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Hypothesis Testing

发布时间:2013-12-07 14:36:40  

Hypothesis Testing: OneSample Tests

Hypothesis Testing
I believe the population mean age is 50 (hypothesis).


Reject hypothesis! Not close.


? ? ? ? ? ?

Random sample

Mean ? ??X = 20

Five Step Model for Hypothesis Tests
Step 1: State null and alternate hypotheses

Step 2: Select a level of significance

Step 3: Identify the test statistic

Step 4: Formulate a decision rule

Step 5: Take a sample, arrive at a decision

Do not reject null

Reject null and accept alternate

Step One: The Null Hypothesis, H0

States the Assumption (numerical) to be tested
e.g. The grade point average of juniors is at least 3.0 (H0: ??? 3.0)


Begin with the assumption that the null hypothesis is TRUE.
(Similar to the notion of innocent until proven guilty)

?Always contains the ? = ? sign

?The Null Hypothesis may or may not be rejected.

The Alternative Hypothesis, H1

Is the opposite of the null hypothesis

e.g. The grade point average of juniors is less than 3.0 (H1: ? < 3.0)
? ?

Never contains the ?=? sign The Alternative Hypothesis may or may not be accepted


Sometimes it is easier to form the alternative hypothesis first.

State the null and alternate hypotheses for population mean H0: ? = μ0 H1: ? = μ0 H0: ? < μ0 H1: ? > μ0 H0: ? > μ0 H1: ? < μ0
3 hypotheses about means

Three possibilities regarding means

The null hypothesis always contains equality.

Step Two: Select a Level of Significance α


The probability of rejecting the null hypothesis when it is actually true; the level of risk in so doing. Designated a (alpha) ? Typical values are 0.01, 0.05, 0.10


Type I Error: Rejecting the null hypothesis when it is actually true (α). Type II Error: Accepting the null hypothesis when it is actually false (β).

Step Two: Select a Level of Significance.
Researcher Accepts Rejects Ho Ho

Null Hypothesis Ho is true

Correct decision Type II Error (b)

Type I error (a) Correct decision

Ho is false

Step Three: Select the test statistic
Test statistic: A value, determined from sample information, used to determine whether or not to reject the null hypothesis.

Examples: z, t, F, c2 z Distribution as a test statistic The z value is based
on the sampling distribution of X, which is normally distributed when the sample is reasonably large (recall Central Limit Theorem).

X ?? z? ?/ n

Step Four: Formulate the decision rule
Critical value: The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected.

Sampling Distribution Of the Statistic z, a Right-Tailed Test (Alternative hypothesis: H1: μ > μ 0 ), 0 .05 Level of Significance

Do not reject [Probability =.95]

Region of rejection [Probability=.05]


Critical value

Decision Rule

Reject the null hypothesis and accept the alternate hypothesis if

Computed -z < Critical -z

Computed z > Critical z

Step Four: Formulate t

he decision rule (using p-value).
p-Value: The probability, assuming that the null hypothesis is
true, of the obtaining the sample results

Decision Rule

If the p-Value is larger than or equal to the significance level, a, H0 is not rejected. If the p-Value is smaller than the significance level, a, H0 is rejected.

Step Five: Make a decision

One-Tailed Tests of Significance
The alternate hypothesis, H1, states a direction (μ > μ0 or μ < μ0 )

Sampling Distribution

of the Statistic z, a Right-Tailed Test (Alternative hypothesis: H1: μ > μ0 ), 0.05 Level of Significance

Reject H0
Do not reject [Probability =.95]

Region of rejection [Probability=.05]


Critical value

Two-Tailed Test of Significance
No direction is specified in the alternate hypothesis H1. (μ ≠μ0 )
Regions of Rejection for a Two-Tailed Test, .05 Level of Significance
Region of rejection [Probability=.025]
Do not reject [Probability =.95]

Region of rejection [Probability=.025]

Critical value


Critical value

Selection of the test statistic for population mean
? Testing for the population mean from a large sample with population standard deviation known
z ? X ?? ? / n

?Testing for the Population Mean from large Sample with population Standard Deviation Unknown. As long as the sample size n > 30, z can be approximated using X ??
z? s/ n

? Testing for a Population Mean: Small Sample, Population Standard Deviation Unknown X ??
t? s/ n

Assumption of t tests: Data come from the normal distribution

Example: Suppose a car manufacturer claims a model gets 25 mpg. A consumer group asks 10 owners of this model to calculate their mpg and the mean value was 22 with a standard deviation of 1.5. Is the manufacturer's claim supported? Write a program to test this claim using R

## Compute the t statistic. Note we assume mu=25 under H_0 > xbar=22;s=1.5;n=10 > t = (xbar-25)/(s/sqrt(n)) >t [1] -6.324555 ## use pt to get the distribution function of t > pt(t,df=n-1) This is a small p-value (0.000068). [1] 6.846828e-05 The manufacturer's claim is suspicious.

One-Sample Mean Test

Test H0 : ? ? ?0 vs H1 : ? ? ?0 under a normal population N(? , ? 2 ) if ? 2 is unknown, the t-test
t? X ? ?0 s/ n

is recommended, where X is the sample mean and n is the sample size

t.test() In R
There is an example concerning daily energy intake in KJ for 11 women. The value are placed in a data vector

Investigate whether the women’s energy intake deviates systematically From a recommended value of 7725KJ. Assuming that data comes from A normal distribution, the object is to test whether this distribution might have mean 7725. This is done with t.test, as follow t.test(daily.intake,mu=7725)

> t.test(daily.intake,mu=7725) One Sample t-test

data: daily.intake t = -2.8208, df = 10, p-value = 0.01814 alternative hypothesis: true m

ean is not equal to 7725 95 percent confidence interval: 5986.348 7520.925 sample estimates: mean of x 6753.636

x<-c(5260,5470,5640,6180,6390,6515,6805,7515,7515,8230,8770) n<-length(x) xbar<-mean(x) s<-sd(x)

tt<-(xbar-7725)/(s/sqrt(n)) # calculate the value of t statistics [1] -2.820754
p.value<-2*pt(tt,df=n-1) # calculate p-value [1] 0.01813724 alpha<-0.05 decision<-ifelse(p.value<alpha,'Reject','Not reject') #Make decision [1] "Reject" t.value<-qt(1-alpha/2,df=n-1) #Critical value decision<-ifelse(abs(tt)>t.value,'Reject','Not reject')

CI<-xbar+c(-t.value*s/sqrt(n),t.value*s/sqrt(n)) #confidence interval
[1] 5986.348 7520.925

Three arguments in function t.test() are relevant in onesample problems
t.test(x, mu=, alternative=“”,conf.level=) x: a numeric vector of data values mu: a number indicating the vaule of the mean under null hypothesis

alternative: a character string specifying the alternative hypothesis, must be one of "two.sided" (default), "greater" or "less".
Conf.level: confidence level of the interval

One-Sample Proportion Test
The fraction or percentage that indicates the part of the population or sample having a particular trait of interest.
Number of successes in thesample p? Number sampled

The sample proportion is p and ? is the population proportion. Test Statistic for Testing a Single Population Proportion


p ??

? (1 ? ? )

Example: NSC
For a Christmas and New Year’s week, the National Safety Council estimated that 500 people would be killed and 25,000 injured on the nation’s roads. The NSC claimed that 50% of the accidents would be caused by drunk driving. A sample of 120 accidents showed that 67 were caused by drunk driving. Use these data to test the NSC’s claim with a = 0.05.

Example: NSC
? ?

Hypothesis H0: p = .5 Test Statistic

H1: p ? 0.5

p0 (1 ? p0 ) .5(1 ? .5) ?p ? ? ? .045644 n 120
z? p ? p0


(67 /120) ? .5 ? ? 1.278 .045644

?Rejection Rule

?Conclusion Do not reject H0. For z = 1.278, the p-value is .201. How to get the p-value using R? (1-Pnorm(1.278))*2

Reject H0 if z < -1.96 or z > 1.96

prop.test() in R
> prop.test(67,120,p=.5,correct =F) 1-sample proportions test without continuity correction data: 67 out of 120, null probability 0.5 X-squared = 1.6333, df = 1, p-value = 0.2012 alternative hypothesis: true p is not equal to 0.5 95 percent confidence interval: 0.4690452 0.6440025 sample estimates: p 0.5583333 X-squared = 1.6333= (1.278)2= z2 prop.test(x, p=, alternative=“”,conf.level=)

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