Note that the text book provides a 80% confidence interval (page 383) which is 3.95,12.05 Compare the interval estimates on ‘confidence’ and ‘width’