ECOSOC MS EXCEL LECTURE SERIES DISTRIBUTIONS

Transcription

1 ECOSOC MS EXCEL LECTURE SERIES DISTRIBUTIONS Module Excel provides probabilities for the following functions: (Note- There are many other functions also but here we discuss only those which will help in hypothesis testing.) Normal Distribution Standard Normal Distribution T distribution Chi distribution NORMDIST and NORMINV NORMSDIST and NORMSINV TDIST and TINV CHIDIST and CHIINV These functions are available in Formulas Tab > More Functions > Statistical. You can know about what each function do by clicking on that function in Statistical tab and then selecting Help on this function. Normal and Standard Normal Distribution: NORMDIST function is used when you have to find P(X<=x) for x given.

2 Example 1: An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bulb life is normally distributed, what is the probability that an Acme light bulb will last at most 365 days? Solution: Go to Formulas > More Functions > Statistical>NORMDIST To find P(X<=365) Type x=365, Mean μ=300, standard deviation σ=50, cumulative=1. Press OK. What you get is P(X<=365). An alternative is to type =NORMDIST(x, μ, σ, 1) in the desired cell. Below is a type of window that opens up. Question: Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10, what is the probability that a person who takes the test will score between 90 and 110? Note: P (90<X<110) = P (X<110) P (X<90) Example 2: Let us suppose the SAT scores nationwide are normally distributed with a mean and standard deviation of 500 and 100, respectively. Answer the following questions based on the given information: A: What is the probability that a randomly selected student score will be less than 600 points? B: What is the probability that a randomly selected student score will exceed 600 points? C: What is the probability that a randomly selected student score will be between 400 and 600? Solution: In the NORMDIST distribution box: i. Enter 600 in X (the value box); ii. Enter 500 in the Mean box; iii. Enter 100 in the Standard deviation box; iv. Type "true" in the cumulative box, then click OK. As you see the value appears, indicating the probability that a randomly selected student's score is below 600 points. Using common sense we can answer part "b" by subtracting from 1. So the part "b" answer is or This is the probability that a randomly selected

3 student's score is greater than 600 points. To answer part "c", use the same techniques to find the probabilities or area in the left sides of values 600 and 400. Since these areas or probabilities overlap each other to answer the question you should subtract the smaller probability from the larger probability. The answer equals that is, NORMINV function is used when you have to find the value x* such that Pr(X <= x*) = p, where p is given. Example 3: The average size of shoes selected is 8 with a variance of 4.8. What will be the minimum size of the top 10% of the shoes (size wise)? Solution: To find x for which P (X<=x) = 0.9 same as P (X>=x) = 0.1 Go to Formulas > More Functions > Statistical > NORMINV Then type as per shown in diagram above. Note standard deviation is square root of variance. An alternative is to type: =NORMINV ( p, mean, σ ) in the desired cell. Question: The average marks obtained by the students in a test are 65 with a standard deviation of % of the students got distinction and 25% of the students failed. What are the pass marks and the minimum marks for distinction? (Standard Normal Distribution is a special type of Normal Distribution having mean=0 and standard deviation=1) NORMSDIST and NORMSINV 1. Find P (X <= 1.9) when x is standard normal (i.e. normal with mean=0 and variance=1). Choose Formulas Tab > More Functions > Statistical > NORMSDIST. Fill in the Function Arguments Tab with Z value of 1.9. This gives result that P (X <= 1.9) =

4 Much simpler is to directly type in the cell = NORMSDIST(1.9) and hit <enter> to get P (X <= 1.9) = Find the value x* such that Pr(X <= x*) = 0.9 when x is standard normal. Choose Formulas Tab > More Functions > Statistical > NORMSINV Fill in the Function Arguments Tab with probability value of 0.9. This gives result that x* = , i.e. Pr (X <= ) = 0.9. Much simpler is to directly type in the cell = NORMSINV (0.9) and hit <enter> to get x* = T Distribution TDIST gives the probability of being in the right tail i.e. P (X > x), or of being in both tails i.e. P ( X > x). TINV considers the inverse of the probability of being in both tails. 1. Find P (X <= 1.9) when x is t-distributed with 9 degrees of freedom. This is 1 P (X > 1.9) where Excel function TDIST gives P (X > 1.9). Choose Formulas Tab > More Functions >Statistical > TDIST. Fill in the Function Arguments Tab as shown below. This gives result that P (X > 1.9) = So P (X <= 1.9) = = Much simpler is to directly type in the cell = 1 - TDIST(1.9, 9, 2) and hit <enter>.

5 2. Find the value x* such that P (X <= x*) = 0.9 when x is t-distributed with 9 degrees of freedom. This is the same value as that for which P ( X >= x*) = 0.2. (Since there is probability 0.1 in the right tail and probability 0.1 in the left tail). (Note: probability associated with two-tailed distribution is entered below.) Choose Formulas Tab > More Functions > Statistical > TINV. Fill in the Function Arguments Tab: This gives result that P ( X > 1.383) = 0.2. So P (X <= 1.383) = 0.9. Much simpler is to directly type in the cell: =TINV (0.2, 9) and hit <enter> to get P ( X > 1.383) = 0.2 Example 4: Suppose scores on an IQ test are normally distributed, with a mean of 100. Suppose 20 people are randomly selected and tested. The standard deviation in the sample group is 15. What is the probability that the average test score in the sample group will be at most 110? Solution: In excel, first we will need to find the t score. Type = ( )/ (15/ sqrt (20)) in any cell. This gives the t-score. Now select the TDIST function and in the functions arguments tab type the t score obtained, degrees of freedom=20-1=19 and Tails=1 Excel will give the result as Hence the required probability: Hence, there is a 99.6% chance that the sample average will be no greater than 110. Or type =1-TDIST(( )/(15/SQRT(20)),19,1).

6 [Note: excel requires you to calculate the t score first, hence instead of that you can also use the T- Distribution Calculator] Link is : In the dropdown box, describe the random variable. Enter a value for degrees of freedom. Enter a value for all but one of the remaining text boxes. Click the Calculate button to compute a value for the blank text box. Describe the random variable Sample mean Degrees of freedom Sample standard deviation Population mean Sample mean (x) Cumulative probability: P(X < x) Calculate Note: The statistics calculator is available for normal distribution and chi-distribution also. Check the link! CHI-DISTRIBUTION CHIDIST Example 5: The Acme Widget Company claims that their widgets last 5 years, with a standard deviation of 1 year. Assume that their claims are true. If you test a random sample of 9 Acme widgets, what is the probability that the standard deviation in your sample will be less than 0.95 years? (Chi-test) Solution: Another shortcoming of excel is that you will have to find the chi-square statistic first. Χ 2 = [( n - 1 ) * s 2 ] / σ 2 Χ 2 = [ ( 9-1 ) * (0.95) 2 ] / (1.0) 2 = 7.22 Now chose CHIDIST function from statistical tab from More Functions in Formulas. Type X=7.22 and degrees of freedom=9-1=8. Hit<enter>

7 Or, we can simply type =CHIDIST ((8*(0.95^2))/1^2,8) Which is =CHIDIST(chi statistic, degrees of freedom) CHIINV Example 6: Find the chi-square critical value, if the probability is 0.75 and the sample size is 25. Solution: Select CHIINV function. Type probability=0.75, degrees of freedom=25-1=24. Hit <enter> Or type =CHIINV(0.75,24)