The following pie chart shows the percentages of total items sold in a month in a certain fast food restaurant. A total of 4900 fast food items were sold during the month. How many were burgers? How many were french fries? Burgers : 4900(.32) = 1568 French Fries : 4900(.18) = 882 4900(.32)=1568 32% or 1,568 burgers were sold during the month. 4900(.18)=882 18% or 882 french fries were sold during the month. Instructor Comments Very good. Answer Key During an hour at a fast food restaurant, the following types of sandwiches are ordered: Turkey Hamburger Cheeseburger Fish Hamburger Turkey Fish Chicken Fish Chicken Turkey Fish Hamburger Fish Cheeseburger FishCheeseburger Hamburger Fish Fish Cheeseburger Hamburger Fish Turkey Turkey Chicken Fish Chicken Cheeseburger Fish Turkey Fish Fish Hamburger Fish Fish Turkey Chicken Hamburger Fish Cheeseburger Chicken Chicken Turkey Fish Chicken Hamburger Chicken Fish Chicken a) Make a frequency distribution for this data. Types of Frequency Sandwiches Turkey 8

b) Is the point 85 above, below, or the same as the mean. How many standard deviations is 85 from the mean. x*65 z=x-u = 85-65 =2 o 10 z=2 The point 85 is above the mean (because it is a positive number), meaning that the data point is two standard deviations above the mean. c) Is the point 57.5 above, below, or the same as the mean. How many standard deviations is 57.5 from the mean. x*65 z=x-u = 57.5-65 =-0.75 o 10 z=-0.75 The point 57.5 is below the mean (because it is a negative number), meaning that the data point is .75 standard deviations below the mean. d) Is the point 107 above, below, or the same as the mean. How many standard deviations is 107 from the mean. x*65 z=x-u = 107-65 =4.2 o 10 z=4.2 The point 107 is above the mean (because it is a positive number), meaning that the data point is 4.2 standard deviations above the mean. Consider the following set of data: {22, 14, 35, 49, 8, 18, 30, 44} a) Find the median. {8,14,18,22,30,35,44,49} Median=22+30 =26 26 b) Find the mode of this set. {8,14,18,22,30,35,44,49} No mode (no number appears more than once). Suppose A and B are two events with probabilities: P(A)=.35,P(B c )=.45,P(A∩B)=.25.

Suppose that 5 out of 11 people are to be chosen to go on a mission trip. In how many ways can these 5 be chosen if the order in which Since we do not want to count all of the possible orderings, we use combinations. If the smoke alarm has lasted 7 years, it would have lasted for 4 years already, so P(EnF)=.10 P(EnF)=.10 P(FlE)=P(EnF) =.10 =.1538 P(E) .65 Suppose that 5 out of 11 people are to be chosen to go on a mission trip. In how many ways can these 5 be chosen if the order in which they are chosen is not important. n=11 r=5 C(n,r)= n! r!(n-r)! C(11,5)= 11! = 11(10)(9)(8)(7)(6) =2772 5!6! 5(4)(3)(2)(1) -1.0 points Instructor Comments You didn't need the 6 on top. It cancelled as part of the 6! on the bottom. Answer Key A factory has eight safety systems. During an emergency, the probability of any one of the safety systems failing is .08. What is the probability that six or more safety systems will fail during an emergency? For six failures: n=8, x=6, p=.08 f(x)= n! p x (1-p) (n-x) = x!(n-x)! f(x)= 8! .08 6 (1-.08) (8-6) =.000006209 6!(8-6)! For seven failures: n=8, x=7, p=.08 f(x)= n! p x (1-p) (n-x) = x!(n-x)! f(x)= 8! .08 7 (1-.08) (7-6) =.000000155 7!(8-7)! For eight failures: n=8, x=8, p=.08 f(x)= n! p x (1-p) (n-x) = x!(n-x)! f(x)= 8! .08 8 (1-.08) (8-8) =.000000002 8!(8-8)!

An archer is shooting arrows at a target. She hits the target 55% of the time. If she takes 14 shots at the target, what is the probability t 14!(14-7)! -2.0 points Instructor Comments You can see in the Answer Key where you are off in the denominator, which throws your answer off. Answer Key Suppose that you take a sample of size 26 from a population that is not normally distributed. Can the sampling distribution of x x be approximated by a normal probability distribution? The sampling distribution of x x cannot be approximated by a normal probability distribution because if the population isn't normally distributed, the sample size has to be at least 30. Suppose that you are attempting to estimate the annual income of 1200 families. In order to use the infinite standard deviation formula, what sample size, n, should you use? n < 0.05= N n < 0.05= 1200 n < 0.05(1200) n < 60 Sample size must be less than 60. Suppose that in a large hospital system, that the average (mean) time that it takes for a nurse to take the temperature and blood pressure of a patient is 175 seconds with a standard deviation of 50 seconds. What is the probability that 35 nurses selected at random will have a mean time of 160 seconds or less to take the temperature and blood pressure of a patient? u=175, o=50, n=35, x x =160 o x x = o √n o x x = 50 = 8.4515 √35 z= x x -u = o x x z= x x -u =160-175 = -1.77 o x x 8.4515 P(Z<-1.77)=.03836 Suppose that in a very large city 7.2 % of the people have more than two jobs. Suppose that you take a random sample of 80 people in that city, what is the probability that 7 % or less of the 80 have more than two jobs?

A shipment of 400 new blood pressure monitors have arrived. Tests are done on 55 of the new monitors and it is found that 10 of the 5 We have a finite population, so we will use Case 2: The proportion of the sample that are defective is 10/55 = .182 so we set P=.182. As we mentioned previously, we estimate p by P. So, p .182± .08 So the proportion of the total that are defective is between .102 and .262. A doctor has a large number of patients and would like to know if his patients prefer to fill in forms electronically or prefer to hand write their forms. So, he surveys 95 patients and finds that 53 prefer electronic forms while 42 prefer hand written forms. Find the 95% confidence limit for the proportion of all patients that prefer the electronic forms. Case 1: P/p=p/n=53/95=.56 n=95 z=1.96 P± z √p(1-p) √n .56± 1.96 √.56(1-.56) √95 .56±.1 .46 to .66 A shipment of 400 new blood pressure monitors have arrived. Tests are done on 55 of the new monitors and it is found that 10 of the 55 give incorrect blood pressure readings. Find the 90% confidence interval for the proportion of all the monitors that give incorrect readings. Case 2: P/p=p/n=10/55=.182 n=55 N= 400 z=1.645 P± z √p(1-p) √N-n √n √N-1 .182± 1.645 √.182(1-.182) √400-55 √55 √400-1 .182±.06191 0.12009 to 0.24391 -2.0 points Instructor Comments You have a math error in your calculation. You can see in the Answer key that the value past the ± should be 0.079587 (or round to 0.08). Answer Key

Suppose we have independent random samples of size n1 = 570 and n2 = 675. The proportions of success in the two samples are p1= .4 From table 6.1, we see that 99% confidence corresponds to z=2.58. Notice that the sample sizes are each greater than 30, so we may us Suppose we have independent random samples of size n 1 = 570 and n 2 = 675. The proportions of success in the two samples are p 1 = .41 and p 2 = .53. Find the 99% confidence interval for the difference in the two population proportions. n 1 = 570, n 2 = 675, p 1 = .41, p 2 = .53, z=2.58 P 1 - P 2 ±z√p 1 (1-p 1 ) + p 2 (1-p 2 ) n 1 n 2 .41-.53±2.58√.41(1-.41) + .53(1-.53) 570 675 -.12±2.58(.0281678) -.12± .072673 The confidence interval is (-.047327,-.192673). Instructor Comments You have your confidence interval written backwards. Answer Key

A new energy drink is supposed to improve a person’s time in the one mile run. The times, in seconds, of eight runners with and withou Find the 98 % confidence interval for mean of the differences, µd. Note that n=8. We will define , di = xi - yi. After doing the appropriate calculations, we find that d =-1.375 sd = 9.365. we look at the student’s t chart for 98% confidence (the 98% is found along the bottom row of the chart) and DOF=8-1=7 (the df=7 is fo t= -6.833 = -6.833/4.22=-1.619 (10.34/√6) We do not reject that null hypothesis due to the fact that t is not less than -t .05 (because -1.619>-2.571). A new energy drink is supposed to improve a person’s time in the one mile run. The times, in seconds, of eight runners with and without the drink are given below: Runner 1 2 3 4 5 6 7 8 x-time (before) 260 273 26 6 26 5 261 247 26 8 27 1 y-time (after) 263 270 28 0 27 9 266 242 26 3 25 9 Find the 98 % confidence interval for mean of the differences, µ d . DOF=n-1 DOF=8-1=7 df=7 so t=2.998 d - =- 1.375 sd= 9.365 d - =d1+d2...dn =-3+3-14-14-5+5+5+12 = -1.375 n 8 -5.0 points Instructor Comments You did not complete the problem. Answer Key Runner 1 2 3 4 5 6 7 8 x-time (before) 260 273 266 265 261 247 26 8 27 1 y-time (after) 263 270 280 279 266 242 26 3 25 9

x x y y Suppose you have 14 data points and you calculate the sample correlation coefficient and find that r = - .58 . Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you answer. n=14, r=-.58, 95% confident? CVCC=.53241 We can be 95% confident that a negative linear relation exists (because the original r is negative) between the variables due to the fact that the absolute of r is │r│=.58 and is greater than .53241 (which we can find on the critical values chart). Suppose you have 40 data points and you calculate the sample correlation coefficient and find that r = - .30 . Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you answer. n=40, r=-.30, 95% confident? CVCC=.31201 We can be 95% confident that a no relation exists between the variables due to the fact that the absolute of r is │r│=.30 and is less than .31201 (which we can find on the critical values chart). Compute the sample correlation coefficient for the following data: Can you be 95% confident that a linear relation exists between the variables? If so, is the relation positive or negative? Justify you answer. n=5 x - =2+3+7+9+12 =6.6 5 y - =3+6+10+14+17 =10 5 s 2 =∑(x-x - ) 2 = (2-6.6) 2 +(3-6.6) 2 +(7-6.6) 2 +(9-6.6) 2 +(12-6.6) 2 =17.3 n-1 5-1 s x =√s 2=√17.3=4.16 s 2 =∑(y-y - ) 2 = (3-10) 2 +(6-10) 2 +(10-10) 2 +(14-10) 2 +(17-10) 2 =32.5 n-1 5-1 s y =√s 2=√32.5=5.7 r=∑(x i -x -)2(y i -y-)2 s x s y n-1 r=(2-6.6 )(3-10 )+(3-6.6 )(6-10 )+(7-6.6 )(10-10 )+(9-6.6 )(14-10 )+(12-6.6)(17-10) ( 4.16) (5.7) (4.16) (5.7) (4.16) (5.7 ) ( 4.16) (5.7) ( 4.16 ) (5.7) =3.96074 =.99. 5-1 4 n=5, Confidence=95%?, r=.99, CVCC=.87834

x x y y Absolute of r is │r│= .99 and is greater than .87834 so a positive linear relation exists. Find the best fit line for the following data: n=5 x - =1+2+5+7+9 =4.8 5 y - =15+12+11+9+7 =10.8 5 s 2 =∑(x -x - ) 2 = (1-4.8) 2 +(2-4.8) 2 +(5-4.8) 2 +(7-4.8) 2 +(9-4.8) 2 =11.2 n-1 5-1 s x =√s 2 =√11.2=3.35 s 2 =∑(y -y - ) 2 = (15-10.8) 2 +(12-10.8) 2 +(11-10.8) 2 +(9-10.8) 2 +(7-10.8) 2 =9.2 n-1 5-1 s y =√s 2 =√9.2=3.03 r=∑(x i -x - ) 2 (y i -y - ) 2 s x s y n-1 r=(1-4.8 )(15-10.8 )+(2-4.8 )(12-10.8 )+(5-4.8 )(11-10.8 )+(7-4.8 )(9-10.8 )+(9-4.8)(7- 10.8) ( 3.35) (3.03) (3.35) (3.03) (3.35) (3.03 ) ( 3.35) (3.03) ( 3.35 ) (3.03)=- 3.8561 =-.964. 5-1 4 n=5, Confidence=95%?, r=-.964, CVCC=.87834 Absolute of r is │r│= .964 and is greater than .87834 so a negative (because original r was negative) linear relation exists. x - =4.8, y - =10.8, S x =3.35, S y =3.03, r=-.964 Slope: m=r S y = -.964 3.03 =-.873 S x 3.35 Y intercept: b=y - -mx - =10.8-(-.873)(4.8)=14.99 Line Equation: y=mx+b y=-.873x+14.99 Find the value of X 2 for 13 degrees of freedom and an area of .100 in the right tail of the chi- square distribution. X 2 =19.812. Find the value of X 2 for 17 degrees of freedom and an area of .050 in the left tail of the chi- square distribution. a=1-.050=.95 X 2 =8.672.

Find the value of X 2 values that separate the middle 80 % from the rest of the distribution for 17 degrees of freedom. a=1-.80=.20/2=.10 in each of the tails. Area to right of first X 2 =.80+.10=.90 X 2 =10.085. Area to right of second X 2 =.10 X 2 = 24.769. Find the critical value of F for DOF=(6,18) and area in the right tail of .01. F=4.01 Numerator= 6, Denominator=18 <---Look up in F Distribution Table in the right tail of .01.

The mayor of a large city claims that 30 % of the families in the city earn more than $ 100,000 per year; 52 % earn between $ 30,000 and $ 100,000 (inclusive); 18 % earn less than $ 30,000 per year. In order to test the mayor’s claim, 285 families from the city are surveyed and it is found that: 90 of the families earn more than $ 100,000 per year; 135 of the families earn between $ 30,000 and $ 100,000 per year (inclusive); 60 of the families earn less $ 30,000. Test the mayor’s claim based on 5 % significance level. H 0 : The mayor's claim is correct. H 1 : The mayor's claim is not correct. More than two possible outcomes: Multinomial experiment-we use the Chi-Square chart here. DOF=3-1=2, Level of Significance=5%=.05 Chi Square=5.991 Expected frequencies= E i =np i n=285 Earn more than $100,000 per year E 1 = 285(.30)=85.5 Earn between $ 30,000 and $ 100,000 per year E 2 =285(.52)=148.2 Earn less than $ 30,0001941 to 1970 E 3 =285(.18)=51.3 Observed Frequencies: Earn more than $100,000 per year=90 Earn between $ 30,000 and $ 100,000 per year=135 Earn less than $ 30,000=60 Test Statistics Calculation: X 2 =∑(O i -E i -) 2=(90-85.5 )2+(135-148.2 )2+(60-51.3 )2= 2.888. E i 85.5 148.2 51.3 X 2 is smaller than the critical value of 5.991 so we do not reject the null hypothesis. A trucking company wants to find out if their drivers are still alert after driving long hours. So, they give a test for alertness to two groups of drivers. They give the test to 475 drivers who have just finished driving 4 hours or less and they give the test to 635 drivers who have just finished driving 8 hours or more. The results of the tests are given below. Passe d Faile d Drove 4 hours or less 365 110 Drove 8 hours or more 440 195 Is there is a relationship between hours of driving and alertness? (Do a test for independence.) Test at the .5 % level of significance.

A trucking company wants to find out if their drivers are still alert after driving long hours. So, they give a test for alertness to two group Is there is a relationship between hours of driving and alertness? (Do a test for independence.) Test at the .5 % level of significance. H0: Driving hours and alertness are independent events. H1: Driving hours and alertness are not independent events. H 0 : There is a relationship between hours of driving and alertness. H 1 : There is not a relationship between hours of driving and alertness. # of Rows= 2, # of Columns=2 DOF=(# of rows-1)(# of columns-1)=(2-1)(2-1)=1 LOS=.005 CSCV=7.879 Passed Failed Row Totals Drove 4 hours or less 365 110 475 Drove 8 hours or more 440 195 635 Column Totals 805 305 1110 Expected Values Calculation: E ij = (TR i )(TC j ) Total number in sample E for drove 4 hours or less and Passed E 11 =(475)(805)/1110=344.48 E for drove 4 hours or less and Failed E 12 =(475)(305)/1110=130.52 E for drove 8 hours or more and Passed E 21 =(635)(805)/1110=460.52 E for drove 8 hours or more and Failed E 22 =(635)(305)/1110=174.48 X2=∑ (O ij -E ij ) 2= (365-344.48) 2+(110-130.52) 2+(440-460.52) 2+(195-174.48) 2=7.776 E ij 344.48 130.52 460.52 174.48 X 2 is less than the CV of 7.879 so we do not reject the null hypothesis. -1.0 points Instructor Comments Your null and alternative hypotheses are switched. The calculations are correct. Answer Key Passe d Faile d Drove 4 hours or less 365 110 Drove 8 hours or more 440 195

We have two rows and three columns, so # of Rows =2 and # of Columns=2. The degrees of freedom are given by: DOF = (# of Rows-1)(# of Columns-1)=(2-1)(2-1)=1. Using this, along with .005 (for the .5 % level of significance) we find in the chi-squa This value is less than the critical value of 7.879. So, we do not reject the null hypothesis.