Sta220 - Statistics Mr. Smith Room 310 Class #19.

Sta220 - Statistics Mr. SmithRoom 310Class #19

Section 6-3 and 6-4 Notes

6-3: Test of Hypothesis about a Population Mean: Normal (z) Statistic

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Procedure


Definition

Example 6.3.1.

The effect of drugs and alcohol on the nervous system has been the subject of considerable research. Suppose a research neurologist is testing the effect of a drug on response time by injecting 100 rats with a unit dose of the drug, subjecting each rat to a neurological stimulus, and recording its response time. The neurologist knows that the mean response time for rats not injected with the drug (the “control” mean) is 1.2 seconds. She wishes to test whether the mean response time for drug-injected rats differs from 1.2 seconds. At α = .01, conduct the test of hypothesis.

Solution

(Mean response time is 1.2 seconds)

(Mean response time is less than 1.2 seconds or greater than 1.2 seconds)

Test Statistic: z

Rejection Region: (α = .01)


Two-tailed rejection region: a = .01

Solution:

You can see that z = -3.0 falls in the lower-tail rejection region, which consists of all values of. Therefore, this sampling experiment provides sufficient evidence to reject and conclude, at the level of significance, that the mean response time for drug-injected rats differs from the control mean of 1.2 seconds.


Definition

Example 6.3.2.Teenage drivers will inevitably be tempted to drive faster, even to exceed the speed limit, by their friends. Did you resist such temptations when you first began to drive? To gain insight into this phenomenon, psychologists from the United Kingdom conducted a survey of 258 student drivers and reported the results in the British Journal of Educational Psychology (Vol. 80, 2010). One of the variables in interest was the response to the question “Are you confident that you can resist your friends’ persuasion to drive faster?” Each response was measured on a 7-point scale, from 1 = “definitely no” to 7 = “definitely yes.” The data was collected 5 months after the students had attended a safe-driver presentation. They psychologist reported a sample mean response of 4.98 and a sample standard deviation of 1.62. Suppose it is known that the true mean response of students who do not attend a safe-driver presentation .

a. Set up the null and alternative hypotheses for testing whether the true mean student-driver response 5 months after a safe-driver presentation is larger than 4.7.

b. Calculate the test statistic for the hypothesis test. c. Find the rejection region for the hypothesis test,

using . d. State the appropriate conclusion, in words of the

problem.

Solution

d. Since the observed value of the test statistic falls in the rejection region, is rejected. There is sufficient evidence to indicate the true mean student-driver response five months after a safe-driver presentation is greater than 4.7 at

6.4: Observed Significance Levels: p-Values

In section 6.2, the rejection region and, correspondingly, the value of are selected prior to conduction the test and the conclusions are stated in terms of rejecting or not rejecting the null hypothesis.

A second method of presenting the results of a statistical test reports the extent to which the test statistic disagrees with the null hypothesis and leaves to the reader the task of deciding whether to reject the null hypothesis.

Recall from Section 6.1

Suppose we test 50 sections of sewer pipe and the mean and standard deviation for these 50 measurements to be

(Pipe does not meet specs)(Pipe meets specs)

Test Statistic: z 2.12

Rejection Region: z > 1.645, which corresponds to .05

The observed significance level (p-value) for this test is

p-value =


Finding the p-value for an upper-tailed test when z = 2.12

The area A in the previous figure is given in Table III as .4830. Therefore, the upper-tail area corresponding to z = 2.12 is

p-value = .5 - .4830 = .0170

.05p-value =.0170

We say that these test results are “statistically significant”; that is, they disagree (rather strongly) with the null hypothesis and favor .

The probability of observing a z-value as large as 2.12 is only 0.0170 if in fact the true value of is 2,400.

Note:

If you choose to select for this test, then you would reject the null hypothesis because the p-value for the test, .0170, is less than .05.

However, if you choose , you would not reject the null hypothesis, because the p-value for the test is larger than .01.


Procedure


Finding the p-value for an one-tailed test


Finding the p-value for an two-tailed test; p-value = 2(p/2)

Go back to Example 6.3.1. Find the observed significance level for the test of the mean response time for drug-injected rats.

(Mean response time is 1.2 seconds)

(Mean response time is less than 1.2 seconds or greater than 1.2 seconds)

Test Statistic: z

Rejection Region:

The observed value of the test statistic was z = -3.0.

p-value = P(z < -3.0 or z > +3.0) = P(|z|> 3.0)

P(z < -3.0) = .5 - .4987 = .0013

Since this is a two tail test, we must double the area.2P(z < - 3.0) = 2(.0013) = .0026

Example 6.4.1

The length of stay (in days) for 100 randomly selected hospital patients, presented in the table below. Suppose we want to test the hypothesis that true mean length of stay (LOS) at the hospital is less than 5 days; that is,

(Mean LOS is 5 days) (Mean LOS is less than 5 days)

Assuming that , use the data in the table to conduct the test of .

Solution

The test statistic, z = -1.28 and the p-value of the test, p = .101.

Since the p-value exceeds our selected value, , we cannot reject the null hypothesis. Hence, there is insufficient evident (at ) to conclude that the true mean LOS at the hospital is less than 5 days.

A hospital administrator, desirous of a mean length of stay less than 5 days, may be tempted to select an level that leads to rejection of the null hypothesis after determining p-value = .101.

There are two reasons one should resist this temptation: First, the administrator would need to select (say ) in order to conclude is true. A Type 1 error rate of 15% is considered to large by most researchers.

Second, and more importantly, such a strategy is considered unethical statistical practice.

Example 6.4.2

In a test of the hypothesis versus , a sample of n = 100 observations possessed mean and standard deviation s = 4.1. Find the p-value for this test.


Procedure

Sta220 - Statistics Mr. Smith Room 310 Class #19.

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