Chapter 9 Homework Problems and Computer Commands Chapter 9 Lecture handout containing homework problems and Minitab computer Commands 39 Total Points for three analysis marked ASSIGNED See Pages 3, 7...

the 3 assigned problems


Chapter 9 Homework Problems and Computer Commands Chapter 9 Lecture handout containing homework problems and Minitab computer Commands 39 Total Points for three analysis marked ASSIGNED See Pages 3, 7 and 10 for Assigned problems Handout To Be Used In Concert with Lecture and Text Testing a mean population variances known-Enter data below into Minitab worksheet. Lecture material not in text: Sigma is Known and Z table may be used Data Display Width of Sheets of Paper in Mil 216.025 216.021 215.978 216.064 216.02 216.012 216.033 215.981 216.013 216.026 215.999 216.00 216.019 216.019 215.972 215.993 215.998 215.955 215.962 216.012 216.037 215.985 216.035 215.996 216.018 216.032 216.006 216.003 215.986 215.99 216.002 216.039 216.039 215.983 215.998 216.002 215.994 215.997 216.007 216.036 216.032 216.031 216.011 216.004 216.011 215.999 215.976 216 216.018 215.981 Sigma is known to be σ = .023mm Problem Definition: Hammermill would like to determine if the width of their paper exceeds 216mm. Paper not meeting these dimensions jams office equipment. Use alpha .05. · Note about setting up your hypotheses: The subscripts 0 on the null and the 1 on the alternative are created using Home>Font commands in Word. · Steps for subscripting: Type the hypotheses (H0 H1) highlight the subscript value and go to HOME>Font and select Subscript, hit ok and the highlighted value will be subscripted. Do the same for the remaining hypothesis. (Notes for Word 2007) H0: µ≤216 H1: µ>216 Decision Rule: If Z test statistic is greater than 1.645 reject the null. TEST STEP Hammermill Paper Test Using Minitab: Enter the data listed above into a Minitab worksheet. Go to Stat>Basic Stat>One sample Z and the box to the right will open. Select samples in columns and select the column containing your data. Enter the standard deviation and test mean as shown. Now go to options and input the confidence level and the direction of the test. Hit ok until your data appears in the session window. One-Sample Z: MM (This is the output from the Session Window) Descriptive Statistics N Mean StDev SE Mean 95% Lower Bound for μ 50 216.007 0.022 0.003 216.002 μ: mean of MM Known standard deviation = 0.023 Test Null hypothesis H₀: μ = 216 Alternative hypothesis H₁: μ > 216 Z-Value P-Value 2.15 0.016 Conclusion: 1) The Z test statistic of 2.15 is greater than the critical value of 1.645. Reject the null hypothesis there is a 5% chance a Type 1 error has been committed. 2) The hypothesized value of Mu=216 does not fall within confidence interval lower bound of 216.002 milliliters of fill. 3) Pvalue .016 is < α="" .05="" interpretation:="" the="" process="" is="" not="" in="" control.="" the="" sheets="" are="" significantly="" larger="" than="" 216="" mm.="" (given="" the="" results="" of="" the="" analysis,="" should="" the="" machine="" be="" serviced="" to="" cut="" paper="" more="" precisely?)="" you="" must="" state="" the="" managerial="" action="" indicated="" as="" a="" result="" of="" the="" analysis.="" assumption:="" randomly="" and="" independently="" selected="" sample.="" sample="" size="" exceeds="" 30="" so="" approximate="" normality="" may="" be="" assumed="" under="" central="" limit="" theorem.="" **assigned:="" use="" the="" six-step="" process="" and="" assumption="" format="" to="" complete="" the="" following:="" please="" reduce="" the="" amount="" of="" information="" in="" problem="" definition="" to="" two="" or="" fewer="" sentences="" and="" try="" to="" avoid="" numbers="" problem="" definition/research="" objective:="" a="" manufacturer="" wishes="" to="" advertise="" a="" new="" environmentally="" friendly="" trash="" bag="" on="" a="" major="" tv="" network.="" the="" manufactures="" claim="" that="" bags="" strength="" has="" been="" significantly="" increased="" compared="" to="" past="" 30-gallon="" bags.="" the="" network="" wants="" significance="" statistical="" evidence="" regarding="" the="" claim="" that="" the="" new="" bag="" is="" stronger.="" agreed="" testing="" will="" measure="" the="" variable="" of="" strength="" in="" terms="" of="" poundage="" of="" content="" placed="" into="" the="" bag="" while="" suspended="" in="" the="" air.="" the="" follow="" data="" of="" n="50" trash="" bags="" represent="" the="" poundage="" of="" trash="" placed="" into="" the="" bag="" at="" the="" time="" the="" bag="" tore.="" the="" new="" bag’s="" µ="" mean="" breaking="" strength="" is="" unknown.="" the="" current="" bag="" on="" the="" market="" has="" a="" known="" mean="" breaking="" strength="" µ="" of="" ≤50="" lbs.="" alpha="" of="" .10="" is="" set="" for="" the="" test.="" the="17" pounds="" dataset="" of="" trash="" bag="" breaking="" strengths="" 85.195="" 37.865="" 146.077="" 64.737="" 65.962="" 118.178="" 61.255="" 22.760="" 55.100="" 62.478="" 10.176="" 109.694="" 47.795="" 97.737="" 71.910="" 43.160="" 55.789="" 103.098="" 120.065="" 66.495="" 72.313="" 60.044="" 98.518="" 16.208="" 77.533="" 34.674="" 33.505="" 106.634="" 14.453="" 35.594="" 37.202="" 96.486="" 16.360="" 31.285="" 36.019="" 14.184="" 22.850="" 25.680="" 66.000="" 90.369="" 114.590="" 40.023="" 61.988="" 56.209="" 19.096="" 50.726="" 82.222="" 14.200="" 84.025="" 37.814="" instructions="" to="" get="" you="" started="" on="" your="" analysis="" of="" trash="" bag="" data="" for="" your="" report="" (prepared="" in="" word)="" include:="" problem="" definition="" hypothesis="" (use="" correct="" symbols="" in="" word="" see="" insert="">Symbols for the correct symbols to use in the hypothesis a Decision Rule (use tables to determine the correct critical value and state the decision referencing the critical value. Test step (Minitab output) Include appropriate output as demonstrated in this handout and during lecture. Including unrelated output will result in point deductions. Conclusion: For the Conclusion section include: · The critical value/critical ratio technique and whether Type 1 or Type 2 errors may have resulted in light of your findings. (If type 1 state alpha value/probability of Type 1. For Type 2 stipulate error by indicating Beta Type 2 error may have occurred. · The confidence interval technique. State whether the bounds of the interval capture the hypothesized parameter. · The p-value technique to test the null hypothesis State the type of error you may have made. **NOTE: Each of the sections described above must show numbers/values for the appropriate technique. For example, you cannot say the confidence interval captures the parameter. You must state the bounds of the interval and value of the parameter or points will be deducted. Enter data from problem into a Minitab worksheet to create a dataset for the trash bag study. Use Minitab to test the hypothesis you formulate for the trash bag tearing strength claim. Present your work in report format using the six-step process demonstrated in class and which is shown on the final page of this handout. For the Interpretation section Also, for Interpretation, provide a clear response to management regarding your results and what they mean to the problem Example: Hammermill hypothesis was rejected which may be interpreted as the process of cutting the paper is not within Hammermill’s established controls. For this problem, be sure to answer whether the bags hold significantly more trash on average and determine their advertisement should be aired on TV. . Assumption: For your Assumption section; address the assumption of normality by referencing the sample size and how the central limit theorem supports normality. Also include the samples was randomly and independently selected. 12 points. Sigma Unknown: Use T table for this section T Test: Margie’s Espresso has learned that hot chocolate is best served at 142 degrees Fahrenheit. Marge tests 24 cups of chocolate randomly selected during business hours. Using an alpha of .10 test to determine if the hot chocolate is the correct temp. Testing A Mean Unknown Population Variance (Hot Chocolate T test) H0: µ=142 H1: µ≠142 Data Display Not in Text HotChoc 140 140 141 145 143 144 142 140 145 143 140 140 141 141 137 142 143 141 142 142 143 141 138 139 Enter the information from the Hot Chocolate example into a Minitab worksheet Stat>Basic Stat>1 sample t and the following dialog box will appear. 1) Position cursor in samples in columns window and select column containing the dataset you created. Double click on the column name in left window to move the dataset to the samples in columns window. 2) Place the assumed population parameter in the test mean box 3) Select options and set the level of confidence and the alternative hypothesis (direction of the test) hit OK when completed and you will be returned to the opening dialog box. 4) Select graphs from the one sample t and confidence interval dialog box, select boxplot>hit OK in each window until test results appear in session window and the boxplot appears. *You will need to copy and paste the boxplot into your finished report to address the assumption of normality. To copy the graph, position your cursor in the white space within the graph area, right click and select copy graph. Open Word, place your cursor where you want to paste the graph, go to edit and select paste special this will allow you resize the graph in Word. To resize, click on graph and six sizing handles will appear around the outer edges click and hold on one of the handles and size the graph. TEST: One-Sample T: DrinkTemp Descriptive Statistics N Mean StDev SE Mean 90% CI for μ 24 141.375 1.996 0.407 (140.677, 142.073) μ: mean of DrinkTemp Test Null hypothesis H₀: μ = 142 Alternative hypothesis H₁: μ ≠ 142 T-Value P-Value -1.53 0.139 Boxplot of DrinkTemp Conclusion: 1) T test stat for critical ratio is -1.53 is less than the T critical value of -1.714. The test stat –1.53 (the number of standard errors between sample mean 141.375 and the parameter 142 is fewer than allowed by the test -1.714 and is thus close to the mean of 142). Fail to reject the null at type 2 error (β beta error) may have been made. 2) The parameter of 142 falls within the interval (140.677, 142.073) and thus the parameter has been captured within the interval. 3) The pvalue of .139 is greater than the alpha (level of significance .10) fail to reject the null. Interpretation: There is no significance difference between the sample mean and hypothesized parameter. The process of preparing hot chocolate is in control and producing hot chocolate relatively close to the
Jul 02, 2021
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