Minitab 14 1
GETTING STARTED To OPEN MINITAB: Click Start>Programs>Minitab14>Minitab14 or Click Minitab 14 on your Desktop The Minitab session will come up like this 2
To SAVE FILE 1. Click File>Save Project As 2. A dialog box will come up like 3
3. Do the followings Click to make new folder Type the folder name and press enter Type the file name and press enter To OPEN A SAVED FILE 1. click File>Open Project 4
2. A new dialog box will be shown on screen like Select the file and press enter or click open To EXIT from Minitab Click File>Exit 5
Using CALCULATOR Sometimes we need to manipulate variables. For example, suppose we are given Height (in inch) of 10 students, but we want it in Feet. Height: 65, 66, 64, 68, 65, 67, 66, 70, 72, 68 That is, we want to convert the variable and get a new one with units in feet. The rules for converting variable is very simple Converting Variable 1. Go to Calculation from Menu-bar 2. Click Calculator 3. Write the variable name in Store result in variable. (you can use the same variable name, then the earlier variable will be replaced by the converted one or you can use different variable name) 4. Then in the Expression, mention what to calculate. Calculation of Height (inch) into Height (ft) 1. Go to Calculation from Menu-bar 2. Click Calculator 3. Write the variable name in Height Ft. (If you use Height, then the values of Height will be replaced by new one, i.e. 1 st value of Height 65 will be replaced by 5.41667 and so on) 4. Then in the Expression, write Height / 12 (There are many other options functions located in the Function scroll menu). Question: The table below is the scores of 50 students in a mathematics course: 62 50 93 71 55 51 19 72 58 47 48 62 99 54 58 56 57 56 61 43 77 50 79 83 62 55 68 71 65 64 90 85 75 22 55 75 49 61 63 77 76 58 61 79 70 59 60 63 41 58 Construct a stem-and-leaf display for all 50 scores Construct a histogram for all 50 scores Calculate the mean, median and variance of all scores 6
Inputting Data In Minitab: 1. Input data in Worksheet Window. 2. Input variables in columns. 3. First row of worksheet window can be used for variables names. Minitab Project Report (a) Steps: 1. Go to the menu "Graph" 2. Select: "Stem-and-leaf" 3. Select the variable "Score" 4. Write the increment = 10 5. Click "OK" Output: Stem-and-leaf of Score N = 50 Leaf Unit = 1.0 1 1 9 2 2 2 2 3 7 4 13789 22 5 001455566788889 (12) 6 011122233458 16 7 01125567799 5 8 35 3 9 039 7
(b) Steps: 1. Go to the menu "Graph" 2. Select: "Histogram" 3. Select the variable "Score" under 'Graph variables' 4. Choose "Bar" under 'Display' and "Graph" under 'For each' 5. Click "Annotation" dropdown menu and select "Title" to write a title of the graph 6. Click "OK" Output: Graphical representation of scores of 50 students using Histogram 20 Frequency 10 0 20 30 40 50 60 70 80 90 100 Score 8
(c) Steps: 1. Go to the menu "Stat" 2. Select "Display Descriptive Statistics" under 'Basic Statistics' 3. Select the variable "Score" 4. Click "OK" Output: Descriptive Statistics: Score Variable N Mean Median TrMean StDev SE Mean Score 50 62.46 61.00 62.70 15.42 2.18 Variable Minimum Maximum Q1 Q3 Score 19.00 99.00 55.00 72.75 Downloading data from the webpage: 1. Go to the webpage 2. Save data in your computer. 3. Click Files>open worksheet> data file name 9
FINDING PROBABILITIES (BINOMIAL DISTRIBUTION) Suppose x is a binomial random variable. Use MINITABto find the following probabilities: (a) P(x = 4) for n = 10, p = 0.3 (b) P(x >= 6) for n = 10, p = 0.3 (c) P(1 < x < 9) for n = 15, p = 0.4 (d) P(x > 5) for n = 15, p = 0.4 (e) P(x<4) for n=15, p=0.4 (f) For n = 10, p = 0.3, obtain the plot of the binomial distribution. Steps: 1. Go to Menu "Calc" > "Probability distribution" > "Binomial" 2. Select: Probability, No. of trials, Probability of success and "Input constant" 1. Click "OK" Solution Figure: Calculation of 3 (a) Find P(x=4) for n=10, p=0.3 P(X = 3) = 0.2668 b) Find P(x>=6) for n=10, p=0.3 P(X >= 6) = P(X = 6) +... + P(X = 10) = 1- P(X = 0) + P(X = 1) +... + P(X = 5) = 1- (0.028248 + 0.121061+ 0.233474 + 0.266828 10
+ 0.200121 + 0.102919) = 1-0.952651 = 0.0473490 c) Find P(1<X<9) for n=15, and p=0.4 P(1<X<9)= P(2<=X<= 8) = P(X=2) +... + P(X=8) = 0.0219+0.0633+0.1267+0.1859+0.2065+0.1770+0.1180 = 0.8997 d) Find P(x>5) for n=15, p=0.4 P(X>5) = 1- P(X<=5) = 1- P(X=0)-...- P(X=5) = 1-0.403216 = 0.596784 e) Find P(x<4) for n=15, p=0.4 P(X <4) = P(X=0) + P(X=1)+ P(X=2)+P(X=3) = 0.0905019 Alternative way : Using Cumulative Probability Steps: 1. Go to Menu "Calc" > "Probability distribution" > "Binomial" 2. Select: Cumulative Probability, No. of trials, Probability of success and "Input constant" 3. Click "OK" b) P(x>=6) = 1- P(x <= 5) For this problem, 1. Go to Menu "Calc" > Prob. dist > Binomial 2. Select Cumulative Probability, type 15 in 'No. of trials' and 0.4 in 'prob. of success' 3. Type 5 in 'Input Constant' and click OK. Output 11
Cumulative Distribution Function Binomial with n = 10 and p = 0.3 x P( X <= x ) 5 0.952651 Thus, P(x >= 6 ) = 1-0.952651 = 0.0473490 (f) Plot of binomial distribution when n=10 and p=0.3 Recall when n=10, the binomial r.v. can take the values as x=0, 1,..., 10. Therefore, we need to compute a table first like this x p(x=x) 0 -- 1 --.... 10 -- Steps: 1. go to calculator>make patterned data>simple set of numbers 2. Do the followings: 12
3. 13
then go to graph> bar chart and do this The binomial graph is then have the shape: 0.30 Chart of C2 vs C1 0.25 0.20 C2 0.15 0.10 0.05 0.00 0 1 2 3 4 5 C1 6 7 8 9 10 14
Finding Probabilities for Normal Distribution Steps: Go to Calc Probability Dist Normal Select Cumulative probability, Mean St. Deviation Select input constant If you type 0.3 in Input Constant, you will have the probability that X lies between (-, 0.3) Figure: Calculation of P(x 0.3), when Mean = 0 and Variance = 1 Output Cumulative Distribution Function Normal with mean = 0 and standard deviation = 1.00000 x P( X <= x ) 0.3000 0.6179 Problem 1: The random variable X has a normal distribution with μ=70 and σ=10. Find the following probabilities: a) P(X<75) b) P(X 90) c) P(60 X 75) 15
Problem 2: Assume that X is a binomial random variable with n=100 and p=0.5. Calculate the exact binomial probability and the approximation obtained by using the normal distribution for a) P(X 48) b) P(50 X 65) c) P(X 70) Hints: For normal approximation: P(X 48) = P(X 48.5)(approx) P(50 X 65)= P(X 65) P(X 49) = P(X 65.5) P(X 49.5) (approx) P(X 70) = 1-P(X<70) = 1-P(X 69) = 1- P(X 69.5) (approx) Then obtain those probabilities from normal distribution using μ = np, σ = npq. Generating Random Samples from Normal Population Steps: Go to Calc Random Data Normal Select Number of random sample, where to store the data (size of random sample, eg. c1-c5 means size 5), Mean and Standard Deviation Click OK The above figure describes drawing 100 random samples (written in Generate rows of data) of size 2 (C1 and C2) from a Normal distribution with Mean 100, Standard Deviation 10. 16
Large Sample Test and Confidence Interval (CI) for Mean Suppose you have a sample of 225 observations of which the known variance is 16 and you want to test: H 0 : μ = 8 vs H 1 : μ 8 Also you want find 90% CI for μ CI: σ ± Z α / (σ known) or n x 2 s ± Zα / (σ unknown) n x 2 Test Statistic: Z = x μ0 σ/ n (σ known) or Z = x μ0 s / n (σ unknown) Calculation in Mintitab 17
The data is located in 1 st Column (C1) OUTPUT Test of mu = 8 vs mu not = 8 The assumed sigma = 4 Variable N Mean StDev SE Mean C1 225 11.565 4.010 0.267 Variable 90.0% CI Z P C1 ( 11.127, 12.004) 13.37 0.000 Large Sample Test and Confidence Interval for Proportion Confidence Interval Confidence Interval Example: 7.4 (Page: 310, Ninth Edition) pˆ ± Z0. 05 pˆqˆ n n = Number of trials: 484, x = number of successes: 257 Calculation in Mintitab 18
Small Sample (n<30) Test and Confidence Interval (CI) for Mean Formula: 100(1-α)% CI for mean s CI: x ± t α / 2 where t α / 2 is based on (n-1) degrees of freedom (df). n Test Statistic: For testing, H 0 : μ = μ0 vs H 1 : μ μ0 x μ0 t =, where t has student s t-distribution with (n-1) df. s / n Example Given a sample 14, 9, 12, 14, 10, 9, 10, 12, 11, 9, 6, 2, 14, 10, 11, 12, 9, 6, 4, 8 Test a) H 0 : = 10 μ vs H 1 : μ 10 H 0 : μ = vs H 1 : μ> 10 H 0 : μ = vs H 1 : μ< 10 b) 10 c) 10 at 5% level of significance, 10% level of significance Find 95% confidence interval for μ Solution: Note that, here the sample size n = 20 (<30), i.e. we have a small sample and also population standard deviation σ is unknown. So we have to use t-distribution to perform the test and to find the confidence interval after estimating σ by the sample standard deviation S. 19
Calculation in Mintitab One-Sample T: OUTPUT Test of mu = 10 vs not = 10 Variable N Mean StDev SE Mean 95% CI T P ex-2 20 9.65000 3.21632 0.71919 (8.14471, 11.15529) -0.49 0.632 In the output T (= -0.49) is the value of the corresponding test statistics and P is the p-value which is 0.632. If the p-value is less than or equal to your level (α ), you can reject H 0. 20
Proportion Test for large sample Test H 0 : p = 0.5 vs H 1 : p < 0.5 pˆ p0 Test Statistic: Z = p0q0, pˆ = 10/ 300, p 0 = 0. 5, q 0 = 1-p 0, n = 300 n α = 0.01 (1%) Using Minitab OUTPUT: Test of p = 0.05 vs p < 0.05 Sample X N Sample p 99.0% Upper Bound Z-Value P-Value 1 10 300 0.033333 0.057443-1.32 0.093 Acknowledgments: This work was mostly done by Taslim S. mallick, Dept. of Mathematics & Statistics, Memorial University of Newfoundland. 21