Chapter

© 2010 Pearson Prentice Hall. All rights reserved

ChapterHypothesis Tests Regarding a Parameter

10

© 2010 Pearson Prentice Hall. All rights reserved

SectionHypothesis Tests for a Population Mean-Population Standard Deviation Known

10.2

© 2010 Pearson Prentice Hall. All rights reserved 10-3

Objectives

1. Explain the logic of hypothesis testing2. Test the hypotheses about a population mean with

known using the classical approach3. Test hypotheses about a population mean with

known using P-values4. Test hypotheses about a population mean with

known using confidence intervals5. Distinguish between statistical significance and

practical significance.


Objective 1

• Explain the Logic of Hypothesis Testing


To test hypotheses regarding the population mean assuming the population standard deviation is known, two requirements must be satisfied:

1. A simple random sample is obtained.2. The population from which the sample is

drawn is normally distributed or the sample size is large (n≥30).

If these requirements are met, the distribution of is normal with mean and standard deviation .

x

n


Recall the researcher who believes that the mean length of a cell phone call has increased from its March, 2006 mean of 3.25 minutes. Suppose we take a simple random sample of 36 cell phone calls. Assume the standard deviation of the phone call lengths is known to be 0.78 minutes. What is the sampling distribution of the sample mean?

Answer: is normally distributed with mean 3.25 and standard deviation .

x

0.78 36 0.13


Suppose the sample of 36 calls resulted in a sample mean of 3.56 minutes. Do the results of this sample suggest that the researcher is correct? In other words, would it be unusual to obtain a sample mean of 3.56 minutes from a population whose mean is 3.25 minutes? What is convincing or statistically significant evidence?


When observed results are unlikely under the assumption that the null hypothesis is true, we say the result is statistically significant. When results are found to be statistically significant, we reject the null hypothesis.


One criterion we may use for sufficient evidence for rejecting the null hypothesis is if the sample mean is too many standard deviations from the assumed (or status quo) population mean. For example, we may choose to reject the null hypothesis if our sample mean is more than 2 standard deviations above the population mean of 3.25 minutes.

The Logic of the Classical Approach


Recall that our simple random sample of 36 calls resulted in a sample mean of 3.56 minutes with standard deviation of 0.13. Thus, the sample mean is

standard deviations above the hypothesized mean of 3.25 minutes.

Therefore, using our criterion, we would reject the null hypothesis and conclude that the mean cellular call length is greater than 3.25 minutes.

z 3.56 3.25

0.132.38


Why does it make sense to reject the null hypothesis if the sample mean is more than 2 standard deviations above the hypothesized mean?


If the null hypothesis were true, then 1-0.0228=0.9772=97.72% of all sample means will be less than

3.25+2(0.13)=3.51.


Because sample means greater than 3.51 are unusual if the population mean is 3.25, we are inclined to believe the population mean is greater than 3.25.


A second criterion we may use for sufficient evidence to support the alternative hypothesis is to compute how likely it is to obtain a sample mean at least as extreme as that observed from a population whose mean is equal to the value assumed by the null hypothesis.

The Logic of the P-Value Approach


We can compute the probability of obtaining a sample mean of 3.56 or more using the normal model.


Recall

So, we compute

The probability of obtaining a sample mean of 3.56 minutes or more from a population whose mean is 3.25 minutes is 0.0087. This means that fewer than 1 sample in 100 will give us a mean as high or higher than 3.56 if the population mean really is 3.25 minutes. Since this outcome is so unusual, we take this as evidence against the null hypothesis.

z 3.56 3.25

0.132.38

P x 3.56 P(Z 2.38) 0.0087.


Assuming that H0 is true, if the probability of getting a sample mean as extreme or more extreme than the one obtained is small, we reject the null hypothesis.

Premise of Testing a Hypothesis Using the P-value Approach


Objective 2

• Test Hypotheses about a Population Mean with Known Using the Classical Approach


Testing Hypotheses Regarding the Population Mean with σ Known Using the

Classical Approach To test hypotheses regarding the population

mean with known, we can use the steps that follow, provided that two requirements are satisfied:

1. The sample is obtained using simple random sampling.

2. The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30).


Step 1: Determine the null and alternative hypotheses. Again, the hypotheses can be structured in one of three ways:


Step 2: Select a level of significance, , based on the seriousness of making a Type I error.


Step 3: Provided that the population from which the sample is drawn is normal or the sample size is large, and the population standard

deviation, , is known, the distribution of the sample mean, , is normal with mean and standard deviation .

Therefore,

represents the number of standard deviations that the sample mean is from the assumed mean. Thisvalue is called the test statistic.

x

0

n

z0 x 0

n


Step 4: The level of significance is used to determine the critical value. The critical region represents the maximum number of standard deviations that the sample mean can be from 0 before the null hypothesis is rejected. The critical region or rejection region is the set of

all values such that the null hypothesis is rejected.


(critical value)

Two-Tailed


(critical value)

Left-Tailed


Right-Tailed

(critical value)


Step 5: Compare the critical value with the test statistic:


Step 6: State the conclusion.


The procedure is robust, which means that minor departures from normality will not adversely affect the results of the test. However, for small samples, if the data have outliers, the procedure should not be used.


Parallel Example 2: The Classical Approach to Hypothesis Testing

A can of 7-Up states that the contents of the can are 355 ml. A quality control engineer is worried that the filling machine is miscalibrated. In other words, she wants to make sure the machine is not under- or over-filling the cans. She randomly selects 9 cans of 7-Up and measures the contents. She obtains the following data.

351 360 358 356 359358 355 361 352

Is there evidence at the =0.05 level of significance tosupport the quality control engineer’s claim? Priorexperience indicates that =3.2ml.Source: Michael McCraith, Joliet Junior College


Solution

The quality control engineer wants to know if the mean content is different from 355 ml. Since the sample size is small, we must verify that the data come from a population that is approximately normal with no outliers.


Assumption of normality appears reasonable.

Normal Probability Plot for Contents (ml)


No outliers.


Solution

Step 1: H0: =355 versus H1: ≠355

Step 2: The level of significance is =0.05.

Step 3: The sample mean is calculated to be 356.667. The test statistic is then

The sample mean of 356.667 is 1.56 standard deviations above the assumed mean of 355 ml.

z0 356.667 355

3.2 91.56


Solution

Step 4: Since this is a two-tailed test, we determine the critical values at the =0.05 level of significance to be -z0.025= -1.96 and z0.025=1.96

Step 5: Since the test statistic, z0=1.56, is less than the critical value 1.96, we fail to reject the null hypothesis.

Step 6: There is insufficient evidence at the =0.05 level of significance to conclude that the mean content differs from 355 ml.


Objective 3

• Test Hypotheses about a Population Mean with Known Using P-values.


A P-value is the probability of observing a sample statistic as extreme or more extreme than the one observed under the assumption that the null hypothesis is true.


Testing Hypotheses Regarding the Population Mean with σ Known Using

P-valuesTo test hypotheses regarding the population mean

with known, we can use the steps that follow to compute the P-value, provided that two requirements are satisfied:

1. The sample is obtained using simple random sampling.

2. The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30).


Step 1: A claim is made regarding the population mean. The claim is used to determine the null and alternative hypotheses. Again, the hypothesis can be structured in one of three ways:


Step 2: Select a level of significance, , based on the seriousness of making a Type I error.


Step 3: Compute the test statistic,

z0 x 0

n


Step 4: Determine the P-value





Step 5: Reject the null hypothesis if the P-value is less than the level of significance, . The comparison of the P-value and the level of significance is called the decision rule.

Step 6: State the conclusion.


Parallel Example 3: The P-Value Approach to Hypothesis Testing: Left-Tailed, Large Sample

The volume of a stock is the number of shares traded in the stock in a day. The mean volume of Apple stock in 2007 was 35.14 million shares with a standard deviation of 15.07 million shares. A stock analyst believes that the volume of Apple stock has increased since then. He randomly selects 40 trading days in 2008 and determines the sample mean volume to be 41.06 million shares. Test the analyst’s claim at the =0.10 level of significance using P-values.


Solution

Step 1: The analyst wants to know if the stock volume has increased. This is a right-tailed test with

H0: =35.14 versus H1: >35.14.

We want to know the probability of obtaining a sample mean of 41.06 or more from a population where the mean is assumed to be 35.14.


Solution


Step 3: The test statistic is

z0 41.06 35.1415.07 40

2.48


Solution

Step 4: P(Z > z0)=P(Z > 2.48)=0.0066.

The probability of obtaining a sample mean of 41.06 or more from a population whose mean is 35.14 is 0.0066.


Solution

Step 5: Since the P-value= 0.0066 is less than the level of significance, 0.10, we reject the null hypothesis.

Step 6: There is sufficient evidence to reject the null hypothesis and to conclude that the mean volume of Apple stock is greater than 35.14 million shares.


Parallel Example 4: The P-Value Approach of Hypothesis Testing: Two-Tailed, Small Sample

A can of 7-Up states that the contents of the can are 355 ml. A quality control engineer is worried that the filling machine is miscalibrated. In other words, she wants to make sure the machine is not under- or over-filling the cans. She randomly selects 9 cans of 7-Up and measures the contents. She obtains the following data.

351 360 358 356 359 358 355 361 352Use the P-value approach to determine if there is evidenceat the =0.05 level of significance to support the quality control engineer’s claim. Prior experience indicates that =3.2ml.Source: Michael McCraith, Joliet Junior College


Solution

The quality control engineer wants to know if the mean content is different from 355 ml. Since we have already verified that the data come from a population that is approximately normal with no outliers, we will continue with step 1.


Solution

Step 1: The quality control engineer wants to know if the content has changed. This is a two-tailed test with

H0: =355 versus H1: ≠355.


Solution


Step 3: Recall that the sample mean is 356.667. The test statistic is then

z0 356.667 355

3.2 91.56


Solution

Step 4: Since this is a two-tailed test,

P-value = P(Z < -1.56 or Z > 1.56) = 2*(0.0594)=0.1188.

The probability of obtaining a sample mean that is more than 1.56 standard deviations away from the assumed mean of 355 ml is 0.1188.


Solution

Step 5: Since the P-value= 0.1188 is greater than the level of significance, 0.05, we fail to reject the null hypothesis.

Step 6: There is insufficient evidence to conclude that the mean content of 7-Up cans differs from 355 ml.


One advantage of using P-values over the classical approach in hypothesis testing is that P-values provide information regarding the strength of the evidence. Another is that P-values are interpreted the same way regardless of the type of hypothesis test being performed. the lower the P-value, the stronger the evidence against the statement in the null hypothesis.


Objective 4

• Test Hypotheses about a Population Mean with Known Using Confidence Intervals


When testing H0: = 0 versus H1: ≠ 0, if a (1-)·100% confidence interval contains 0, we do not reject the null hypothesis. However, if the confidence interval does not contain 0, we have sufficient evidence that supports the statement in the alternative hypothesis and conclude that ≠ 0 at the level of significance, .


Parallel Example 6: Testing Hypotheses about a Population Mean Using a Confidence Interval

Test the hypotheses presented in Parallel Examples 2 and 4 at the =0.05 level of significance by constructing a 95% confidence interval about , the population mean can content.


Solution

Lower bound:

Upper bound:

We are 95% confident that the mean can content is between 354.6 ml and 358.8 ml. Since the mean stated in the null hypothesis is in this interval, there is insufficient evidence to reject the hypothesis that the mean can content is 355 ml.

356.667 1.963.2

9354.58

356.667 1.963.2

9358.76


Objective 5

• Distinguish between Statistical Significance and Practical Significance


When a large sample size is used in a hypothesis test, the results could be statistically significant even though the difference between the sample statistic and mean stated in the null hypothesis may have no practical significance.


Practical significance refers to the idea that, while small differences between the statistic and parameter stated in the null hypothesis are statistically significant, the difference may not be large enough to cause concern or be considered important.


Parallel Example 7: Statistical versus Practical Significance

In 2003, the average age of a mother at the time of her first childbirth was 25.2. To determine if the average age has increased, a random sample of 1200 mothers is taken and is found to have a sample mean age of 25.5. Assuming a standard deviation of 4.8, determine whether the mean age has increased using a significance level of =0.05.


SolutionStep 1: To determine whether the mean age has

increased, this is a right-tailed test with

H0: =25.2 versus H1: >25.2.


Step 3: Recall that the sample mean is 25.5. The test statistic is then

z0 25.5 25.24.8 1200

2.17


SolutionStep 4: Since this is a right-tailed test,

P-value = P(Z > 2.17) = 0.015.

The probability of obtaining a sample mean that is more than 2.17 standard deviations above the assumed mean of 25.2 is 0.015.

Step 5: Because the P-value is less than the level of significance, 0.05, we reject the null hypothesis.


SolutionStep 6: There is sufficient evidence at the 0.05

significance level to conclude that the mean age of a mother at the time of her first childbirth is greater than 25.2.

Although we found the difference in age to be significant, there is really no practical significance in the age difference (25.2 versus 25.5).

Large sample sizes can lead to statistically significant results while the difference between the statistic and parameter is not enough to be considered practically significant.

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