© 2010 pearson prentice hall. all rights reserved chapter hypothesis tests regarding a parameter 10

© 2010 Pearson Prentice Hall. All rights reserved

Chapter

Hypothesis Tests Regarding a Parameter

10


Section

The Language of Hypothesis Testing

10.1

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

1. Determine the null and alternative hypotheses

2. Explain Type I and Type II errors

3. State conclusions to hypothesis tests

Objectives


Objective 1

• Determine the Null and Alternative Hypotheses


A hypothesis is a statement regarding a characteristic of one or more populations.

In this chapter, we look at hypotheses regarding a single population parameter.


Examples of Claims Regarding a Characteristic of a Single Population

• In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.

Source: ReadersDigest.com poll created on 2008/05/02



• According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.




• According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher wonders if the mean length of a call has increased since then.

• Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased.



CAUTION!

We test these types of statements using sample data because it is usually impossible or impractical to gain access to the entire population. If population data are available, there is no need for inferential statistics.


Hypothesis testing is a procedure, based on sample evidence and probability, used to test statements regarding a characteristic of one or more populations.


1. A statement is made regarding the nature of the population.

2. Evidence (sample data) is collected in order to test the statement.

3. The data is analyzed to assess the plausibility of the statement.

Steps in Hypothesis Testing


The null hypothesis, denoted H0, is a statement to be tested. The null hypothesis is a statement of no change, no effect or no difference. The null hypothesis is assumed true until evidence indicates otherwise. In this chapter, it will be a statement regarding the value of a population parameter.


The alternative hypothesis, denoted H1, is a statement that we are trying to find evidence to support. In this chapter, it will be a statement regarding the value of a population parameter.


In this chapter, there are three ways to set up the null and alternative hypotheses:

1. Equal versus not equal hypothesis (two-tailed test)H0: parameter = some value

H1: parameter ≠ some value





2. Equal versus less than (left-tailed test)H0: parameter = some value

H1: parameter < some value





2. Equal versus less than (left-tailed test)H0: parameter = some value

H1: parameter < some value

3. Equal versus greater than (right-tailed test)H0: parameter = some value

H1: parameter > some value


“In Other Words”

The null hypothesis is a statement of “status quo” or “no difference” and always contains a statement of equality. The null hypothesis is assumed to be true until we have evidence to the contrary. The claim that we are trying to gather evidence for determines the alternative hypothesis.


For each of the following claims, determine the null and alternative hypotheses. State whether the test is two-tailed, left-tailed or right-tailed.

a) In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.

b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher wonders if the mean length of a call has increased since then.

c) Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased.

Parallel Example 2: Forming Hypotheses



The hypothesis deals with a population proportion, p. If the percentage participating in charity work is no different than in 2008, it will be 0.62 so the null hypothesis is H0: p=0.62.

Since the researcher believes that the percentage is different today, the alternative hypothesis is a two-tailed hypothesis: H1: p≠0.62.

Solution


b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.

The hypothesis deals with a population mean, . If the mean call length on a cellular phone is no different than in 2006, it will be 3.25 minutes so the null hypothesis is H0: =3.25.

Since the researcher believes that the mean call length has increased, the alternative hypothesis is: H1: > 3.25, a right-tailed test.

Solution


c) Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased.

The hypothesis deals with a population standard deviation, . If the standard deviation with the new equipment has not changed, it will be 0.23 ounces so the null hypothesis is H0: = 0.23.

Since the quality control manager believes that the standard deviation has decreased, the alternative hypothesis is: H1: < 0.23, a left-tailed test.

Solution


Objective 2

• Explain Type I and Type II Errors


1. We reject the null hypothesis when the alternative hypothesis is true. This decision would be correct.

Four Outcomes from Hypothesis Testing



2. We do not reject the null hypothesis when the null hypothesis is true. This decision would be correct.





3. We reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a Type I error.





3. We reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a Type I error.

4. We do not reject the null hypothesis when the alternative hypothesis is true. This decision would be incorrect. This type of error is called a Type II error.



For each of the following claims, explain what it would mean to make a Type I error. What would it mean to make a Type II error?



Parallel Example 3: Type I and Type II Errors



A Type I error is made if the researcher concludes that p≠0.62 when the true proportion of Americans 18 years or older who participated in some form of charity work is currently 62%.

A Type II error is made if the sample evidence leads the researcher to believe that the current percentage of Americans 18 years or older who participated in some form of charity work is still 62% when, in fact, this percentage differs from 62%.

Solution



A Type I error occurs if the sample evidence leads the researcher to conclude that >3.25 when, in fact, the actual mean call length on a cellular phone is still 3.25 minutes.

A Type II error occurs if the researcher fails to reject the hypothesis that the mean length of a phone call on a cellular phone is 3.25 minutes when, in fact, it is longer than 3.25 minutes.

Solution


= P(Type I Error)

= P(rejecting H0 when H0 is true)

= P(Type II Error)

= P(not rejecting H0 when H1 is true)


The probability of making a Type I error, , is chosen by the researcher before the sample data is collected.

The level of significance, , is the probability of making a Type I error.


“In Other Words”

As the probability of a Type I error increases, the probability of a Type II error decreases, and vice-versa.


Objective 3

• State Conclusions to Hypothesis Tests


CAUTION!

We never “accept” the null hypothesis because without having access to the entire population, we don’t know the exact value of the parameter stated in the null. Rather, we say that we do not reject the null hypothesis. This is just like the court system. We never declare a defendant “innocent”, but rather say the defendant is “not guilty”.


According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.

• Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion.

a) Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion.

Parallel Example 4: Stating the Conclusion


a) Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion.

The statement in the alternative hypothesis is that the mean call length is greater than 3.25 minutes. Since the null hypothesis (=3.25) is rejected, we conclude that there is sufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes.

Solution


b) Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion.

Since the null hypothesis (=3.25) is not rejected, we conclude that there is insufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. In other words, the sample evidence is consistent with the mean call length equaling 3.25 minutes.

Solution


Section

Hypothesis Tests for a Population Mean-Population Standard Deviation Known

10.2


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:


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 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.



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 35.14

15.07 402.48


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


351 360 358 356 359 358 355 361 352

Use 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.


Solution

Step 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 25.2

4.8 12002.17


Solution

Step 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.


Solution

Step 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.


Section

Hypothesis Tests for a Population Mean- Population Standard Deviation Unknown

10.3


Objectives

1. Test the hypotheses about a population mean with unknown


To test hypotheses regarding the population mean assuming the population standard deviation is unknown, we use the t-distribution rather than the Z-distribution. When we replace with s,

follows Student’s t-distribution with n-1 degrees of freedom.

x s

n


1. The t-distribution is different for different degrees of freedom.

Properties of the t-Distribution



2. The t-distribution is centered at 0 and is symmetric about 0.




2. The t-distribution is centered at 0 and is symmetric about 0.

3. The area under the curve is 1. Because of the symmetry, the area under the curve to the right of 0 equals the area under the curve to the left of 0 equals 1/2.



4. As t increases (or decreases) without bound, the graph approaches, but never equals, 0.



4. As t increases (or decreases) without bound, the graph approaches, but never equals, 0.

5. The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution because using s as an estimate of introduces more variability to the t-statistic.



6. As the sample size n increases, the density curve of t gets closer to the standard normal density curve. This result occurs because as the sample size increases, the values of s get closer to the values of by the Law of Large Numbers.



Objective 1

• Test hypotheses about a population mean with unknown


Testing Hypotheses Regarding a Population Mean with σ Unknown

To test hypotheses regarding the population mean with unknown, we use the following steps, provided that:


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. The hypotheses can be structured in one of three ways:


Step 3: Compute the test statistic

which follows Student’s t-distribution with n-1 degrees of freedom.

t0 x 0

s

n


Step 4: Use Table VI to determine the critical value using n-1 degrees of freedom.

Classical Approach


Classical Approach

Two-Tailed


Classical Approach

Left-Tailed


Classical Approach

Right-Tailed



Classical Approach


Step 4: Use Table VI to estimate the P-value using n-1 degrees of freedom.

P-Value Approach


P-Value Approach

Two-Tailed


P-Value Approach

Left-Tailed


P-Value Approach

Right-Tailed


Step 5: If the P-value < , reject the null hypothesis.

P-Value Approach


Parallel Example 1: Testing a Hypothesis about a Population Mean, Large Sample

Assume the resting metabolic rate (RMR) of healthy males in complete silence is 5710 kJ/day. Researchers measured the RMR of 45 healthy males who were listening to calm classical music and found their mean RMR to be 5708.07 with a standard deviation of 992.05.

At the =0.05 level of significance, is there evidence to conclude that the mean RMR of males listening to calm classical music is different than 5710 kJ/day?


Solution

We assume that the RMR of healthy males is 5710 kJ/day. This is a two-tailed test since we are interested in determining whether the RMR differs from 5710 kJ/day.

Since the sample size is large, we follow the steps for testing hypotheses about a population mean for large samples.


Solution

Step 1: H0: =5710 versus H1: ≠5710


Step 3: The sample mean is = 5708.07 and the sample standard deviation is s=992.05. The test statistic is

t0 5708.07 5710

992.05 45 0.013

x


Solution: Classical Approach

Step 4: Since this is a two-tailed test, we determine the critical values at the =0.05 level of significance with n-1=45-1=44 degrees of freedom to be approximately -t0.025= -2.021 and

t0.025=2.021.

Step 5: Since the test statistic, t0=-0.013, is between

the critical values, we fail to reject the null hypothesis.


Solution: P-Value Approach

Step 4: Since this is a two-tailed test, the P-value is the area under the t-distribution with n-1=45-1=44 degrees of freedom to the left of -t0.025= -0.013

and to the right of t0.025=0.013. That is, P-value

= P(t < -0.013) + P(t > 0.013) = 2 P(t > 0.013). 0.50 < P-value.

Step 5: Since the P-value is greater than the level of significance (0.05<0.5), we fail to reject the null hypothesis.


Solution

Step 6: There is insufficient evidence at the =0.05 level of significance to conclude that the mean RMR of males listening to calm classical music differs from 5710 kJ/day.


Parallel Example 3: Testing a Hypothesis about a Population Mean, Small Sample

According to the United States Mint, quarters weigh 5.67 grams. A researcher is interested in determining whether the “state” quarters have a weight that is different from 5.67 grams. He randomly selects 18 “state” quarters, weighs them and obtains the following data.

5.70 5.67 5.73 5.61 5.70 5.67

5.65 5.62 5.73 5.65 5.79 5.73

5.77 5.71 5.70 5.76 5.73 5.72

At the =0.05 level of significance, is there evidence to conclude that state quarters have a weight different than 5.67 grams?


Solution

We assume that the weight of the state quarters is 5.67 grams. This is a two-tailed test since we are interested in determining whether the weight differs from 5.67 grams.

Since the sample size is small, we must verify that the data come from a population that is normally distributed with no outliers before proceeding to Steps 1-6.


Assumption of normality appears reasonable.


No outliers.


Solution

Step 1: H0: =5.67 versus H1: ≠5.67


Step 3: From the data, the sample mean is calculated to be 5.7022 and the sample standard deviation is s=0.0497. The test statistic is

t0 5.7022 5.67

.0497 182.75



Step 4: Since this is a two-tailed test, we determine the critical values at the =0.05 level of significance with n-1=18-1=17 degrees of freedom to be -t0.025= -2.11 and t0.025=2.11.

Step 5: Since the test statistic, t0=2.75, is greater than

the critical value 2.11, we reject the null hypothesis.



Step 4: Since this is a two-tailed test, the P-value is the area under the t-distribution with n-1=18-1=17 degrees of freedom to the left of -t0.025= -2.75

and to the right of t0.025=2.75. That is, P-value

= P(t < -2.75) + P(t > 2.75) = 2 P(t > 2.75). 0.01 < P-value < 0.02.

Step 5: Since the P-value is less than the level of significance (0.02<0.05), we reject the null hypothesis.


Solution

Step 6: There is sufficient evidence at the =0.05 level of significance to conclude that the mean weight of the state quarters differs from 5.67 grams.


Summary

Which test to use?

Provided that the population from which the sample is drawn is normal or that the sample size is large,

• if is known, use the Z-test procedures from Section 10.2

• if is unknown, use the t-test procedures from this Section.


Section

Hypothesis Tests for a Population Proportion

10.4


Objectives

1. Test the hypotheses about a population proportion

2. Test hypotheses about a population proportion using the binomial probability distribution.


Objective 1

• Test hypotheses about a population proportion.


Recall:

• The best point estimate of p, the proportion of the population with a certain characteristic, is given by

where x is the number of individuals in the sample with the specified characteristic and n is the sample size.

ˆ p x

n


Recall:

• The sampling distribution of is approximately normal, with mean and standard deviation

provided that the following requirements are satisfied:

1. The sample is a simple random sample.2. np(1-p) ≥ 10.3. The sampled values are independent of each

other.

ˆ p

ˆ p p

ˆ p p(1 p)

n


Testing Hypotheses Regarding a Population Proportion, p

To test hypotheses regarding the population proportion, we can use the steps that follow, provided that:

1. The sample is obtained by simple random sampling.

2. np0(1-p0) ≥ 10.3. The sampled values are independent of each

other.



Note: We use p0 in computing the standard error rather than . This is because, when we test a hypothesis, the null hypothesis is always assumed true.

z0 ˆ p p0

p0(1 p0)

n

ˆ p


Step 4: Use Table V to determine the critical value.

Classical Approach


Classical Approach

Two-Tailed

(critical value)


Classical Approach

Left-Tailed

(critical value)


Classical Approach

Right-Tailed

(critical value)



Classical Approach


Step 4: Use Table V to estimate the P-value.

P-Value Approach


P-Value Approach

Two-Tailed


P-Value Approach

Left-Tailed


P-Value Approach

Right-Tailed



P-Value Approach


Parallel Example 1: Testing a Hypothesis about a Population Proportion: Large Sample Size

In 1997, 46% of Americans said they did not trust the media “when it comes to reporting the news fully, accurately and fairly”. In a 2007 poll of 1010 adults nationwide, 525 stated they did not trust the media. At the =0.05 level of significance, is there evidence to support the claim that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997?

Source: Gallup Poll


Solution

We want to know if p>0.46. First, we must verify the requirements to perform the hypothesis test:

1. This is a simple random sample.

2. np0(1-p0)=1010(0.46)(1-0.46)=250.8>10

3. Since the sample size is less than 5% of the population size, the assumption of independence is met.


Solution

Step 1: H0: p=0.46 versus H1: p>0.46


Step 3: The sample proportion is .

The test statistic is then

ˆ p 525

10100.52

z0 0.52 0.46

0.46(1 0.46)

1010

3.83



Step 4: Since this is a right-tailed test, we determine the critical value at the =0.05 level of significance to be

z0.05= 1.645.

Step 5: Since the test statistic, z0=3.83, is

greater than the critical value 1.645, we reject the null hypothesis.



Step 4: Since this is a right-tailed test, the P- value is the area under the standard normal distribution to the right of the test statistic z0=3.83. That is, P-value =

P(Z > 3.83)≈0.

Step 5: Since the P-value is less than the level of significance, we reject the null

hypothesis.


Solution

Step 6: There is sufficient evidence at the =0.05 level of significance to

conclude that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997.


Objective 2

• Test hypotheses about a population proportion using the binomial probability distribution.


For the sampling distribution of to be approximately normal, we require np(1-p) be at least 10. What if this requirement is not met?

ˆ p


Parallel Example 4: Hypothesis Test for a Population Proportion: Small Sample Size

In 2006, 10.5% of all live births in the United States were to mothers under 20 years of age. A sociologist claims that births to mothers under 20 years of age is decreasing. She conducts a simple random sample of 34 births and finds that 3 of them were to mothers under 20 years of age. Test the sociologist’s claim at the = 0.01 level of significance.



Approach:

Step 1: Determine the null and alternative hypotheses

Step 2: Check whether np0(1-p0) is greater than or equal to 10, where p0 is the proportion stated in the null hypothesis. If it is, then the sampling distribution of is approximately normal and we can use the steps for a large sample size. Otherwise we use the following Steps 3 and 4.

ˆ p



Approach:

Step 3: Compute the P-value. For right-tailed tests, the P-value is the probability of obtaining x

or more successes. For left-tailed tests, the P- value is the probability of obtaining x or fewer successes. The P-value is always computed with the proportion given in the null hypothesis.

Step 4: If the P-value is less than the level of significance, , we reject the null hypothesis.


Solution

Step 1: H0: p=0.105 versus H1: p<0.105

Step 2: From the null hypothesis, we have p0=0.105. There were 34 mothers sampled, so np0(1-p0)=3.57<10. Thus, the sampling distribution of is not approximately normal.

ˆ p


Solution

Step 3: Let X represent the number of live births in the United States to mothers under 20

years of age. We have x=3 successes in n=34 trials so = 3/34= 0.088. We want to determine whether this result is unusual

if the population mean is truly 0.105. Thus,

P-value = P(X ≤ 3 assuming p=0.105 )

= P(X = 0) + P(X =1)

+ P(X =2) + P(X = 3)

= 0.51

ˆ p


Solution

Step 4: The P-value = 0.51 is greater than the level of significance so we do not

reject H0. There is insufficient evidence to conclude that the percentage of live births in the United States to mothers under the age of 20 has decreased below the 2006 level of 10.5%.


Section

Hypothesis Tests for a Population Standard Deviation

10.5


Objective

1. Test hypotheses about a population standard deviation


Chi-Square Distribution

If a simple random sample of size n is obtained from a normally distributed population with mean and standard deviation , then

where s2 is a sample variance has a chi-square distribution with n-1 degrees of freedom.

2 (n 1)s2

2


Characteristics of the Chi-Square Distribution

1. It is not symmetric.2. The shape of the chi-square distribution

depends on the degrees of freedom, just as with Student’s t-distribution.

3. As the number of degrees of freedom increases, the chi-square distribution becomes more nearly symmetric.

4. The values of 2 are nonnegative, i.e., the values of 2 are greater than or equal to 0.


Objective 1

• Test hypotheses about a population standard deviation.


Testing Hypotheses about a Population Variance or Standard Deviation

To test hypotheses about the population variance or standard deviation, we can use the following steps, provided that


2. The population is normally distributed.



02

(n 1)s2

02


Step 4: Use Table VII to determine the critical value using n-1 degrees of freedom.

Classical Approach


Classical Approach

Two-Tailed


Classical Approach

Left -Tailed


Classical Approach

Right-Tailed



Classical Approach


Step 4: Use Table VII to determine the approximate P-value for a left-tailed or right-tailed test

by determining the area under the chi- square distribution with n-1 degrees of freedom to the left (for a left-tailed test) or right (for a right-tailed test) of the test statistic. For two-tailed tests, we recommend the use of technology

or confidence intervals.

P-Value Approach


P-Value Approach

Left-Tailed


P-Value Approach

Right-Tailed



P-Value Approach


CAUTION!

The procedures in this section are not robust.


Parallel Example 1: Testing a Hypothesis about a Population Standard Deviation


351 360 358 356 359358 355 361 352

In section 9.2, we assumed the population standard deviationwas 3.2. Test the claim that the population standarddeviation, , is greater than 3.2 ml at the =0.05level of significance.


Solution

Step 1: H0: =3.2 versus H1: >3.2.

This is a right-tailed test.


Step 3: From the data, the sample standard deviation is computed to be s=3.464.

The test statistic is

02

(9 1)(3.464)2

3.22 9.374



Step 4: Since this is a right-tailed test, we determine the critical value at the =0.05 level of significance with 9-1=8 degrees of freedom to be 2

0.05= 15.507.

Step 5: Since the test statistic, , is less than the critical value, 15.507, we fail to reject the null hypothesis.

02 9.374



Step 4: Since this is a right-tailed test, the P- value is the area under the 2-distribution with 9-1=8 to the right of the test statistic

. That is, P-value = P(2 > 9.37)>0.10.

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

hypothesis.

02 9.374


Solution:

Step 6: There is insufficient evidence at the =0.05 level of significance to conclude

that the standard deviation of the can content of 7-Up is greater than 3.2 ml.


Section

Putting It Together: Which Method Do I Use?

10.6


Objective

1. Determine the appropriate hypothesis test to perform


Objective 1

• Determine the Appropriate Hypothesis Test to Perform


Section

The Probability of a Type II Error and the Power of the Test

10.7


Objective

1. Determine the probability of making a Type II error

2. Compute the power of the test


Objective 1

• Determine the Probability of Making a Type II Error


The Probability of a Type II Error

Step 1: Determine the sample mean that separates the rejection region from the nonrejection region.


Step 2: Draw a normal curve whose mean is a particular value from the alternative hypothesis with the sample mean(s) found in Step 1 labeled.


Step 3:


Parallel Example 1: Computing the Probability of a Type II Error

In Section 10.2, we tested the hypothesis that the mean trade volume of Apple stock was greater than 35.14 million shares, H0: =35.14 versus H1: >35.14, based upon a random sample of size n=40 with the population standard deviation, , assumed to be 15.07 million shares at the =0.1 level of significance. Compute the probability of a type II error given that the population mean is =40.62.


Solution

Step 1: Since z0.9=1.28, we let Z=1.28,

0=35.14, =15.07, and n=40 to find the sample mean that separates the rejection region from the nonrejection region:

For any sample mean less than 38.19, we do not reject the null hypothesis.

1.28 x 35.14

15.07

40

x 38.19


Solution

Step 2:


Solution

Step 3: P(Type II error)

= P(do not reject H0 given H1 is true) = P( < 38.19 given that =40.62)

=

= 0.1539

x

P Z 38.19 40.62

15.07

40

P(Z 1.02)


Objective 2

• Compute the Power of the Test


The complement of , 1- , is the probability of rejecting the null hypothesis when the alternative hypothesis is true. The value of 1- is referred to as the power of the test. The higher the power of the test, the more likely the test will reject the null when the alternative hypothesis is true.


Parallel Example 2: Computing the Power of the Test

Compute the power for the Apple stock example.


Solution

The power = 1- =1-0.1539 = 0.8461. There is a 84.61% chance of rejecting the null hypothesis when the true population mean is 40.62.

© 2010 pearson prentice hall. all rights reserved chapter hypothesis tests regarding a parameter 10

Documents