Student’s t-distributions

Student’s t-Model:

Family of distributions similar to the Normal model but changes based on degrees-of-freedom.

Degrees-of-freedom = sample size (n) – 1

Differs because we need to estimate the mean and standard deviation of a population using our sample data

The graph of Student’s t is generally wider than the normal model.

The larger the sample size the closer Student’s t-model approach the Normal model.

http://www.stat.tamu.edu/~jhardin/applets/signed/T.html

One Sample t-intervalConditions:Randomization10% ConditionNearly Normal – Two possible graphs to check:

1. Histogram: unimodal and symmetric, with no outliers

2. Normal Probability Plot: a straight line, no curve

**It can be skewed if the sample size is large.

Generally n > 30. CLT kicks in at that point.

All conditions have been met to use Student’s t-model for a one sample t-interval

Normal Probability PlotFinds the z-score of each value compared to the

sample mean and standard deviation. Plots these against the values themselves.

No curve in the plot shows a unimodal and symmetric distribution

Mechanics:Calculate sample mean, , and sample standard

deviation, sFind the degrees of freedom, df = n – 1Calculate critical value using given confidence level

and degrees of freedom,

One Sample t-interval

x

dft

CI : df

sx t

n

Example: Coffee MachineA coffee vending machine dispenses coffee into a

paper cup. You’re supposed to get 10 ounces of coffee, but the amount varies slightly from cup to cup. Below are the amounts measured in a random sample of 20 cups. Is there evidence that the machine is shortchanging customers? Construct a 95% confidence interval.

9.9 9.7 10.0 10.1 9.99.6 9.8 9.8 10.0 9.59.7 10.1 9.9 9.6 10.29.8 10.0 9.9 9.5 9.9

Conditions:Randomization: Stated as a random sample10% Condition: 20 cups is less than 10% of all cups

dispensed by the coffee machineNearly Normal Condition: The histogram is unimodal

and approximately symmetric. Or

The normal probability plot shows an

approximately linear pattern.All conditions have been met to use

Student’s t-model for a one sample

t-interval.

Mechanics:

(9.752, 9.938)

We are 95% confident that the true mean amount of coffee from the coffee machine is between 9.752 to 9.938 ounces.

9.845x 0.199s 20n

. . 20 1 19d f

19 2.093t 0.199

CI : 9.845 2.09320

df

sx t

n

1 Sample t-TestInference about the mean of a population, µ.

Hypotheses:

H0: µ = µ0 The population mean is _____

HA: µ <;>;≠ µ0

Conditions:

1.Randomization

2.10% Condition

3.Nearly Normal (draw a picture)

Mechanics:Calculate sample mean, , and sample standard

deviation, sFind the degrees of freedom, df = n – 1State alpha valueCalculate test statistic, Find P-Value

Conclusion:Compare P-Value to alpha.State conclusion in context

x

0. .d f

xt

s

n

Hypothesis:

H0: µ = 10; The mean amount of coffee dispensed by the coffee machine is 10 ounces.

HA: µ < 10; The mean amount of coffee dispensed by the coffee machine is less than 10 ounces.

Conditions:

1.Randomization: Random sample

2.10% Condition: 20 cups is less than 10% of all cups dispensed.

3.Nearly Normal: The histogram is

unimodal and approximately symmetric

with no outliers.

Example: Coffee Machine

Mechanics:

Conclusion:

Since the P-Value is less than alpha (0.0012 < 0.05) we reject the null hypothesis. There is statistically significant evidence that the mean amount of coffee from the coffee machine is less than 10 ounces.

9.845x 0.199s

20n . . 20 1 19d f

0.05

019

xt

s

n

9.845 100.199

20

3.49

19( 3.49) 0.0012P Value P t