Chapter 9

Part C

III. One-Tailed TestsB. P-values

• Using p-values is another approach to conducting a hypothesis test, yielding the same result.

• In general: a p-value is the probability of obtaining a sample result that is at least as unlikely as what is observed.

• For example, suppose you calculate a z-score of negative 2.80.

A diagram

The standard normal probability table tells us that there is a .0026 probability of observing a sample mean less than or equal to what was observed.

Z0Z=-2.80

P-value is

P=.0026

.4974

Z0Z=-2.80

P-value is

P=.0026

How do you use it?

•If you’re testing the hypothesis at the 99% level, then =.01.•If p-value< , reject Ho.

Z=-2.33

=.01

Result of the test.

• So in this example, .0026<.01, so we reject Ho.• Had we been using our previous methodology of

comparing the test statistic to the critical value, we would get the same decision.

• Since Z=-2.80 is greater (absolute value) than Z=.01=-2.33, we would reject Ho.

An Example

Ho: 10

Ha: < 10

n=36, =8, s=5, =.01, Z=.01=-2.33

Test statistic, Z=(-2)/(5/6)=-2.4

p-value for a Z=-2.4 is .0082 and since this is less than =.01, reject the null Ho.

MAKE SURE YOU CAN DO THIS!

x

IV. Two-Tailed Tests about :(large sample)

• The big important gold star difference when you do a two-tailed test is that the rejection range is split equally in each tail of the sampling distribution.

• If you get a test statistic that is either too low, or too high, you will reject Ho.

A. An Example

• A lathe is set to cut bars of steel into perfect lengths of 6 centimeters. If the bars are anything but 6 cm, our customers will incur substantial costs in resizing the bars and will find other suppliers.

Ho: = 6 cm

Ha: 6 cm

The rejection range

If you are testing this hypothesis at the =.05 level of significance, then .025 (/2) is in each tail.

= 6 x.4750 .4750

.025 .025

Critical Z and Test Statistic

• Find the z-score that is associated with .4750, and you’ll find Z=.025=±1.96.

• In a sample of 121 bars, you find a sample mean of 6.08 cm with a standard deviation of .44 cm.

• With a test statistic of 2.0, you would reject Ho and conclude, with 95% confidence, that the bars are not exactly 6 cm.

0.2

121/44.

608.6

Z

B. P-values for 2-tailed tests

• Because the significance level () is split between each tail of the sampling distribution, so is the p-value.

• The p-value is reported as 2 times the area in one tail.

• If 2*(area in tail) < , reject Ho.

Example

• In the previous example, Z=2.0. The area in the tail, beyond Z=2.0 is (.5-4772)=.0228

• The p-value = 2(.0228) = .0456, and since this is less than =.05, reject Ho.

Z0

p=.0228/2=.025/2=.025

C. Confidence Intervals

Recall, that with a (1- ) confidence coefficient, a confidence interval is constructed as:

If you have a 2-tailed hypothesis test like:

Ho: = 0

Ha: 0

x zn

/2

If you take a sample, and your confidence interval includes the hypothesized value of 0, then you cannot reject the null.

If your 0 is outside of the confidence interval, reject the null.

Back to our example

Our sample mean was 6.08 cm, with a standard error of (.44/11)=.04.

So our 95% confidence interval would look like:

or a range between 6.0016 and 6.1584. Since 6 cm is NOT in the interval, we reject Ho.

)04(.96.108.6