Two sample z test minitab

Right tailed test for Paper manufacturing example 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 Format>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. Hammermill Paper Test: 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.

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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: papermm Test of mu = 216 vs > 216

The assumed standard deviation = 0.023

95%

Lower

Variable N Mean StDev SE Mean Bound Z P

papermm 50 216.007 0.022 0.003 216.002 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.

**Assigned: Use the six step process and assumption format to complete the following: Problem: Superior Carpet Cleaning Service Inc. is considering acquiring a store in a new market region. Management is interested in determining if the average cost of cleaning a 12’ by 18’ wall-to-wall carpet is greater than $50.95. To test this hypothesis, management randomly selects 23 customers who have recently had a 12’ by 18’ wall-to-wall carpet cleaned and inquires as to the price they were charged. Assume the population is normally and distributed and the population standard deviation is $3.49. Test this claim using a .10 level of significance.

Sampled Cleaning Prices 52.50 56.97 52.46 52.35 56.37 50.97 53.27 50.91 50.12 56.75

51.27 52.91 49.12 52.01 49.57 56.37 54.91 57.25 51.23 48.56

51.35 48.65 48.46

Instructions to get you started: Enter data from problem into a Minitab worksheet to create a dataset for the Cleaning Prices study. Use Minitab to test the hypothesis you formulate for the Cleaning study. 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

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For your report (prepared in Word) you will need a Problem Definition, a 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, for the Conclusion section include the critical value/critical ratio technique, the confidence interval technique the p-value technique to test the null hypothesis State the type of error you may have made. For the Interpretation section state whether the air quality has improved or has stayed the same. Note: Be sure to include the following in your conclusion: Crtical value/Critical Ratio technique Confidence interval p-values technique. All three techniques should agree. Also, for step six, Interpretation, provide a clear response to management regarding your results and what they mean to the problem (significant or insignificant). See last page of this handout for notes on the six step process.

Sigma Unknown: Use T table for this section T Test: Margies 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

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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 handle will appear around the outer edges click and hold on one of the handles and size the graph.

One-Sample T: TEMP Test of mu = 142 vs not = 142

Variable N Mean StDev SE Mean 90% CI T P

TEMP 24 141.375 1.996 0.407 (140.677, 142.073) -1.53 0.139

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 desired temp.

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Assumption Section: In your Assumption section describe how the boxplot supports or indicates the lack of normality (The data was sampled from a normally distributed population).

Boxplot for Hot Chocolate Assumption Section

145144143142141140139138137136

X _

Ho

Choc

Boxplot of Choc (with Ho and 90% t-confidence interval for the mean)

*Assigned Problem Kabul: According to data released by World Bank, the mean PM10 (particulate matter) for the city of Kabul Afghanistan in 2012 was 46. Suppose that because of efforts to improve air quality in Kabul, increases in modernization and efforts to establish environmentally friendly businesses, city leaders believe PM rates have decreased. To test this notion, city leaders randomly sample 12 readings over a one year

period. The readings are shown below. Set alpha to .01

Kabul PM Data (enter into Minitab column) 31 44 35 53 57 47 32 40 31 38 53 45

Instructions to get you started: Enter data from problem into a Minitab worksheet to create a dataset for the Kabul PM data. Use Minitab to test the hypothesis you formulate for the Kabul study. 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 your report (prepared in Word) you will need a Problem Definition, a 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, for the Conclusion section include the critical value/critical ratio technique, the confidence interval technique the p-value technique to test the null hypothesis State the type of error you may have made. For the Interpretation section state whether the air quality has improved or has stayed the same. For you Assumption section; address the assumption of normality by referencing the boxplot of the data. Be sure to cut and paste the graph form Minitab to your report. Resize the graph to fit neatly within your report in the section you label Assumptions. Include a brief discussion of how normality is or is not satisfied according to what you see in the graph. At minimum, be sure to reference the median and mean and central tendency or your discussion of the graph. 7 points

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Approximate binomial distribution using normal distribution: Use Z table Testing a Proportion Page 354: Assumption of normality (does the normal distribution approximate the binomial)? np≥5 and n(1-p)≥5. Make sure this condition holds or the test is invalid and the binomial distribution will have to be used to develop the probabilities. Problem Definition: Retailers are concerned with the costs associated with serial returners. The individuals make a purchase, wear the garment and then return for a refund. Many of the garments are damaged and not re-saleable. To identify and stop serial returners a chain of stores has purchase software called Trakker. The sales manager wants to test to see if the practice has been reduced as a result of utilizing the software. Originally, the return rate was 13%. Use the following information to determine if the proportion of returns has fallen below 13%. Set alpha at .05.

H0:  ≥ .13 (returns are equal to or greater than .13)

H1:  < .13 (return rate has fallen)

Decision Rule: If Z test statistic is less than -1.645 reject the null hypothesis. Dialog box for test (from Minitab) Select Stat>Basic Stat>1 Proportion and the following dialog box will appear Select summarized data radio button>enter number of trials (n) and success (x) in the appropriate boxes. *Note: The problem provides the sample proportion for the survey. In order to convert the proportion of events to the number of events, x, multiply the sample proportion 22 by sample size of 250 (the number of trails n) to get number of events. Select Options> enter the correct confidence level and the test proportion given (assumed proportion) in problem> set the alternative hypothesis>select the test based on normal distribution box>Ok>Ok Data will appear in session window. You will not have a graph for the proportions problem.

Step 4 Output from Minitab Session window: Test and CI for One Proportion Test of p = 0.13 vs p < 0.13

95%

Upper

Sample X N Sample p Bound Z-Value P-Value

1 22 250 0.088000 0.117471 -1.97 0.024

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Conclusion: The critical ratio (test statistic from Minitab) is less than the critical value of -1.645. Reject the null hypothesis there is a .05 chance that a type 1 error has occurred. The upper bound of the confidence interval is less than .13 meaning a revealing a reject scenario. The pvalue is less than the alpha level of .05 which supports a reject scenario. Interpretation: The sample data provides evidence that the return rate has fallen significantly.

**Assigned: In some states the law requires drivers to turn on their headlights when driving in the rain. A highway patrol officer believes that less than one-quarter (.25) of all drivers follow this rule. As a test, she randomly samples 200 cars driving in the rain and counts the number whose headlights are turned on. She finds this number of cars with headlights on to be 41. Does the officer have enough evidence at 10% level of significance to support her belief that fewer than .25 of all cars follow the rules? One tailed test to the left. Required in Conclusion section of six step process: Critical ratio/Critical value of Z, Confidence interval and p-value. Use examples in this handout to correctly format the test. In Assumption section: Include computations to satisfy normal approximation of binomial distribution by solving np≥5 and n(1-p)≥5. You will have to compute assumption by hand and type the results of each equation onto your homework. (5 points)

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Six step process for hypothesis testing homework

Problem Definition: Read problem and develop problem definition Hypothesis: Determine if test is one or two tailed test is required. Use Format>Font in Word to set subscripts on hypotheses as indicated H0 H1 Use the symbol map in Word (Insert>Symbol) for the appropriate Greek symbols. You will need the correct symbols for

beta , alpha  and Mu . Decision Rule: Determine t critical value for given alpha and degree of freedom and compose your decision rule. Test: Cut required material from session window in Minitab and paste into Word. Conclusion: State only reject or (fail to reject) null hypothesis and the type error associated with your decision. Include critical value/critical ratio, pvalue, and confidence interval. Interpretation: Briefly report your findings in layman’s terms. Remember if you fail to reject the null the sampling error is caused by chance alone. If you reject the null chance and some other factor(s) are causing the sampling error to be larger than acceptable. Notes regarding assumptions: For samples containing less than 30 items be sure to address the assumption of normality using the boxplot from Minitab. When the graph appears in Minitab, position your cursor in the white space within the graph, right click and select copy graph. Open Word, place your cursor where you want to paste the graph, go to edit on the toolbar and select paste special (paste special is not available on the right click menu).To resize, click on graph and six sizing handles will appear around the outer edges of the graph. For best results click and hold the sizing handle in the lower right corner and move diagonally to the upper left hand corner until size desired is achieved. The graph sizes in this handout are acceptable for your report but make sure you can read the verbiage within the graph.

Satisfy Assumptions: If needed (sample is normally distributed, approx of binomial with normal etc)