Section 10.2

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Section 10.2

Hypothesis Testing for Population Means ( ) s Known

With valuable content added by D.R.S., University of Cordele






Objectives

o Use the rejection region to draw a conclusion. o Use the p-value to draw a conclusion.

There are two ways to get the answer and we learn BOTH.• The “Critical Value” method, comparing your z Test Statistic

to the z Critical Value, which was determined by the chosen α Level of Significance.

• The “p-Value method”, in which your z Test Statistic leads to a “p-Value”, an area under the normal curve, which is compared to the chosen α Level of Significance.






Hypothesis Testing for Population Means (s Known)

Test Statistic for a Hypothesis Test for a Population Mean (s Known)

When (1) the population standard deviation is known, (2) the sample taken is a simple random sample, and (3) either the sample size is at least 30 or the population distribution is approximately normal, THEN COMPUTE: the test statistic for a hypothesis test for a population mean is given by

xz

n







THEN COMPUTE: the test statistic for a hypothesis test for a population mean is given by

xz

n

My sample mean is how far away from the H0 Null Hypothesis mean?

Gauged by the standard deviation for a sample as defined in the Central Limit Theorem.







Test Statistic for a Hypothesis Test for a Population Mean (s Known) (cont.)

where x̄� is the sample mean, is the presumed value of the population mean from the null hypothesis, is the population standard deviation, and n is the sample size.






Rejection Regions

Value Left-tailed testValue Right-tailed testValue Two-tailed test

Alternative Hypothesis, Type of Hypothesis TestaH

The “rejection region” is the one-tail area of size α.

The “rejection region” consists of two tails, each of them of size α/2, left and right.

And the “Fail to Reject H0” region is the big area of size 1 – α.






Rejection Regions

Decision Rule for Rejection Regions Reject the null hypothesis, H0, if the test statistic calculated from the sample data falls within the rejection region.

Refer to the next three pictures. They illustrate the Left-Tailed, Right-Tailed, and Two-Tailed cases.

Be able to draw the right kind of picture for each Hypothesis Test you perform.






A Left-Tailed Hypothesis Test: Ha < μ

The critical z value is the negative z value which separates the left tail of area α.






A Right-Tailed Hypothesis Test: Ha > μ

The critical z value is the positive z value which separates the right tail of area α.






A Two-Tailed Hypothesis Test: Ha ≠ μ

The critical z values are the +/- z values which separate the two tails, area α/2 each.

Commonly-occurring critical values

Area in tail(s),

α, alpha(and c too)

One-Tailed Test

Two-Tailed Test

0.10 (0.90) 1.28 ±1.645

0.05 (0.95) 1.645 ±1.96

0.02 (0.98) 2.05 ±2.33

0.01 (0.99) 2.33 ±2.575

• These values for the α Level Of Significance (or c Level Of Confidence) are typical. You should already know how to find them using tables and using invNorm(). It is convenient to have this special table for reference.






Example 10.10: Using a Rejection Region in a Hypothesis Test for a Population Mean (Right Tailed, ‑ s Known)

The state education department is considering introducing new initiatives to boost the reading levels of fourth graders. The mean reading level of fourth graders in the state over the last 5 years was a Lexile reader measure of 800 L. (A Lexile reader measure is a measure of the complexity of the language that a reader is able to comprehend.) The developers of a new program claim that their techniques will raise the mean reading level of fourth graders by more than 50 L.






Example 10.10: Using a Rejection Region in a Hypothesis Test for a Population Mean (Right Tailed, ‑ s Known) (cont.)

To assess the impact of their initiative, the developers were given permission to implement their ideas in the classrooms. At the end of the pilot study, a simple random sample of 1000 fourth graders had a mean reading level of 856 L. It is assumed that the population standard deviation is 98 L. Using a 0.05 level of significance, should the findings of the study convince the education department of the validity of the developers’ claim?







Solution Step 1: State the null and alternative hypotheses.

The developers want to show that their classroom techniques will raise the fourth graders’ mean reading level to more than 850 L. This is written mathematically as > 850, and since it is the research hypothesis, it will be Ha. The mathematical opposite is ≤ 850. Thus, we have the following hypotheses.

0: 850: 850a

HH







Step 2: Determine which distribution to use for the test statistic, and state the level of significance. Note that the hypotheses are statements about the population mean, is known, the sample is a simple random sample, and the sample size is at least 30. Thus, we will use a normal distribution, which means we need to use the z-test statistic.







In addition to determining which distribution to use for the test statistic, we need to state the level of significance. The problem states that a = 0.05.

Step 3: Gather data and calculate the necessary sample statistics. At the end of the pilot study, a simple random sample of 1000 fourth graders had a mean reading level of 856 L. The population standard deviation is 98 L.







Thus, the test statistic is calculated as follows.

856 850

1.9

9800

41 0

xz

n







Step 4: Draw a conclusion and interpret the decision. Remember that we determine the type of test based on the alternative hypothesis. In this case, the alternative hypothesis contains “>,” which indicates that this is a right tailed test. ‑To determine the rejection region, we need a z-value so that 0.05 of the area under the standard normal curve is to its right. If 0.05 of the area is to the right then 1 0.05 = 0.95 is the area to the left.







If we look up 0.9500 in the body of the cumulative z-table, the corresponding critical z-value is 1.645. Alternately, we can look up c = 0.95 in the table of critical z-values for rejection regions. Either way, the rejection region is z ≥ 1.645.







The z-value of 1.94 falls in the rejection region. So the conclusion is to reject the null hypothesis. Thus the evidence collected suggests that the education department can be 95% sure of the validity of the developers’ claim that the mean Lexile reader measure of fourth graders will increase by more than 50 points.






p-Values

p-valueA p-value is the probability of obtaining a sample statistic as extreme or more extreme than the one observed in the data, when the null hypothesis, H0, is assumed to be true.

We’ll give examples of how to find a p-value by hand.But usually you’ll get the p-value for free from the TI-84’s ZTest or TTest feature.

How to calculate the p value for your z Test Statistic

You’ve calculated your z = Test Statistic using the formula.Here’s how to define your p-value:• If it’s a left-tailed test, what’s the area to the left of

your z ?• If it’s a right tailed test, what’s the area to the right

of your z?• If it’s a two-tailed test, what’s 2x the area to the

left/right of your neg/pos z?






Example 10.11: Calculating the p-Value for a z-Test Statistic for a Left-Tailed Test

Calculate the p-value for a hypothesis test with the following hypotheses. Assume that data have been collected and the test statistic was calculated to be z = 1.34.

0: 0.15: 0.15a

HH






Example 10.11: Calculating the p-Value for a z-Test Statistic for a Left-Tailed Test (cont.)

Solution The alternative hypothesis tells us that this is a left-tailed test. Therefore, the p-value for this situation is the probability that z is less than or equal to 1.34, written p-value = P(z −1.34). Use a table or appropriate technology to find the area under the standard normal curve to the left of z = 1.34. Thus, the p-value ≈ 0.0901.

normalcdf( ________, ________ )






Example 10.11: Calculating the p-Value for a z-Test Statistic for a Left-Tailed Test (cont.)






Example 10.12: Calculating the p-Value for a z-Test Statistic for a Right-Tailed Test


0: 0.43: 0.43a

HH






Example 10.12: Calculating the p-Value for a z-Test Statistic for a Right-Tailed Test (cont.)

Solution The alternative hypothesis tells us that this is a right-tailed test. Therefore, the p-value for this situation is the probability that z is greater than or equal to 2.78, written p-value = P(z 2.78). Use a table or appropriate technology to find the area under the standard normal curve to the right of z = 2.78. Thus, the p-value ≈ 0.0027.

normalcdf( ________, ________ )






Example 10.12: Calculating the p-Value for a z-Test Statistic for a Right-Tailed Test (cont.)






Example 10.13: Calculating the p-Value for a z-Test Statistic for a Two-Tailed Test


0: 0.78: 0.78a

HH






Example 10.13: Calculating the p-Value for a z-Test Statistic for a Two-Tailed Test (cont.)

Solution The alternative hypothesis tells us that this is a two-tailed test. Thus, the p-value for this situation is the probability that z is either less than or equal to 2.15 or greater than or equal to 2.15, which is written mathematically as Use a table or appropriate technology to find the area under the standard normal curve to the left of z1 = −2.15. The area to the left of z1 = −2.15 is 0.0158.

-value 2.15 .p P z







Since the standard normal curve is symmetric about its mean, 0, the area to the right of z2 = 2.15 is also 0.0158. Thus, the p-value is calculated as follows.

-value 0.01

0.0316

58 2p

normalcdf( ________, ________ )*2






Example 10.14: Calculating the p-Value for a z-Test Statistic Using a TI-83/84 Plus Calculator

Enter normalcdf(856,1û99, 850, 98/ð (1000)) and press .

They demonstrate here how to find the p-value using normalcdf and “x” values.Recall that the sample mean was bar_x = 856, it was a right tailed-test, the null hypothesis said the mean score was 850, the population standard deviation was 98, and the sample size was 1000.

But in practice, we’re probably going to use TI-84 Z-Test instead to give us the p-values, so don’t put a lot of emphasis on this slide.






Example 10.14: Calculating the p-Value for a z-Test Statistic Using a TI-83/84 Plus Calculator (cont.)

The p-value returned is approximately 0.0264, as shown in the screenshot.

But again, Z-Test is a much betterway !!!!

Z-Test for this problem.

STAT, TESTS, 1:Z-Test. Since they gave us summary statistics, we pick “Stats” (not “Data”, which is for a problem where we have the individuals’ scores).μ0 comes from H0

σ is the “known” stdev.bar-x from our samplen is our sample sizeGive the Ha inequality

Highlight Calculate and press ENTER.

Z-Test for this problem.

STAT, TESTS, 1:Z-Test. Here are the results.It reminds you of what you told it was the alternative hypothesis. Have a look and make sure it’s what you wanted.It tells you the Test Statistic.It tells you the p-Value.It reminds you of the sampledata, the sample mean andsample size you told it.






p-Values

Conclusions Using p-Values • If p-value ≤ a, then reject the null hypothesis. • If p-value > a, then fail to reject the null

hypothesis.






Example 10.15: Determining the Conclusion to a Hypothesis Test Using the p-Value

For a certain hypothesis test, the p-value is calculated to be p-value = 0.0146. a. If the stated level of significance is 0.05, what is the conclusion to the hypothesis test? b. If the level of confidence is 99%, what is the conclusion to the hypothesis test? Solution a. The level of significance is a = 0.05. Next, note that 0.0146 < 0.05. Thus, p-value ≤ a, so we reject the

null hypothesis.






Example 10.15: Determining the Conclusion to a Hypothesis Test Using the p-Value (cont.)

b. The level of confidence is 99%, so the level of significance is calculated as follows.

Next, note that 0.0146 > 0.01. Thus, p-value > a, so we fail to reject the null hypothesis.

a

11 0.990.01

c






Example 10.16: Performing a Hypothesis Test for a Population Mean (Right-Tailed, s Known)

A researcher claims that the mean age of women in California at the time of a first marriage is higher than 26.5 years. Surveying a simple random sample of 213 newlywed women in California, the researcher found a mean age of 27.0 years. Assuming that the population standard deviation is 2.3 years and using a 95% level of confidence, determine if there is sufficient evidence to support the researcher’s claim.

Example 10.16: TI-84 Z-Test solution

You still have to write the hypotheses:• Null Hypothesis: ________ Alternative: ________What is the α Level Of Significance? α = _______What inputs do you give to the TI-84 Z-Test?• _______, ________• _______, ________• : ≠ or < or >The TI-84 Z-Test output gives us a p-value of ________.Therefore we { reject or fail to reject } H0.






Example 10.16: Performing a Hypothesis Test for a Population Mean (Right-Tailed, s Known) (cont.)


The researcher’s claim is investigating the mean age at first marriage for women in California. Therefore, the research hypothesis, Ha, is that the mean age is greater than 26.5, > 26.5. The logical opposite is ≤ 26.5. Thus, the null and alternative hypotheses are stated as follows.

0: 26.5: 26.5a

HH







Step 2: Determine which distribution to use for the test statistic, and state the level of significance. Note that the hypotheses are statements about the population mean, the population standard deviation is known, the sample is a simple random sample, and the sample size is at least 30. Thus, we will use a normal distribution and calculate the z-test statistic.







We will draw a conclusion by computing the p-value for the calculated test statistic and comparing the value to a. For this hypothesis test, the level of confidence is 95%, so the level of significance is calculated as follows.

a

11 0.950.05

c







Step 3: Gather data and calculate the necessary sample statistics. From the information given, we know that the presumed value of the population mean is = 26.5, the sample mean is x̄� = 27.0, the population standard deviation is = 2.3, and the sample size is n = 213.








27.0 26.52.3

131

23. 7

xz

n







Step 4: Draw a conclusion and interpret the decision.

The alternative hypothesis tells us that we have a right-tailed test. Therefore, the p-value for this test statistic is the probability of obtaining a test statistic greater than or equal to z = 3.17, written as

To find the p-value, we need to find the area under the standard normal curve to the right of z = 3.17.

-value 3.17 .p P z







Using a normal distribution table or appropriate technology, we find that the area is p-value ≈ 0.0008. Comparing this p-value to the level of significance, we see that 0.0008 < 0.05, so p-value ≤ a. Thus, the conclusion is to reject the null hypothesis. Therefore, we can say with 95% confidence that there is sufficient evidence to support the researcher’s claim that the mean age at first marriage for women in California is higher than 26.5 years.






Example 10.17: Performing a Hypothesis Test for a Population Mean (Two-Tailed, s Known)

A recent study showed that the mean number of children for women in Europe is 1.5. A global watch group claims that German women have a mean fertility rate that is different from the mean for all of Europe. To test its claim, the group surveyed a simple random sample of 128 German women and found that they had a mean fertility rate of 1.4 children. The population standard deviation is assumed to be 0.8. Is there sufficient evidence to support the claim made by the global watch group at the 90% level of confidence?

Example 10.17: TI-84 Z-Test solution

You still have to write the hypotheses:• Null Hypothesis: ________ Alternative: ________What is the α Level Of Significance? α = _______What inputs do you give to the TI-84 Z-Test?• _______, ________• _______, ________• : ≠ or < or >The TI-84 Z-Test output gives us a p-value of ________.Therefore we { reject or fail to reject } H0.






Example 10.17: Performing a Hypothesis Test for a Population Mean (Two-Tailed, s Known) (cont.)


The watch group is investigating whether the mean fertility rate for German women is

different from the mean for all of Europe. Thus, they need to find evidence that the mean fertility rate is not equal to 1.5.







So the research hypothesis, Ha, is that the mean does not equal 1.5, ≠ 1.5. The logical opposite is = 1.5. Thus, the null and alternative hypotheses are stated as follows.

0: 1.5: 1.5a

HH







Step 2: Determine which distribution to use for the test statistic, and state the level of significance.

Note that the hypotheses are statements about the population mean, is known, the sample is a simple random sample, and the sample size is at least 30. Thus, we will use a normal distribution and calculate the z-test statistic.







We will draw a conclusion by computing the p-value for the calculated test statistic and comparing the value to a. For this hypothesis test, the level of confidence is 90%, so the level of significance is calculated as follows.

a

11 0.900.10

c







Step 3: Gather data and calculate the necessary sample statistics. From the information given, we know that the presumed value of the population mean is = 1.5, the sample mean is the

population standard deviation is = 0.8, and the sample size is n = 128.

1.4,x








1.4 1.50.81

12

.48

1

xz

n







Step 4: Draw a conclusion and interpret the decision. The alternative hypothesis tells us that we have a two-tailed test. Therefore, the p-value for this test statistic is the probability of obtaining a test statistic that is either less than or equal to z1 = −1.41 or greater than or equal to z2 = 1.41, which is written mathematically as To find the p-value, we need to find the sum of the areas under the standard normal curve to the left of z1 = −1.41 and to the right of z2 = 1.41.

-value 1.41 .p P z







By looking up z1 = −1.41 in the cumulative normal distribution table, we find that the area to the left is equal to 0.0793. Since the standard normal curve is symmetric about its mean, 0, the area to the right of z2 = 1.41 is also 0.0793. Thus the p-value is calculated as follows.

-value 0.07

0.

2

6

93

158

p







Comparing the p-value to the level of significance, we see that 0.1586 > 0.10, so p-value > a. Thus, the conclusion is to fail to reject the null hypothesis. This means that, at a 90% level of confidence, the evidence does not support the watch group’s claim that the fertility rate of German women is different from the mean for all of Europe.






Example 10.18: Performing a Hypothesis Test for a Population Mean Using a TI-83/84 Plus Calculator (Two-Tailed, s Known)

A recent study showed that the mean number of children for women in Europe is 1.5. A global watch group claims that German women have a mean fertility rate that is different from the mean for all of Europe. To test its claim, the group surveyed a simple random sample of 128 German women and found that they had a mean fertility rate of 1.4 children. The population standard deviation is assumed to be 0.8. Is there sufficient evidence to support the claim made by the global watch group at the 90% level of confidence? Use a TI-83/84 Plus calculator to answer the question.






Example 10.18: Performing a Hypothesis Test for a Population Mean Using a TI-83/84 Plus Calculator (Two-Tailed, s Known) (cont.)

Solution Even when working with technology, Steps 1 and 2 are the same, so we will just reiterate the important information for these two steps from Example 10.17. Step 1: State the null and alternative hypotheses.

The hypotheses are stated as follows.

0: 1.5: 1.5a

HH







Step 2: Determine which distribution to use for the test statistic, and state the level of significance. Since is known, the sample is a simple random sample, and the sample size is at least 30, we use the normal distribution and z-test statistic for this hypothesis test for the population mean. The level of significance is a = 0.10.







Step 3: Gather data and calculate the necessary sample statistics. This is where we begin to use a TI-83/84 Plus calculator. Let’s start by writing down the information from the problem, as we will need to enter these values into the calculator. We know that the presumed value of the population mean is = 1.5, the sample mean is x̄� = 1.4, the population standard deviation is = 0.8, and the sample size is n = 128.







Press , scroll to TESTS, and choose option 1:Z-Test. Since we know the sample statistics, choose Stats instead of Data. For Ã€, enter the value from the null hypothesis, thus enter 1.5 for Ã€. Enter the rest of the given values, as shown in the following screenshot on the left. Choose the alternative hypothesis øÃ€. Highlight Calculate and press .







The output screen, shown on the right, displays the alternative hypothesis, calculates the z-test statistic and the p-value, and then reiterates the sample mean and sample size that were entered.







Step 4: Draw a conclusion and interpret the decision. The p-value given by the calculator is approximately 0.1573. It is not identical to the one we found when using the table because there were several intermediate steps when we found the p-value by hand that reduced the accuracy of the p-value. However, the conclusion is the same since the two p-values are extremely close.







Since p-value > a, the conclusion is to fail to reject the null hypothesis. This means that at a 90% level of confidence, the evidence does not support the watch group’s claim that the fertility rate of German women is different from the mean for all of Europe.

Excel’s Z.TEST(data array, μ, σ) function

• Give it the cells (“range”) of the sample data values.• And the μ mean from the null hypothesis.• And the σ population standard deviation.• It gives you back the one-tailed p-value.

• So it calculates a z = test statistic based on the sample.

• And then it calculates the area in the tail beyond that z

• Either left tail beyond negative z• Or right tail beyond positive z, it knows which one.

Excel’s Z.TEST(data array, μ, σ) function

• Give it the cells (“range”) of the sample data values.• And the μ mean from the null hypothesis.• And the σ population standard deviation.• It gives you back the one-tailed p-value.• You compare that p-value to the α level of

significance that was pre-selected in the design.• If p-value < α, then reject H0.

• If p-value > α, then fail to reject H0.

Excel’s Z.TEST(data array, μ, σ) if you have a Two-Tailed Situation

• Recall the two-tailed situation: Ha has a ≠.• So the rejection regions are two tails of area α/2.• Excel Help says to do this to get the p-value:

2 * MIN(Z.TEST(data , μ, σ), 1 – Z.TEST(data , μ, σ))(whereas we tell TI-84 it’s two-tailed and TI-84 takes care of doubling it transparently, thank you.)

• And then decide as usual:• If p-value < α, then reject H0.

• If p-value > α, then fail to reject H0.

Section 10.2

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