Statistical Inference with Small Samples

Transcription

1 Statistical Inference with Small Samples

2 Administrivia o Homework 5 due Friday Nov 10 2

3 Previously on CSCI 3022 Statistical inference for population mean when data is normal and n is large and Z - TEST is known: ( Ife ) - Mai ) I±z' a. Eu is unknown: ( * - µ Thru ) - Non ) 3

4 Previously on CSCI 3022 Statistical inference for population mean when data is NOT normal and n is large and is known: c III ) Enron ) is unknown: - I µ ( ) CLT - Mon ) 4

5 Previously on CSCI 3022 Statistical inference for population mean when data is normal and n is small and is known: (a) * ~ Nloil ) is unknown:??????? 5

6 The Story so Far for Means Thus far, we ve talked about Hypothesis Testing / Confidence Intervals for the mean of a population in the following cases: Normal Data / Known Normal Data / Unknown Non-Normal Data / Known n 30 n<30 " " " " " " 1%1/1111/111 4/111111/1/ ' Non-Normal Data / Unknown #t Z - TEST 1 BOOTSTRAP Ex t - TEST 6

7 Small-Sample Tests for µ o When n is small we cannot invoke the Central Limit Theorem o When n is small and the variance is unknown we need to do something else When mean X µ is the sample mean of a random sample of size n from a normal distribution with, the random variable skin follows a probability distribution called a t-distribution with parameter degrees of freedom. = n 1 7

8 The t-distribution The following figure shows the pdf of some members of the family of t-distributions 8

9 t Properties of t-distributions Let denote the t-distribution with parameter degrees of freedom = n -1 t o Each curve is bell-shaped and centered at 0 t o Each -curve is more spread out than the standard normal distribution o As increases, the spread of the corresponding -curve decreases t!1 t o As the sequence of -curves approaches the standard normal curve 9

10 The t-critical Value We can extend all of our inferential mechanics to the small-sample case by introducing the so-called t-critical value, which we denote t, Def: The t-critical value,, is the point such that the area under the -curve to the right of is equal to t, t, t * a t 0.05,6 taw Example: is the t-critical value that captures the upper-tail area of 0.05 under the t curve with 6 degrees of freedom. 10

11 The t-confidence Interval for the Mean x Let and s be the sample mean and sample standard deviation computed from the results of a random sample with of size n from a normal population with mean. µ 100(1 )% Then a t-confidence interval for the mean is given by: µ - th, n. I Fu [I I + them ], Or, more compactly: I ± txn, ni In 11

12 t-confidence Interval Example Example: Suppose the GPAs for 23 students have a histogram that looks as follows: I t 05,22 5=0.308 n=23 Shot,o5 qq.es#0g=.0642t.os,2z=stats.t.ppf/ The sample mean of the data is and the sample standard deviation is Find a 90% confidence interval for the mean GPA.. 95,227=1.717 a * 0642 [ 3.033, ] 12

13 The t-test, Critical Regions and P-Values Alternative Hypothesis Critical Region Level Test H 1 : > 0 H 1 : < 0 t t apple t, t, H 1 : 6= 0 t apple t /2, or t t /2, STUDENT TEST IZED STATISTIC t.x.us/in 13

14 The t-test, Critical Regions and P-Values Alternative Hypothesis P-Value Level Test H 1 : > 0 G- P (T t H 0 ) apple H 1 : < 0 H 1 : 6= 0 P (T apple t H 0 ) apple 2 min(p (T apple t H 0 ),P(T t H 0 )) apple 14

15 t-test Example Example: Suppose the GPAs for 23 students have a histogram that looks as follows: I Ho :µ= 3.30 g- n= He : M 3.30 The sample mean of the data is and the sample standard deviation is Determine if there is sufficient evidence to conclude at the 0.10 significance level that the mean GPA is not equal to = =

16 t-test Example ( w/p - values ) than PNAIUE zx stats.t. Cd f ( ,227=

17 t-test Example Example: Suppose the GPAs for 23 students have a histogram that looks as follows: The sample mean of the data is and the sample standard deviation is Determine if there is sufficient evidence to conclude at the 0.10 significance level that the mean GPA is not equal to

18 t-test Example 42,22 t.os.az = = stats ppf ( 0,95.,.tk#p%EeMztsion RE Ecton ) Region t= 0 t. 22 ) = 1,71 S ) NLE t= < - t. 05,22= WE Reject Ho 18

19 Inference for Variances We ve talked about estimating confidence intervals for the variance of a population using the Bootstrap But if your population is normally distributed, we have some theory which gives us a better confidence interval and works for both large and small sample sizes Question: What does the sampling distribution of the variance look like when the population is normally distributed? 19

20 The Chi-Squared Distribution The chi-squared ( 2 The pdfs of the family of ) distribution is also parameterized by degrees of freedom 2 distributions are gross, so lets just draw them. = n 1 20

21 A Confidence Interval for the Variance X 1,X 2,...,X n Let be a random sample from a normal distribution with mean and standard deviation. Define the sample variance in the usual way as s ' - ±, I (n 1) S 2 / 2, lxi. - est 2 n 1 Then the random variable follows the distribution. µ Then it follows that P ( k 2 < Xin, Xian n., ) = I - a 21

22 The Chi-Squared Dist is Non-Symmetric Because the distribution is non-symmetric, we need to use two different critical values. yng STATS. CHIZ,ppF( ( - Liz ) 2/2 ah 1 t t 22

23 A Confidence Interval for the Variance For a 100(1 )% confidence interval we choose the two critical values 1 /2,n 1 and 2 /2,n 1 which attributes /2 probability to each tail. Then, with 100(1 )% confidence we can say that 2, =t P ( Xian,.fm# < XI.,n. Inches. 'x nm ( n - 1) s In. 2, < r ' < he X. a. nt 23

24 A Confidence Interval for the Variance For a 100(1 )% confidence interval we choose the two critical values 1 /2,n 1 and 2 /2,n 1 which attributes /2 probability to each tail. Then, with 100(1 )% confidence we can say that 2 (n 1)S 2 2 /2,n 1 < 2 < (n 1)S2 2 1 /2,n 1 Question: How can we use this to get a 100(1 )% confidence interval for the standard deviation? ' En I t iii. 24

25 Variance CI Example Example: A large candy manufacturer produces packages of candy targeted to weight 52g. The weight of the packages of candy is known to be normally distributed, but a QC engineer is concerned that the variation in the produced packages is larger than acceptable. In an attempt to estimate the variance she selects, n=10 bags at random and weighs them. The sample yields a sample variance of 4.2g. Find a 95% confidence interval for the variance and a 95% confidence interval for the standard deviation. X=, 05 Xm = 025 n=10 SE 4.2 X3h. XIE,n X2.gs - in,g,n -1 = X ( 10-1) = Stats.< Hlz.?os,g= stats. CHLZ. ppfl ppf( ( 10 - ' ,02 = 1,9g ' % CI For 02 :[ 1.99, 14.0 ], - 05,9 ) = ,9 ) = = 14,0 95% CI Fort :[ 1.41, 3.74 ] 25

26 OK! Let s Go to Work! Get in groups, get out laptop, and open the Lecture 19 In-Class Notebook Let s: o Do some stuff! 26

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