introduction to the t-statistic introduction to statistics chapter 9 oct 13-15, 2009 classes #16-17

Introduction to the t-statistic

Introduction to StatisticsChapter 9

Oct 13-15, 2009Classes #16-17

The Problem with Z-scores Z-scores have a shortcoming as an inferential

statistic: The computation of the standard error requires

knowing the population standard deviation ().  In reality, we rarely know the value of . When we don’t know , can’t compute standard

error. Therefore, we use the t statistic, rather than the

Z-score, for hypothesis testing.

How to test? Still need to know about standard deviation of the

population To figure this out…

Use formula learned back in chapter 4:

Use this to calculate estimated standard error (sM) We calculate t statistic similarly to how we calculated the z-

statistic, but now we will use the estimated standard error of the mean (rather than the standard error of the mean)

1

)( 2

n

XXS

The t Statistic Use the sample variance (s2) to estimate the population

variance s2 = SS/df = SS/(n-1)

Use variance s2 in the formula to get the estimated standard error: Provides an estimate of the standard distance between M

and when is unknown

estimated standard error = sM = s n

= s2

n

The t Statistic Finally, we replace the standard error in the

z-score formula with the estimated standard error to get the t statistic formula:

t = t = MM-- sM

Illustration

M = n

t = t = MM-- sM

sM = s n

Z = Z = MM-- M

In chapter 8 for Z-scores we used:

Population SD

Now we are using an estimate of standard error by using the sample SD

plugging it into z-score formula:

plugging it into t statistic formula:

Another change… Up until this chapter we have been using

formulas that used the standard deviation as part of the standard error formula

Now, we shift our focus to the formula based on variance

On page, 234 the book gives reasoning for this.

Main reason: sample variance (s2) provides an accurate and unbiased estimate of the population variance (²)

One more change… Although the definitional formula for sum of squares

is the most direct for computing SS it has its problems When the mean is not a whole number the deviations will

contain decimals and thus calculations become more difficult leading to rounding error

Therefore, from now on we will use the SS computational formula which can be found on page 93 in the textbook “From now on” means all future tests and final exam

Sample Variance Therefore, we will use sample variance rather

than sample standard deviation to compute s

Sample standard deviation is a descriptive statistic rather than a inferential statistic

Sample variance will provide the most accurate way to estimate the standard error

We are now using variance-based formula in these equations

Why?  Inferential purpose, rather than descriptive Drawing inferences about the population

estimated standard error = s M = s n

= s2

n

t statistic Definition:

Used to test hypotheses about an unknown population mean when the value of is unknown.

The formula for the t statistic has the same structure as the z-score formula, except the t statistic uses the estimated standard error in the denominator. t = t = MM--

sM

Z-score vs. T-Score

Z-distribution stays the same, regardless of sample size

T-distribution changes, depending on how many pieces of information you have: degrees of freedom here, df = n-1

Everything else stays the same

Have an alpha level Have one-tailed and two-tailed tests Determine boundaries of critical region Determine whether t-statistic falls in critical

region If it does, reject null and know that p<alpha

Degrees of Freedom How well does s approximate ?

Depends on the size of the sample. The larger the sample, the better the

approximation.

Degrees of Freedom (df) = n-1 Measures the number of scores that are free to vary

when computing SS for sample data. The value of df also describes how well a t

statistic estimates a normal curve.

Degrees of Freedom Degrees of Freedom = df = n-1 As df (sample size) gets larger, 3 things

result: 1) s2 (sample variance) better represents 2

(population variance). 2) t better approximates z. 3) in general, the sample better represents the

population.

The t-distribution

T-distribution  The set of all possible t statistics obtained by selecting

all possible samples of size n from a given population  How well the t distribution approximates a normal distribution

is determined by the df.  In general, the greater n (and df), the more normal the t

distribution becomes.  t distribution more variable and flatter than normal z-score

distribution – why is this the case? Both mean and standard error can vary in t-distribution – only the mean varies in the z-distribution

Distributions of the t statistic

The Versatility of the t test You do not need to know when testing

with t The t test permits hypothesis testing in

situations in which is unknown All you really need to compute t is a

sensible null hypothesis and a sample drawn from the unknown population

Hypothesis Testing with t (two tails)

Same four steps, with a few differences: Now estimating the standard error, so

compute t rather than z Consult t-distribution table rather than

Unit Normal Table to find critical value for t (this will involve the calculation of the df)

Hypothesis Testing w/ t-statistic Instead of the Unit Normal Table, we now have

the t-table p. 531-532 Similar in form to the Unit Normal Table Pay attention to the df column!!

Let’s think about this table for a minute Looking at the two-tail, p=0.05 column:

What is value at 10 df? What is value at 20 df? What is value at 30 df? What is value at 120 df?

A portion of the t-distribution table

Hypothesis Testing with t (two tails) Step 1: State the hypotheses. Step 2: Set and locate the critical region.

You will need to calculate the df to do this, and use the t distribution table.

Step 3: Graph (shade) the critical region. Step 4: Collect sample data and compute t.

This will involve 3 calculations, given SS, n, , and M: a) the sample variance (s2) b) the estimated standard error (sM) c) the t statistic

Hypothesis Testing with t (two tails)

Step 5: Go back to graph and see if tcalc falls in the critical region

Step 6: Make a decision. Compare t computed in Step 3 tCALC with tCRIT found in the t table:

If tCALC > tCRIT (ignoring signs) Reject HO

If tCALC < tCRIT (ignoring signs) Fail to Reject HO

One-Tailed Hypothesis Testing with t Same as with z, only steps 1 and 2 change. Step 1:

Now use directional hypotheses. H0: = ? and H1: ? (predicts decrease) OR H0: = ? and H1: ? (predicts increase).

Step 2: Now the critical region located in only one tail of the

distribution (sign of tCRIT represents the direction of the predicted effect).

You will have to use a different column on the t distribution table.

Example1 Do eye-spot patterns affect behavior?

If eye-spots do affect behavior, birds should spend more or less time in chamber w/ eye-spots painted on the walls.

Sample of n=16 birds. Allowed to wander between the 2 chambers for 60 minutes. If eye-spots do not affect behavior, we’d expect they’d

spend about 30 minutes in each chamber.

We’re told the sample mean =39, SS = 540.

Example1 Step 1: State the hypotheses

Ho: µplain side = 30 min. H1: µplain side ≠ 30 min. Two-tailed Alpha = 0.05

Step 2: Locate the critical region Based on df. What are df here? What is the critical value?

Step 3: Shade in critical region

Example 1 Step 4: Calculate the t-statistic.

First calculate the sample variance s2 = SS/n-1 , 540/15 = 36.

Next use the sample variance (s2) to calculate the estimated standard error

sM

Finally, compute the t-statistic:

50.125.216

362

n

s

650.1

9

50.1

3039

Ms

Mt

Example 1 Step 5: Make a decision.

T-calculated = 6.00 t-critical = + 2.131 We observe that our t-value is in the region of

rejection. We conclude that eye-spots have an effect on

predatory behavior.

Example 2A teacher was trying to see whether a new teaching method

would increase the Test of English as Foreign Language (TOFEL) scores of students. She received a report which included a partial list of previous scores on the exam. Unfortunately, most of the records were burned in a fire that occurred in the school’s Records’ Department. From the available data, students taught by old methods had = 580. She tested her method in a class of 20 students and got a mean of 595 and variance of 225. Is this increase statistically significant at the level of 0.05 in a 2-tailed test?

Example 2 Step 1:

Step 2:

Example 2: Step 3

Example 2 Step 4:

Step 5:

Example 3 A researcher believes that children in poverty-stricken regions

are undernourished and underweight. Past studies show the mean weight of 6-year olds is normally distributed with a 20.9 kg. However, the exact mean and standard deviation of the population is not available. The researcher collects a sample of 9 children from these poverty-stricken regions, with a sample mean of 17.3 kg & s = 2.51 kg.

Using a one-tailed test and a 0.01 level of significance, determine if this sample is significantly different from what would be expected for the population of 6-year olds.

Example 3 Step 1

Step 2

Example 3: Step 3

Example 3 Step 4

Step 5

Example 4 A researcher has developed a new formula (Sunblock

Extra) that she claims will help protect against the harmful rays of the sun. In a recent promotion for the new formula she is quoted as saying she is sure her new formula is better than the old one (Sunscreen).

Her prediction: The “improved” Sunblock Extra will score higher than the previous Sunscreen score of 12?

She decides to use the .05 significance level to test for differences. To the right are the Sunblock Extra scores for participants in her study.

In notation form: H0: HA: Determine if there is a significant difference between the

new product and the old one (make your decision and interpret).

X 2X 12 144 13 169

6 36 11 121 12 144

8 64 11 121

7 49 10 100 16 256 10 100

7 49 14 196 15 225 16 256

168 2030

Example 4 Step 2:

Example 4: Step 3

Example 4 Step 4

Step 5

Example 5 Scientists believe that the “Monstro Motors”

new model will get the highest gas mileage of any car on their lot. Although, not much data is available on the older cars, from a review of previous models they estimate that the best of the rest of their cars achieved 67 m.p.g. They using an alpha level α = .01.

H0: HA:

Determine if there is a significant difference between the MPG of the new car and the best old model on their lot (make your decision and interpret).

X 2X 65 4225 76 5776 69 4761 71 5041 74 5476 78 6084 77 5929 68 4624 72 5184 75 5625 74 5476 64 4096 69 4761 63 3969 82 6724

1077 77751

Example 5 Step 2:

Example 5: Step 3

Example 5 Step 4

Step 5

Steps Step 2?

Step 3?

Step 4?

Effect Size estimated d

r squared

s

Md

deviation standard sample

differencemean estimated

dft

tr

2

22

Effect Size

s

Md

deviation standard sample

differencemean estimated

-0.24

25.3

8.

25.3

122.11

s

Md

Example 4

Small effect (small to medium)

Example 5 Large effect

0.87

49.5

8.4

49.5

678.71

s

Md

Credits http://myweb.liu.edu/~nfrye/psy53/ch9.ppt#9 http://homepages.wmich.edu/~malavosi/Chapt9PPT_S_05.ppt#2 http://faculty.plattsburgh.edu/alan.marks/Stat%20206/Introduction%20to%

20the%20t%20Statistic.ppt#4 http://home.autotutor.org/hiteh/Stats%20S04/Statistics04onesamplet-test1.p

pt#7