Chapter 10

HAWKES LEARNING SYSTEMS

Students Matter. Success Counts.

Copyright © 2013 by Hawkes Learning

Systems/Quant Systems, Inc.

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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 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.






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.






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.






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.

0: 82: 82a

HH







Next, note that we have the following values from the problem heading.

0 82581450.05

xn

s







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 øÃ€. Select Calculate and press .







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.







The output screen for this test tells us that our alternative hypothesis is ≠ 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 = 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.






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.






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.

0: 4: 4a

HH







So we have a right-tailed t-test. The following values are given to us in the problem heading.

0 44.500.85100.10

xsn







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 >Ã€. Select Calculate and press .







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.







The output screen for this test tells us that our alternative hypothesis is > 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 = 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.






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.






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.

0: 0.65: 0.65a

H pH p







So we have a left-tailed one-proportion z-test. The following values are given to us in the problem heading.

0 0.6527500.01

pxn







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 .







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.







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 = 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.

ˆ 0.54.p






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.






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

ATM UsageNumber of Times Used

Monday 38Tuesday 33

Wednesday 41Thursday 25

Friday 22Saturday 38Sunday 45







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.







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.

0 1 2 3 4 5 6 7: : There is some difference amongst the probabilities.a

H p p p p p p pH







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).

1 242242

7 7i iE np







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, = 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.







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 .







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 = 0.05 and 0.0553 > 0.05, we have p-value > , 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.






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.

Observed Sample of 13,660 VotersObama McCain Other Total

Male 3455 2764 65 6284Female 3541 3762 73 7376

Total 6996 6526 138 13,660






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.

0: Gender and voting preference are independent.: Gender and voting preference are not independent.a

HH







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.







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.







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.







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‐test statistic, the p value, ‑and the number of degrees of freedom.







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.







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 ‑ = 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.






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?






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.

0: 90: 90a

HH







So we have a left-tailed z-test with the following values.

88.1 (Enter in cell B1.)90 (Enter in cell B2.)3.5 (Enter in cell B3.)33 (Enter in cell B4.)

x

n

s







Now use Excel to calculate the test statistic and p-value. Remember that the test statistic is given by the following formula.

xz

n

-

s







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.







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.







The p-value is approximately 0.0009, which is less than the level of significance, 0.01. Thus, we have p-value ≤ , so the null hypothesis is rejected. With 99% confidence, Ted can conclude that Fred’s handicap is below 18 strokes.






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.






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.

0: 9.8: 9.8a

HH







So we have a right-tailed t-test with the following values.

10.6 (Enter in cell B1.)9.8 (Enter in cell B2.)1.2 (Enter in cell B3.)10 (Enter in cell B4.)

x

sn







Next, use Excel to calculate the test statistic and critical value. The test statistic is given by the following formula.

xt

sn

-







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 = 0.05 in the right tail, which means that the area to the left of the critical value is 1 - 0.05 = 0.95.







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.







Since t ≈ 2.108 and t = t0.05 1.833, we have t ≥ t, so we reject the null hypothesis and conclude that the district produces more trash than is anticipated. At a confidence level of 95%, we can support the sanitation engineer’s claim that the population mean for this district is more than 9.8 tons of garbage per truck.






Example T.9: Using Microsoft Excel to Perform a Hypothesis Test for a Population Proportion (Two Tailed)‑

Randy has heard that 10% of the population is left-handed. To test this proposition, he takes a simple random sample and records the results. Out of the 65 people surveyed, 10 are left handed. Using ‑ = 0.10, determine whether there is sufficient evidence to refute the claim that 10% of the population is left handed.‑






Example T.9: Using Microsoft Excel to Perform a Hypothesis Test for a Population Proportion (Two Tailed) (cont.)‑

Solution Immediately, we recognize that we need to perform a z test for a population proportion. Note that the ‐sample size is n = 65, and the hypothesized value of the population proportion is p = 0.10. Since np = 65(0.10) = 6.5 5 and n(1 − p) = 65(1 - 0.10) = 58.5 5, the conditions are met for utilizing this test. The null and alternative hypotheses are as follows.

0: 0.10: 0.10a

H pH p







First, note that so enter =10/65 in cell B1

to calculate the sample proportion. Then enter the hypothesized value of the population proportion, 0.10, in cell B2 and the sample size, 65, in cell B3. Next, use Excel to calculate the test statistic and p-value. Recall that the test statistic is given by the following formula.

10ˆ ,65

xp

n

ˆ

1p p

zp p

n

-

-







Enter =(B1-B2)/SQRT(B2*(1-B2)/B3) in cell B5 to calculate the test statistic, z ≈ 1.45. Since the test is two tailed, the ‑ p-value is the sum of the areas under the standard normal curve to the left of z1 ≈ −1.45 and to the right of z2 ≈ 1.45. The NORM.S.DIST function calculates the area to the left of the z-value entered, so enter =2*(1-NORM.S.DIST(B5, TRUE)) in cell B6 to produce the p-value.







The p-value is approximately 0.1479, which is greater than = 0.10, so we fail to reject the null hypothesis. Thus, at the 0.10 level of significance, this evidence is not sufficient to refute the claim that 10% of the population is left handed.‑






Example T.10: Using Microsoft Excel to Perform a Chi-Square Test for Goodness of Fit

A manager at a clothing store believes that pink shirts do not sell as well as other colors. Despite this observation, her buyer continues to supply the store with equal numbers of red, green, blue, and pink shirts. The manager tallies up last month’s sales by shirt color. The results are shown below.

Last Month’s Sales by Shirt ColorRed Green Blue Pink

Sales 27 25 26 17Expected Sales 23.75 23.75 23.75 23.75






Example T.10: Using Microsoft Excel to Perform a Chi-Square Test for Goodness of Fit (cont.)

Can the manager prove with a confidence level of 90% that the numbers of red, green, blue, and pink shirts sold are not equal?







Solution First, state the null and alternative hypotheses as follows.

Next, enter the data into cells A1 through E3, as shown in the screenshot at the end of this example, and then calculate the test statistic. The latter is actually a two-part task.

0 1 2 3 4: : There is some difference amongst the probabilities.a

H p p p pH







The first part requires finding the addends that compose the test statistic using the calculation

for each shirt color. The

second part is computing the sum of these four calculations. First, enter =(B2-B3)^2/B3 in cell B4, and then copy this formula and paste it into cells C4, D4, and E4. Next, find the sum by entering =SUM(B4:E4) in cell B6. This sum is the test statistic, c2 ≈ 2.642.

2Sales Expected SalesExpected Sales-







The Excel formula for the value of c2 corresponding to a given area in the right tail is =CHISQ.INV.RT(probability, deg_freedom), where probability is the area in the right tail and deg_freedom is the number of degrees of freedom. Thus, the critical value for this test is calculated by entering =CHISQ.INV.RT(0.10, 3) into cell B7, where = 1 - 0.90 = 0.10, which is the area in the right tail of the chi square distribution with ‑ k - 1 = 3 degrees of freedom. This critical value is 2 2

0.10 6.251.c c







Then, since 2.642 < 6.25, we have and thus we fail to reject the null hypothesis. Therefore, at the 90% confidence level, the evidence does not support the manager’s claim that the numbers of red, green, blue, and pink shirts sold are not equal.

2 2c c






Example T.11: Using Microsoft Excel to Perform a Chi-Square Test for Association

The US Census Bureau reported that Charleston County, SC had the following population distribution in 2010.

Source: American FactFinder. “Profile of General Population and Housing Characteristics: 2010 Demographic Profile Data.” US Census Bureau. 2010. http://quickfacts.census.gov/qfd/states/45/45019lk.html (31 Jan. 2012).

Observed Sample of 350,209 ResidentsUnder 18 18 to 44 45 to 64 65 and Over Total

Male 33,022 74,659 43,116 19,074 169,871Female 35,722 71,044 47,925 25,647 180,338

Total 68,744 145,703 91,041 44,721 350,209






Example T.11: Using Microsoft Excel to Perform a Chi-Square Test for Association (cont.)

Is there sufficient evidence at the 1% level to say that age group is related to gender? Solution Our null hypothesis is that age group and gender are independent of one another.

0: Gender and age group are independent.: Gender and age group are not independent.a

HH







After entering the given data into Excel, the next step is calculating the expected value for each cell in the contingency table. Enter the given contingency table of observed values in cells A1 through F4 as shown in the following screenshot. The expected values are calculated by using the formula

Enter =B$4*$F2/$F$4 into cell B7 and then copy and paste this formula into cells B7 through E8.

row total column total.iE n







Note that the $ symbol is used to “lock” the column or row that follows the $ when a formula is copied from one cell and pasted into another. The only remaining step is calculating the p value for the test. The Excel formula used ‑to calculate the p value for a chi square test for ‑ ‑association is =CHISQ.TEST(actual_ range, expected_range), where actual_range is the range of cells containing the observed values and expected_range is the range of cells containing the expected values. Thus, the p value for this test is found by entering ‑=CHISQ.TEST(B2:E3, B7:E8) into cell B9.







The p value is approximately 0.0000. Since this ‑ p value ‑is less than = 0.01, we reject the null hypothesis. Therefore, at the 0.01 level of significance, the evidence supports the claim that age group is related to gender for residents of Charleston County, SC.






Example T.12: Using MINITAB to Perform a Hypothesis Test for a Population Proportion (Right Tailed) ‑

Alfredo thinks he is a better golfer than Tiger Woods was in 2008. He backs this up with the fact that in a simple random sample of 180 holes that he has played in the past year, he hit 133 greens in regulation. In 2008, Tiger had the highest greens in regulation percentage on the PGA Tour, 71.39%. Using = 0.10, test Alfredo’s claim of golfing superiority. Source: PGATOUR.com. “PGA TOUR – Performance Stats: Tiger Woods – 2008 Stats.” 2010. http://www.pgatour.com/golfers/008793/tiger-woods/ performance stats/#uber (1 Feb. 2012). ‑






Example T.12: Using MINITAB to Perform a Hypothesis Test for a Population Proportion (Right Tailed) (cont.) ‑

Solution First, we identify the test as a hypothesis test for a population proportion. Before we perform the hypothesis test, we must verify that the necessary conditions are satisfied. Note that the sample size is n = 180. Since Alfredo claims that his greens in regulation percentage is better than Tiger’s, we will use Tiger’s percentage in the hypotheses. Thus, the hypothesized value of the population proportion is p = 0.7139.







Since np = 180(0.7139) = 128.502 5 and n(1 - p) = 180(1 – 0.7139) = 51.498 5, the conditions are met for utilizing the z test statistic. We have the following ‑null and alternative hypotheses.

This test will use the following parameters.

0: 0.7139: 0.7139a

H pH p

1331800.7139

xnp







In Minitab, go to Stat Basic Statistics 1 Proportion► ► . Choose Summarized data and enter 133 for the number of events and 180 for the number of trials. Then select Perform hypothesis test and enter 0.7139 for the hypothesized proportion. Click on Options to enter 90 for the confidence level; then choose greater than for the alternative hypothesis. Since we are using the z test statistic for this hypothesis test, select the ‑option Use test and interval based on normal distribution.







Click OK on the options window and OK on the main dialog window and the following results will be displayed in the Session window.







The last column in the results displays the p-value. With a p-value of 0.229, we fail to reject the null hypothesis, so the evidence does not support Alfredo’s claim that he is a better golfer than Tiger Woods.






Example T.13: Using MINITAB to Perform a Chi-Square Test for Goodness of Fit

A local library is open 20 hours per week. The librarian wants to know whether the proportions of books checked out are the same as the proportions of hours the library is open for every day of the week. She records the number of books checked out as well as the number of hours the library was open on each day of one randomly selected week. The results are as follows.

Library Books and HoursMon. Tues. Wed. Thurs. Fri. Sat. Sun.

Number of Books 101 165 162 180 117 82 closedNumber of Hours Open 2 4 4 5 3 2 closed






Example T.13: Using MINITAB to Perform a Chi-Square Test for Goodness of Fit (cont.)

Calculate the c2 -test statistic using MINITAB.Solution First, enter the given data into columns C1, C2, and C3 in the MINITAB worksheet, as shown in the following screenshot. Next, we need to calculate the expected number of books checked out for each day.







We will calculate the expected number of books for each day using the formula E = npi, where n is the total number of books checked out for the week, and pi is the proportion of hours that the library is open on that day. Go to Calc Calculator► and enter the following expression: SUM(C2)*(C3/SUM(C3)). Choose to store the result in column C4 and click OK. The expected number of books checked out for each day will then be calculated and stored in column C4.







The next step is calculating the addends that compose

the c2-test statistic using the formula

Again, go to Calc Calculator► . Enter the expression (C2-C4)**2/C4. This time you want to store the result in column C5. Click OK and the values will be calculated and stored in column C5. The completed worksheet will appear as shown in the following screenshot.

2

.i i

i

O EE-







Finally, go to Calc Column Statistics► . Select Sum and enter C5 as the input variable; then click OK. The resulting sum that is shown in the Session window is our test statistic, c2 ≈ 7.690.






Example T.14: Using MINITAB to Perform a Chi-Square Test for Association

Gretchen wants to know whether gender has a relationship to beverage choice among her coworkers. She defines beverage choice as the first beverage consumed upon arrival at work and selects a random sample of her coworkers. She records her coworkers’ consumption habits one morning and the results are shown in the following table. Is there sufficient evidence at the 0.05 level to say that gender and beverage choice are related?






Example T.14: Using MINITAB to Perform a Chi-Square Test for Association (cont.)

Solution First, we state our hypotheses, letting the null hypothesis be that gender and beverage choice are independent.

Observed Sample of 90 CoworkersCoffee Tea Water Other Total

Male 10 7 18 9 44Female 8 12 11 15 46

Total 18 19 29 24 90







Now enter the data into the worksheet as shown in the following screenshot. Do not enter the total column or the total row into the spreadsheet.

0: Gender and beverage choice are independent.: Gender and beverage choice are not independent.a

HH







Next, go to Stat Tables Chi-Square Test (Table in ► ►Worksheet). Enter C2-C5 as the columns containing the table, as shown in the following dialog box.







Click OK and the results in the Session window should appear as follows.






Example T.14: Using MINITAB to Perform a Chi‐Square Test for Association (cont.)

The expected values, c2-test statistic, number of degrees of freedom, and p value are displayed in the ‑results. Since the p-value = 0.196, we fail to reject the null hypothesis at = 0.05. Thus, there is not sufficient evidence at the 0.05 level to say that gender and beverage choice are related among Gretchen’s coworkers.

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