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## 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**Use the rejection region to draw a conclusion. 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**Hypothesis Testing for Population Means (s**Known) THEN COMPUTE: the test statistic for a hypothesis test for a population mean is given by 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.**Hypothesis Testing for Population Means (s**Known) Test Statistic for a Hypothesis Test for a Population Mean (s Known) (cont.) where x̄ is the sample mean, m is the presumed value of the population mean from the null hypothesis, s is the population standard deviation, and n is the sample size.**Rejection Regions**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**• 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?**Example 10.10: Using a Rejection Region in a Hypothesis Test**for a Population Mean (Right‑Tailed, s Known) (cont.) 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 m > 850, and since it is the research hypothesis, it will be Ha. The mathematical opposite is m ≤ 850. Thus, we have the following hypotheses.**Example 10.10: Using a Rejection Region in a Hypothesis Test**for a Population Mean (Right‑Tailed, s Known) (cont.) 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, s 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.**Example 10.10: Using a Rejection Region in a Hypothesis Test**for a Population Mean (Right‑Tailed, s Known) (cont.) 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.**Example 10.10: Using a Rejection Region in a Hypothesis Test**for a Population Mean (Right‑Tailed, s Known) (cont.) Thus, the test statistic is calculated as follows.**Example 10.10: Using a Rejection Region in a Hypothesis Test**for a Population Mean (Right‑Tailed, s Known) (cont.) 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.**Example 10.10: Using a Rejection Region in a Hypothesis Test**for a Population Mean (Right‑Tailed, s Known) (cont.) 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.**Example 10.10: Using a Rejection Region in a Hypothesis Test**for a Population Mean (Right‑Tailed, s Known) (cont.)**Example 10.10: Using a Rejection Region in a Hypothesis Test**for a Population Mean (Right‑Tailed, s Known) (cont.) 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-value A 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.**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 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 = 2.78.**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 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 = -2.15.**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.**Example 10.13: Calculating the p-Value for a z-Test**Statistic for a Two-Tailed Test (cont.) 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. normalcdf( ________, ________ )*2**Example 10.13: Calculating the p-Value for a z-Test**Statistic for a Two-Tailed Test (cont.)**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 . 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. 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.**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 better way !!!!**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 sample n is our sample size Give 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 rejectthe null hypothesis.**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.) Solution Step 1: State the null and alternative hypotheses. 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, m > 26.5. The logical opposite is m ≤ 26.5. Thus, the null and alternative hypotheses are stated as follows.**Example 10.16: Performing a Hypothesis Test for a Population**Mean (Right-Tailed, s Known) (cont.) 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.**Example 10.16: Performing a Hypothesis Test for a Population**Mean (Right-Tailed, s Known) (cont.) 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.**Example 10.16: Performing a Hypothesis Test for a Population**Mean (Right-Tailed, s Known) (cont.) 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 m = 26.5, the sample mean is x̄ = 27.0, the population standard deviation is s = 2.3, and the sample size is n = 213.**Example 10.16: Performing a Hypothesis Test for a Population**Mean (Right-Tailed, s Known) (cont.) Thus, the test statistic is calculated as follows.**Example 10.16: Performing a Hypothesis Test for a Population**Mean (Right-Tailed, s Known) (cont.) 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.**Example 10.16: Performing a Hypothesis Test for a Population**Mean (Right-Tailed, s Known) (cont.)**Example 10.16: Performing a Hypothesis Test for a Population**Mean (Right-Tailed, s Known) (cont.) 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.