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Chapter 10

Chapter 10. Technology. Example T.1: Using a TI‑83/84 Plus Calculator to Find the Area to the Left of a Given t ‑Value. Find the area under the curve to the left of t = - 1.638 for a t ‑distribution with 3 degrees of freedom. Solution

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Chapter 10

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  1. Chapter 10 Technology

  2. Example T.1: Using a TI‑83/84 Plus Calculator to Find the Area to the Left of a Given t‑Value Find the area under the curve to the left of t = -1.638 for a t‑distribution with 3 degrees of freedom. Solution Press and then to access the DISTR menu; then choose option 6:tcdf(. Since we want the area to the left, the lower bound is -∞, which we will enter as -1û99. Enter tcdf(-1û99,-1.638,3), as shown in the screenshot below.

  3. Example T.1: Using a TI‑83/84 Plus Calculator to Find the Area to the Left of a Given t‑Value (cont.) Thus, the area under the curve to the left of t = -1.638 for a t‑distribution with 3 degrees of freedom is approximately 0.1000.

  4. Example T.2: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Two‑Tailed, s Known) The mean reading score for third graders in one state is believed to be 82. An educational researcher claims that this figure is wrong. A simple random sample of 45 third graders in the state is chosen and the mean reading score is calculated to be 81. The population standard deviation is believed to be 5. Do the data gathered support the researcher’s claim? Use a 0.05 level of significance to draw your conclusion. Assume that the population distribution is approximately normal.

  5. Example T.2: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Two‑Tailed, s Known) (cont.) Solution First, we notice that this will be a test for a population mean with s known. This is a z‑test, which is option 1 under the TESTS menu. We also need to state the null and alternative hypotheses.

  6. Example T.2: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Two‑Tailed, s Known) (cont.) Next, note that we have the following values from the problem heading.

  7. Example T.2: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Two‑Tailed, s Known) (cont.) Since we have these values and not the raw data, we want to enter the values directly. Press ; then scroll to the TESTS menu and choose option 1:Z‑Test. Choose Stats and enter the values as shown in the screenshot on the left below. Choose the alternative hypothesis øÀ. Select Calculate and press .

  8. Example T.2: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Two‑Tailed, s Known) (cont.) The output screen, shown above on the right, displays the alternative hypothesis, the z‑test statistic, and the p‑value, and then reiterates the sample mean and sample size that were entered.

  9. Example T.2: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Two‑Tailed, s Known) (cont.) The output screen for this test tells us that our alternative hypothesis is m ≠ 82, our test statistic is z ≈ -1.34, and our p‑value is approximately 0.1797. Since our p-value is 0.1797 and it is greater than a = 0.05, we fail to reject the null hypothesis. Therefore, the evidence does not support the educational researcher’s claim that the mean reading score for third graders in the state is not 82.

  10. Example T.3: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) The mean number of phone calls per night that teenage girls make is believed to be no more than 4. A concerned parent in one local neighborhood claims that the mean is really more than 4. A simple random sample of ten teenage girls in the neighborhood were asked how many phone calls they made last night. The mean number of phone calls they made was 4.50 with a standard deviation of 0.85. Does this information support the concerned parent’s claim? Use a 0.10 level of significance to draw your conclusion. Assume that the population distribution is approximately normal.

  11. Example T.3: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) (cont.) Solution The population standard deviation is unknown and the population distribution is approximately normal, so we will use a t‑test. The t‑test is option 2 under the TESTS menu. Next, we state the null and alternative hypotheses.

  12. Example T.3: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) (cont.) So we have a right-tailed t-test. The following values are given to us in the problem heading.

  13. Example T.3: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) (cont.) Press ; then scroll to the TESTS menu and choose option 2:T‑Test. Choose Stats and enter the values as shown in the screenshot on the left below. Choose the alternative hypothesis >À. Select Calculate and press .

  14. Example T.3: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) (cont.) The output screen, shown above on the right, displays the alternative hypothesis, the t‑test statistic, and the p‑value, and then reiterates the sample mean, sample standard deviation, and sample size that were entered.

  15. Example T.3: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) (cont.) The output screen for this test tells us that our alternative hypothesis is m > 4, our test statistic is t ≈ 1.860, and our p‑value is approximately 0.0479. Since our p-value is 0.0479 and that is less than a = 0.10, we reject the null hypothesis. Therefore, the evidence supports the parent’s claim that the mean number of phone calls per night that teenage girls make is more than 4.

  16. Example T.4: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Proportion (Left‑Tailed) Let’s consider Example 10.24 from Section 10.4. A local politician believes that less than 65% of his constituents will vote in favor of a tax increase to pay for a new school. His staff asked a simple random sample of 50 of his constituents whether they favor the tax increase and 27 said that they would vote in favor of the tax increase. Does this information support the politician’s claim? Use a 0.01 level of significance to draw your conclusion.

  17. Example T.4: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Proportion (Left‑Tailed) (cont.) Solution This example deals with a population proportion, and we already verified that the necessary conditions are satisfied to allow us to use the normal distribution and the z‑test statistic in Example 10.24. So we will use a z‑test for a population proportion. This particular z‑test is option 5 under the TESTS menu. Next, we state the null and alternative hypotheses.

  18. Example T.4: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Proportion (Left‑Tailed) (cont.) So we have a left-tailed one-proportion z-test. The following values are given to us in the problem heading.

  19. Example T.4: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Proportion (Left‑Tailed) (cont.) Press ; then scroll to the TESTS menu and choose option 5:1‑PropZTest. Enter the values as shown in the screenshot on the left below. Choose the alternative hypothesis <p€. Select Calculate and press .

  20. Example T.4: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Proportion (Left‑Tailed) (cont.) The output screen, shown above on the right, displays the alternative hypothesis, the z‑test statistic, the p‑value, and the sample proportion, and then repeats the sample size we entered.

  21. Example T.4: Using a TI‑83/84 Plus Calculator to Perform a Hypothesis Test for a Population Proportion (Left‑Tailed) (cont.) The output screen for this test tells us that our alternative hypothesis is p < 0.65, our test statistic is z ≈ -1.63, our p‑value is approximately 0.0515, and our sample proportion is Since our p-value is 0.0515, which is greater than a = 0.01, we fail to reject the null hypothesis. This means that, at the 0.01 level of significance, this evidence does not sufficiently support the politician’s claim that less than 65% of his constituents favor a tax increase to pay for a new school.

  22. Example T.5: Using a TI‑84 Plus Calculator to Perform a Chi-Square Test for Goodness of Fit Let’s consider Example 10.30 from Section 10.6. A local bank wants to evaluate the usage of its ATM. Currently the bank manager assumes that the ATM is used consistently throughout the week, including weekends. She decides to use statistical inference with a 0.05 level of significance to test a customer’s claim that the ATM is much busier on some days of the week than it is on other days. During a randomly selected week, the bank recorded the number of times the ATM was used on each day. The results are listed in the following table.

  23. Example T.5: Using a TI‑84 Plus Calculator to Perform a Chi-Square Test for Goodness of Fit (cont.)

  24. Example T.5: Using a TI‑84 Plus Calculator to Perform a Chi-Square Test for Goodness of Fit (cont.) Solution As we determined in Example 10.30, the chi‑square test for goodness of fit is the appropriate choice for this scenario. On newer TI‑84 Plus calculators, the chi‑square test for goodness of fit is option D:c2GOF‑Test under the TESTS menu. Before running the test on the calculator, we must determine the hypotheses and calculate the expected values.

  25. Example T.5: Using a TI‑84 Plus Calculator to Perform a Chi-Square Test for Goodness of Fit (cont.) When stating the hypotheses to be tested, we take the null hypothesis to be that the proportions of customers are the same for every day of the week. Thus, we can write the null and alternative hypotheses as follows.

  26. Example T.5: Using a TI‑84 Plus Calculator to Perform a Chi-Square Test for Goodness of Fit (cont.) In order to compute the c2 -test statistic, we first need to calculate the expected value for each day of the week. Since we are assuming that the number of times the ATM is used does not vary for each day, then the probability will be the same for every day, so the expected number of customers for each day of the week is calculated as follows. Note that n = 38 + 33 + 41 + 25 + 22 + 38 + 45 = 242 (the total number of times the ATM was used).

  27. Example T.5: Using a TI‑84 Plus Calculator to Perform a Chi-Square Test for Goodness of Fit (cont.) Now we are ready to use a TI‑84 Plus calculator to compute the c2 -test statistic as well as the p‑value. We will compare the p‑value to the given level of significance, a = 0.05, to make a decision. Begin by pressing and then choose option 1:Edit. Enter the observed values in L1 and the expected values in L2. Next, press ; then scroll to the TESTS menu and choose option D:c2GOF‑Test. The calculator will prompt you for the following: Observed, Expected, and df, as shown in the screenshot on the left.

  28. Example T.5: Using a TI‑84 Plus Calculator to Perform a Chi-Square Test for Goodness of Fit (cont.) Enter L1 for Observed since that is the list where you entered the observed values. Enter L2 for Expected, and enter the number of degrees of freedom for df. The number of degrees of freedom for this test is df= 7 - 1 = 6, so enter 6 for df. Finally, select Calculate and press .

  29. Example T.5: Using a TI‑84 Plus Calculator to Perform a Chi-Square Test for Goodness of Fit (cont.) The output screen, shown above on the right, displays the c2 -test statistic and the p‑value and reiterates the number of degrees of freedom that was entered. We see that c2 ≈ 12.314 and the p-value ≈ 0.0553. Since a = 0.05 and 0.0553 > 0.05, we have p-value > a, so the conclusion is to fail to reject the null hypothesis. Thus, there is not sufficient evidence at the 0.05 level of significance to support the customer’s claim that the ATM is used significantly more on any particular day of the week.

  30. Example T.6: Using a TI‑83/84 Plus Calculator to Perform a Chi-Square Test for Association Let’s consider Example 10.32 from Section 10.7. Suppose that the following data were collected in a poll of 13,660 randomly selected voters during the 2008 presidential election campaign.

  31. Example T.6: Using a TI‑83/84 Plus Calculator to Perform a Chi-Square Test for Association (cont.) Is there evidence at the 0.05 level to say that gender and voting choice were related for this election? Solution Let’s begin by stating the null and alternative hypotheses. We let the null hypothesis be that gender and voting preference are independent of one another.

  32. Example T.6: Using a TI‑83/84 Plus Calculator to Perform a Chi-Square Test for Association (cont.) As noted in Example 10.32, we can safely assume that the necessary conditions have been met for this example, so we will perform a chi‑square test for association. To calculate c2 on a TI-83/84 Plus calculator, we need to enter the observed values in a matrix. The calculator will create and store the matrix of expected values as part of the c2-test. Begin by pressing and then to access the MATRIX menu. Scroll over to EDIT and choose the matrix you want to edit.

  33. Example T.6: Using a TI‑83/84 Plus Calculator to Perform a Chi-Square Test for Association (cont.) Before entering values into the matrix, you must enter the number of rows and then the number of columns. We do not enter the “Total” row or column. Excluding the totals, the table of observed values has 2 rows and 3 columns, so the size of the matrix is 2 × 3. The matrix of observed values is shown in the screenshot.

  34. Example T.6: Using a TI‑83/84 Plus Calculator to Perform a Chi-Square Test for Association (cont.) Next, press and scroll over to the TESTS menu. Choose option C:c2‑Test. If the observed values are in matrix [A] and we want the calculator to store the expected values in matrix [B], we do not need to change the default settings. If you need to change the name of the matrix entered for either Observed or Expected, press and then to access the MATRIX menu. Then choose the name of the matrix needed.

  35. Example T.6: Using a TI‑83/84 Plus Calculator to Perform a Chi-Square Test for Association (cont.) Your screen should appear as shown in the screenshot on the left below. After selecting Calculate and pressing , the output screen, shown below on the right, displays the c2‐teststatistic, the p‑value, and the number of degrees of freedom.

  36. Example T.6: Using a TI‑83/84 Plus Calculator to Perform a Chi-Square Test for Association (cont.) After running the test, we can view the matrix of expected values calculated by the TI‑83/84 Plus. Press and then to access the MATRIX menu. Then scroll over to EDIT and choose option 2:[B]. The matrix of expected values for this test is shown in the screenshot.

  37. Example T.6: Using a TI‑83/84 Plus Calculator to Perform a Chi-Square Test for Association (cont.) The output screen for this test tells us that the value of our test statistic is c2 ≈ 67.276. It also tells us that the p‑value is approximately 0.0000, which is less than a = 0.05, so we reject the null hypothesis that the two variables are independent. Based on the given data, we can conclude that gender and voting choice were related for the 2008 presidential election.

  38. Example T.7: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Left‑Tailed, s Known) Fred claims he has an 18-stroke golf handicap. This means he averages 18 strokes over par on a par‑72 course. Ted thinks he is not telling the truth and collects a simple random sample of 33 of Fred’s recent scorecards from par‑72 courses. From these, he calculates the sample mean to be 88.1 strokes. Assume that the population standard deviation is known to be 3.5 strokes. Can he claim with 99% confidence that Fred has a handicap below 18 strokes?

  39. Example T.7: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Left‑Tailed, s Known) (cont.) Solution First, we recognize that the desired method is a z-test for a population mean since the population standard deviation is known and the sample size is at least 30. If Ted has an 18‑stroke handicap as he claims, then his mean score on a par‑72 course is 72 + 18 = 90 strokes. Therefore, we state the null and alternative hypotheses as follows.

  40. Example T.7: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Left‑Tailed, s Known) (cont.) So we have a left-tailed z-test with the following values.

  41. Example T.7: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Left‑Tailed, s Known) (cont.) Now use Excel to calculate the test statistic and p-value. Remember that the test statistic is given by the following formula.

  42. Example T.7: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Left‑Tailed, s Known) (cont.) Compute the test statistic by entering =(B1‑B2)/(B3/SQRT(B4)) in cell B6. Recall from Chapter 6 that the Excel formula, =NORM.S.DIST(z, TRUE), calculates the area under the standard normal distribution to the left of the z-value entered.

  43. Example T.7: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Left‑Tailed, s Known) (cont.) Since this is a left-tailed test, the p-value is produced by entering =NORM.S.DIST(B6, TRUE). The Excel spreadsheet is shown in the screenshot.

  44. Example T.7: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Left‑Tailed, s Known) (cont.) The p-value is approximately 0.0009, which is less than the level of significance, 0.01. Thus, we have p-value ≤ a, so the null hypothesis is rejected. With 99% confidence, Ted can conclude that Fred’s handicap is below 18 strokes.

  45. Example T.8: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) A sanitation engineer believes that one waste management district is producing more garbage than usual. He believes the population mean is now more than 9.8 tons per truck. He chooses a simple random sample of 10 trucks that will be serving the district and weighs them before and after pickup. He calculates the sample mean to be 10.6 tons of garbage per truck with a standard deviation of 1.2 tons. Can his claim be substantiated at a confidence level of 95%? Assume that the population distribution of weights is approximately normal.

  46. Example T.8: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) (cont.) Solution First, recognize that the desired test is a t-test for a population mean since the population standard deviation is unknown and the population distribution is approximately normal. Next, state the null and alternative hypotheses as follows.

  47. Example T.8: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) (cont.) So we have a right-tailed t-test with the following values.

  48. Example T.8: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) (cont.) Next, use Excel to calculate the test statistic and critical value. The test statistic is given by the following formula.

  49. Example T.8: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) (cont.) The test statistic is produced by entering =(B1‑B2)/(B3/SQRT(B4)) in cell B6. The Excel formula for the value of t corresponding to a given area in the left tail is =T.INV(probability, deg_freedom), where probability is the area in the left tail and deg_freedom is the number of degrees of freedom. Since this is a right‑tailed test with a confidence level of 95%, we want the critical value of t with an area of a = 0.05 in the right tail, which means that the area to the left of the critical value is 1 - 0.05 = 0.95.

  50. Example T.8: Using Microsoft Excel to Perform a Hypothesis Test for a Population Mean (Right‑Tailed, s Unknown) (cont.) Enter =T.INV(0.95, 9) to find the critical value for this test. Note that n- 1 degrees of freedom are used. The Excel spreadsheet is shown in the screenshot.

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