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each Questions: Work question, using Excel where appropriate. Remember, each gestation period is its own normal distribution. Thus, you will need to change the "mean" and "standard deviation" to reflect the question you are answering. 2. What percent of the babies born with each gestation period have a low birth weight (under 5.5 pounds)? a. under 28 weeks b. 32 to 35 weeks c. 37 to 39 weeks d. 42 weeks and over Answer: 3. Describe the weights of the top 10% of the babies born with each gestation period. a. 37 to 39 weeks b. 42 weeks and over Answer: 4. For each gestation period, what is the probability that a baby will weigh between 6 and 9 pounds at birth? a. 32 to 35 weeks b. 37 to 39 weeks c. 42 weeks and over Answer: 5. A birth weight of less than 3.3 pounds is classified by the NCHS as a "very low birth weight." What is the probability that a baby has a very low birth weight for each gestation period? a. under 28 weeks b. 32 to 35 weeks c. 37 to 39 weeks Answer: Gestation Period Under 28 Weeks 28 to 31 Weeks 32 to 35 Weeks 36 Weeks 37 to 39 Weeks 40 Weeks 41 Weeks 42 Weeks and over Question 2 Question 3 Question 4 Question 5 Note: To solve these questions for the oth change to the correct mean and standard d NORMDIST(x,mean,stddev,true) = returns the given mean and standard deviation. NORMINV(leftarea,mean,stddev) = returns "leftarea" for the given mean and standard Gestation Period Under 28 Weeks 28 to 31 Weeks 32 to 35 Weeks 36 Weeks 37 to 39 Weeks 40 Weeks 41 Weeks 42 Weeks and over Mean Standard Birth Deviation Weight 1.88 1.19 4.07 1.87 5.73 1.48 6.46 1.2 7.33 1.09 7.72 1.05 7.83 1.08 7.65 1.12 Work each question, using Excel where appropriate. Remember, each gestation period is its own normal distribution. Thus, you will need to change the "mea 2. What percent of the babies born with each gestation period have a low birth weight (under 5.5 pounds a. under 28 weeks b. 32 to 35 weeks c. 37 to 39 weeks d. 42 weeks and over Answer: mean a Under 28 weeks standard deviation 1.88 1.19 1.88 1.19 6 3.04 =(5.5-1.88)/1.19 3.04 is= 5.50 is Prob(Z)= 1.48 5.73 1.48 6 -0.16 =(5.5-5.73)/1.48 -0.16 is= 5.50 is Prob(Z)= 1.09 7.33 1.09 6 -1.68 =(5.5-7.33)/1.09 -1.68 is= 5.50 is Prob(Z)= 1.12 7.65 1.12 6 -1.92 =(5.5-7.65)/1.12 -1.92 is= 5.50 is Prob(Z)= Mean== Standard deviation == x= z=(x-)/= Cumulative Probability corresponding to z= Or Probability corresponding to x< b 32 to 35 weeks 5.73 Mean== Standard deviation == x= z=(x-)/= Cumulative Probability corresponding to z= Or Probability corresponding to x< c 37 to 39 weeks 7.33 Mean== Standard deviation == x= z=(x-)/= Cumulative Probability corresponding to z= Or Probability corresponding to x< d 42 weeks and over 7.65 Mean== Standard deviation == x= z=(x-)/= Cumulative Probability corresponding to z= Or Probability corresponding to x< 3. Describe the weights of the top 10% of the babies born with each gestation period. a. 37 to 39 weeks b. 42 weeks and over Significance level = No of tails= Z value corresponding to 0.1 significance level and 1 tails= Answer: mean a. 37 to 39 weeks Mean== Standard deviation == z=(x-)/= x=+z Answer: more than 8.73 pounds 7.65 1.12 7.65 1.12 1.2816 9.09 =7.65+(1.2816*1.12) 9.09 pounds standard deviation 7.33 1.09 7.33 1.09 1.2816 8.73 =7.33+(1.2816*1.09) 0.1 1 1.2816 b. 42 weeks and over Mean== Standard deviation == z=(x-)/= x=+z Answer: more than 4. For each gestation period, what is the probability that a baby will weigh between 6 and 9 pounds at bir a. 32 to 35 weeks b. 37 to 39 weeks c. 42 weeks and over Answer: mean standard deviation a. 32 to 35 weeks 5.73 1.48 5.73 pounds Mean== Standard deviation == 1.48 pounds x1= 6 pounds x2= 9 pounds z1=(x1-)/= 0.18 =(6-5.73)/1.48 z2=(x2-)/= 2.21 =(9-5.73)/1.48 Cumulative Probability corresponding to z1= 0.18 is= Cumulative Probability corresponding to z2= 2.21 is= Therefore probability that the value of x will be between x1= is = 41.40% =98.64%-57.24% Answer: 41.40% b. 37 to 39 weeks 7.33 1.09 7.33 pounds Mean== Standard deviation == 1.09 pounds x1= 6 pounds x2= 9 pounds z1=(x1-)/= -1.22 =(6-7.33)/1.09 z2=(x2-)/= 1.53 =(9-7.33)/1.09 Cumulative Probability corresponding to z1= -1.22 is= Cumulative Probability corresponding to z2= 1.53 is= Therefore probability that the value of x will be between x1= is = 82.61% =93.73%-11.12% Answer: 82.61% 7.65 1.12 c. 42 weeks and over 7.65 pounds Mean== Standard deviation == 1.12 pounds x1= 6 pounds x2= 9 pounds z1=(x1-)/= -1.47 =(6-7.65)/1.12 z2=(x2-)/= 1.21 =(9-7.65)/1.12 Cumulative Probability corresponding to z1= -1.47 is= Cumulative Probability corresponding to z2= 1.21 is= Therefore probability that the value of x will be between x1= is = 81.57% =88.6%-7.03% Answer: 81.57% 5. A birth weight of less than 3.3 pounds is classified by the NCHS as a "very low birth weight." What is probability that a baby has a very low birth weight for each gestation period? a. under 28 weeks b. 32 to 35 weeks c. 37 to 39 weeks Answer: mean a. under 28 weeks standard deviation 1.88 1.19 1.88 pounds 1.19 pounds 3.3 pounds 1.19 =(3.3-1.88)/1.19 1.19 is= 3.30 is Prob(Z)= Mean== Standard deviation == x= z=(x-)/= Cumulative Probability corresponding to z= Or Probability corresponding to x< Answer: 88.36% b. 32 to 35 weeks 5.73 1.48 5.73 pounds 1.48 pounds 3.3 pounds -1.64 =(3.3-5.73)/1.48 -1.64 is= 3.30 is Prob(Z)= Mean== Standard deviation == x= z=(x-)/= Cumulative Probability corresponding to z= Or Probability corresponding to x< Answer: 5.03% c. 37 to 39 weeks 7.33 1.09 7.33 pounds 1.09 pounds 3.3 pounds -3.7 =(3.3-7.33)/1.09 -3.7 is= 3.30 is Prob(Z)= Mean== Standard deviation == x= z=(x-)/= Cumulative Probability corresponding to z= Or Probability corresponding to x< Answer: 0.01% Mean Birth Weight 1.88 4.07 5.73 6.46 7.33 7.72 7.83 7.65 Answer Standard Deviation 1.19 1.87 1.48 1.2 1.09 1.05 1.08 1.12 o solve these questions for the other gestation periods, you will need to o the correct mean and standard deviation. IST(x,mean,stddev,true) = returns the area (probability) to the left of x for mean and standard deviation. V(leftarea,mean,stddev) = returns the x value that has an area to the left of for the given mean and standard deviation. you will need to change the "mean" and "standard deviation" to reflect the question you are answering. ow birth weight (under 5.5 pounds)? 1 =NORMSDIST(F70) 1 0r= 99.88% 0.44 =NORMSDIST(F79) 0.44 0r= 43.83% 0.05 =NORMSDIST(F90) 0.05 0r= 4.66% 0.03 =NORMSDIST(F101) 0.03 0r= 2.75% station period. =NORMSINV(1-G109/G110) igh between 6 and 9 pounds at birth? 0.57 0r= 0.99 0r= 6 and x2= 57.24% =NORMSDIST(F146) 98.64% =NORMSDIST(F147) 9 pounds 0.11 0r= 0.94 0r= 6 and x2= 11.12% =NORMSDIST(F163) 93.73% =NORMSDIST(F164) 9 pounds 0.07 0r= 0.89 0r= 6 and x2= 7.03% =NORMSDIST(F179) 88.60% =NORMSDIST(F180) 9 pounds "very low birth weight." What is the riod? 0.88 0.88 0r= 0.88 88.36% 0.05 0.05 0r= 0.05 5.03% 0 0 0r= 0 0.01% Age Distribution in the United States, Page 277 Answer each of the 6 questions. The age distribtuion and sample means is already given below. Sample means below were obtained from random samples of size n = 40. Sample Means 28.14 31.56 36.86 32.37 36.12 39.53 36.19 39.02 35.62 36.3 34.38 32.98 36.41 30.24 34.19 44.72 38.84 42.87 38.9 34.71 34.13 38.25 38.04 34.07 39.74 40.91 42.63 35.29 35.91 34.36 36.51 36.47 32.88 37.33 31.27 35.8 Age 0-4 5-9 10 - 14 15 - 19 20 - 24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 - 59 60 - 64 65 - 69 70 - 74 75 - 79 80 - 84 85 - 89 90 - 94 95 - 99 Relative Midpoint Frequency 2 6.7% 7 6.8% 12 7.4% 17 7.2% 22 7.0% 27 6.2% 32 6.8% 37 7.3% 42 8.1% 47 7.6% 52 6.6% 57 5.5% 62 4.2% 67 3.4% 72 3.0% 77 2.6% 82 1.9% 87 1.0% 92 0.5% 97 0.2% mid*relfreq mid^2*relfreq 0.13 0.27 0.48 3.33 0.89 10.66 1.22 20.81 1.54 33.88 1.67 45.2 2.18 69.63 2.7 99.94 3.4 142.88 3.57 167.88 3.43 178.46 3.14 178.7 2.6 161.45 2.28 152.63 2.16 155.52 2 154.15 1.56 127.76 0.87 75.69 0.46 42.32 0.19 18.82 509.18 22.57 Population Mean = 36.48 Population Variance = Population Standard Deviation = Questions: Remember, each gestation period is its own normal distribution. Thus, you will need to change the "mean" and "standard deviation" to reflect the question you are answering. 1. Enter the age distribution of the United States into a technology tool. Use the the tool to find the mean age in the United States. Answer: 2. Enter the set of sample means into a technology tool. Find the mean of the set of sample means. How does it compare with the mean age in the United States? Does this agree with the results by central Limit theorem? Answer: 3. Are the ages of people in the United States normally distributed? Explain your reasoning. Answer: 4. Sketch a relative frequency histogram for hte 36 sampl means. Use 9 classes. Is the histogram approximately bell shaped and symmetic? Does this agree with the results predicted by the Central mean of the set of sample means. How does it compare with the mean age in the United States? Does this agree with the results by central Limit theorem? Answer: 3. Are the ages of people in the United States normally distributed? Explain your reasoning. Answer: 4. Sketch a relative frequency histogram for hte 36 sampl means. Use 9 classes. Is the histogram approximately bell shaped and symmetic? Does this agree with the results predicted by the Central Limite Theorem? Answer: 5. Use technology tool to find the standard deviation of the set of 36 samoke means. How does it compare with the standard deviation of the ages? Does this agree with the result predicted by the Central Limit Theorem? Answer: 1. Enter the age distribution of the United States into a technology tool. Use the the tool to find the mean age in the United States. Answer: (Calculation has already been done above) 2. Enter the set of sample means into a technology tool. Find the mean of the set of sample means. How does it compare with the mean age in the United States? Does this agree with the results by central Limit theorem? Answer: Mean= 36.21 =AVERAGE(H8:H43) Population mean= 36.48 (calculated above) Thus the result agrees with the Central Limit theorem Central limit theorem says that mean of all the sample means = Population mean 3. Are the ages of people in the United States normally distributed? Explain your reasoning. Answer: The histogram is shown below; it does not appear to be the bell shaped normal distribution Age 0-4 5-9 10 - 14 Relative Frequency 6.7% 6.8% 7.4% Relative Frequency 9.0% 8.0% 7.0% 6.0% 5.0% 4.0% 3.0% Relative Frequency 15 - 19 20 - 24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 - 59 60 - 64 65 - 69 70 - 74 75 - 79 80 - 84 85 - 89 90 - 94 95 - 99 7.2% 7.0% 6.2% 6.8% 7.3% 8.1% 7.6% 6.6% 5.5% 4.2% 3.4% 3.0% 2.6% 1.9% 1.0% 0.5% 0.2% 9.0% 8.0% 7.0% 6.0% 5.0% 4.0% 3.0% 2.0% 1.0% 0.0% 5 - 9 15 - 19 25 - 29 35 - 39 45 - 49 55 - 59 65 - 69 75 - 79 85 - 89 95 0 - 4 10 - 14 20 - 24 30 - 34 40 - 44 50 - 54 60 - 64 70 - 74 80 - 84 90 - 94 4. Sketch a relative frequency histogram for hte 36 sampl means. Use 9 classes. Is the histogram approximately bell shaped and symmetic? Does this agree with the results predicted by the Central Limite Theorem? Answer: Sample Means 28.14 31.56 36.86 32.37 36.12 39.53 36.19 39.02 35.62 36.3 34.38 32.98 36.41 30.24 34.19 44.72 38.84 42.87 38.9 34.71 34.13 38.25 38.04 Number of classes= Minimum= Maximum= class width= Class 28.14 to 29.99 to 31.84 to 33.69 to 35.54 to 37.39 to 39.24 to 41.09 to 42.94 to Table Class 28.14 - 29.99 29.99 - 31.84 - 31.84 33.69 33.69 - 35.54 35.54 - 37.39 9 28.14 44.72 1.85 =(44.72-28.14)/9 29.99 31.84 33.69 35.54 37.39 39.24 41.09 42.94 44.79 relative frequency Frequency 0.028 1 0.083 3 0.083 3 0.194 7 0.306 11 34.07 39.74 40.91 42.63 35.29 35.91 34.36 36.51 36.47 32.88 37.33 31.27 35.8 37.39 - 39.24 39.24 - 41.09 41.09 - 42.94 42.94 - 44.79 Total 0.139 0.083 0.056 0.028 1.000 5 3 2 1 36 0.350 0.300 0.250 0.200 0.150 0.100 0.050 0.000 29.99 - 31.84 33.69 - 35.54 37.39 - 39.24 4 28.14 - 29.99 31.84 - 33.69 35.54 - 37.39 39.24 - 41.09 Histogram is approximately bell shaped and symmetic and agrees with the results predicted by the Central Limite Theorem which says that whatever the population shape, the sampling distribution is approximately normal 5. Use technology tool to find the standard deviation of the set of 36 samoke means. How does it compare with the standard deviation of the ages? Does this agree with the result predicted by the Central Limit Theorem? Answer: Standard deviation= 3.5518 =STDEV(H8:H43) Standard deviation of population= 22.57 Standard deviation of sample mean=Standard deviation / sqauare root of sample size sample size= 40 Therefore Standard deviation of sample mean= 3.57 These two values are close; thus the result predicted by central limit theorem is true s below were random ze n = 40. - 69 75 - 79 85 - 89 95 - 99 70 - 74 80 - 84 90 - 94 9 - 39.24 41.09 - 42.94 39.24 - 41.09 42.94 - 44.79 Confidence Level 90% 95% 98% 99% Alpha 0.100 0.050 0.020 0.010 Area to Left 0.95 0.98 0.99 1 Z-Score Z-Score (Rounded) 1.64 1.65 1.96 1.96 2.33 2.33 2.58 2.58 Questions: 1. Using Excel, find the z-score that corresponds to the following Confidence Levels: a. b. c. d. 80% 85% 92% 97% The table above illustrates how Excel can find the z-score, using NORMINV, that corresponds to various levels of confidence. Note: Our textbook rounds the 99% confidence level to 2.575. The Confidence Level is the area in the "center" of the normal distribution. This leaves two tails. Alpha is the area in the tails. Thus, alpha = (100% - Confidence Level) and is expressed in decimal form. The Area to Left is then 1 - alpha/ 2. Table with "formulas" is shown at far right. 2. Using Excel, find the t-score that corresponds to the following Confidence Levels and Sample Sizes: a. b. c. d. 95% with n = 25 96% with n = 15 97% with n = 21 91% with n = 10 3. Suppose we wish to estimate the population mean using a confidence interval. When is it appropriate to us a z-score? When is it appropriate to use a t-score? Confidence Level 90% 95% 98% 99% Area in Two Tails 0.100 0.050 0.020 0.010 Sample Size 25 14 12 27 t-Score t-Score (Rounded) 1.71 1.71 2.16 2.16 2.72 2.72 2.78 2.78 The table above illustrates how Excel can find the t-score, using TINV, that corresponds to various levels of confidence. The Confidence Level is the area in the "center" of the tdistribution. This leaves two tails. Area in Two Tails is simply the total area in both of the tails. Note that Area in Two Tails = 100% - Confidence Level and is expressed in decimal form. Area in Two Tails = Alpha. The format of TINV is TINV(probability,degrees of freedom) whre probability = Area in Two Tails and degrees of freedom = Sample Size - 1 Table with "formulas" is shown at far right. 1. Using Excel, find the z-score that corresponds to the following Confidence Levels: a. 80% Confidence level= or Significance level = No of tails= Z value corresponding to 0.2 significance level and 2 tails= 80% 20.00% =1-0.8 2 1.2816 =NORMSINV(1-H48/H49) b. 85% Confidence level= or Significance level = No of tails= Z value corresponding to 0.15 significance level and 2 tails= c. 92% Confidence level= or Significance level = No of tails= Z value corresponding to 0.08 significance level and 2 tails= d. 97% Confidence level= or Significance level = No of tails= Z value corresponding to 0.03 significance level and 2 tails= 97% 3.00% =1-0.97 2 2.1701 =NORMSINV(1-H70/H71) 92% 8.00% =1-0.92 2 1.7507 =NORMSINV(1-H62/H63) 85% 15.00% =1-0.85 2 1.4395 =NORMSINV(1-H55/H56) 2. Using Excel, find the t-score that corresponds to the following Confidence Levels and Sample Sizes: a. 95% with n = 25 Confidence level= 95% Or Significance level = 0.05 n= 25 degrees of freedom=n-1= 24 =25- 1 No of tails= 2 t value= 2.06 =TINV(E80,E82) (corresponding to 24 degrees of freedom and 0.05 significance level ) b. 96% with n = 15 Confidence level= 96% Or Significance level = 0.04 n= 15 degrees of freedom=n-1= 14 =15- 1 No of tails= 2 t value= 2.26 =TINV(E91,E93) (corresponding to 14 degrees of freedom and 0.04 significance level ) c. 97% with n = 21 Confidence level= Or Significance level = 97% 0.03 n= 21 degrees of freedom=n-1= 20 =21- 1 No of tails= 2 t value= 2.34 =TINV(E103,E105) (corresponding to 20 degrees of freedom and 0.03 significance level ) d. 91% with n = 10 Confidence level= 91% Or Significance level = 0.09 n= 10 degrees of freedom=n-1= 9 =10- 1 No of tails= 2 t value= 1.9 =TINV(E114,E116) (corresponding to 9 degrees of freedom and 0.09 significance level ) 3. Suppose we wish to estimate the population mean using a confidence interval. When is it appropriate to use score? When is it appropriate to use a t-score? When the Population standard deviation is known Sample size is larger than 30 When the Population standard deviation is not known Normal distribution, (z Normal distribution, (z table) table) Sample size is 30 or Normal distribution, (z t distribution, (t table) less and we assume table) that the population is normal or approximately so core that corresponds to the : Confidence Level 90% 95% 98% 99% Alpha Area to Left =(1-N2)/2 =1-O2 =(1-N3)/2 =1-O3 =(1-N4)/2 =1-O4 =(1-N5)/2 =1-O5 ore that corresponds to the and Sample Sizes: mate the population mean When is it appropriate to use priate to use a t-score? Confidence Level 90% 95% 98% 99% Area in Two Tails =1-N21 =1-N22 =1-N23 =1-N24 Sample Size 25 14 12 27 : =NORMSINV(1-H48/H49) =NORMSINV(1-H55/H56) =NORMSINV(1-H62/H63) =NORMSINV(1-H70/H71) and Sample Sizes: hen is it appropriate to use a z- Z-Score =NORMINV(P2,0,1) =NORMINV(P3,0,1) =NORMINV(P4,0,1) =NORMINV(P5,0,1) Z-Score (Rounded) =ROUND(Q2,3) =ROUND(Q3,3) =ROUND(Q4,3) =ROUND(Q5,3) t-Score =TINV(O21,P21-1) =TINV(O22,P22-1) =TINV(O23,P23-1) =TINV(O24,P24-1) t-Score (Rounded) =ROUND(Q21,3) =ROUND(Q22,3) =ROUND(Q23,3) =ROUND(Q24,3) Dollars Spent 75 74 80 68 79 85 77 82 79 67 90 72 76 75 69 85 78 79 82 66 75 85 90 76 85 67 89 82 69 79 82 80 84 79 78 81 77 84 80 76 Bob loves making candy, especially varieties of caramel, including plain, chocolate dipped dipped caramels with pecans. Bob has received lots of compliments from his friends and encouraged him to start his own candy making business. After several days of research, Bob finds that the national average amount of money spen type of specialty candy is $75. Bob believes that the citizens in his area spend more than whether or not this is true could help Bob make a wise decision regarding his future busine Bob wants to use statistics to support his claim, and to help him obtain a small business lo estimate of the true amount of money local citizens do spend on this type of specialty cand Bob randomly selects several people from his local phone book and asks the person that a they typically spend per year on candy like he will make. He obtains the following results ( 85, 77, 82, 79, 67, 90, 72, 76, 75, 69, 85, 78, 79, 82, 66, 75, 85, 90, 76, 85, 67, 89, 82, 69, 84, 80, 76. Based upon these results, Bob is hoping his area has a good customer base for his new b bank is impressed with his use of statistics and will grant him the loan he needs to start it! Questions: 1. Find the sample mean and sample standard deviation of the amount citizens spend per 2. When finding a confidence interval for the true mean spent of ALL citizens, should we u a t-score? Why? 3. Find the z/t-values (as appropriate) for a 95% confidence interval and a 92% confidenc 4. Find a 95% and a 92% confidence interval for the true mean amount that citizens spen 5. What do you think the lowest possible mean amount spent per year is? Why? 6. Do you think Bob has a good customer base for his new business? Explain. Questions: 1. Find the sample mean and sample standard deviation of the amount citizens spend per year. Sample mean= 78.40 =AVERAGE(A2:A41) Sample standard deviation= 6.2050 =STDEV(A2:A41) 2. When finding a confidence interval for the true mean spent of ALL citizens, should we use a z-score or a t score? Why? Since sample size= 40 which is greater than 30; we will use z distribution 3. Find the z/t-values (as appropriate) for a 95% confidence interval and a 92% confidence interval. Confidence level= or Significance level = No of tails= Z value corresponding to 0.05 significance level and 2 tails= Confidence level= or Significance level = No of tails= Z value corresponding to 0.08 significance level and 2 tails= 95% 5.00% 2 1.9600 =NORMSINV(1-H59/H60) 92% 8.00% 2 1.7507 =NORMSINV(1-H64/H65) 4. Find a 95% and a 92% confidence interval for the true mean amount that citizens spend per year. 95% Confidence limits $78.40 Mean== Standard deviation == 6.2050 40 sample size=n= x=standard error of mean=/n= 0.98 = ( 6.205 / 40) 95% Confidence level= 5% =100% -95% Therefore Significance level== No of tails= 2 This is a 2 tailed test because we are calculating the confidence interval Z at the 0.05 level of significance 2 tailed test = Upper confidence limit= Lower confidence limit= +z*x= -z*x= 1.96 80.32 =78.4+1.96*0.9811 76.477 =78.4-1.96*0.9811 95% Confidence limit: (rounding off the values) Upper limit= $80.32 $76.48 Lower limit= 92% Confidence limits $78.40 Mean== Standard deviation == 6.2050 40 sample size=n= x=standard error of mean=/n= 0.98 = ( 6.205 / 40) 92% Confidence level= 8% =100% -92% Therefore Significance level== No of tails= 2 This is a 2 tailed test because we are calculating the confidence interval Z at the 0.08 level of significance 2 tailed test = Upper confidence limit= Lower confidence limit= +z*x= -z*x= 1.75 80.12 =78.4+1.7507*0.9811 76.682 =78.4-1.7507*0.9811 92% Confidence limit: (rounding off the values) Upper limit= $80.12 $76.68 Lower limit= 5. What do you think the lowest possible mean amount spent per year is? Why? The lowest possible mean amount would be more than $75 6. Do you think Bob has a good customer base for his new business? Explain. More research needs to be done before it can be determined that Bob has a good customer base Compliments from friends and neighbors is not enough to establish a good customer base luding plain, chocolate dipped caramels and chocolate pliments from his friends and neighbors, and several have erage amount of money spent annually per person on this ns in his area spend more than that per year. Knowing ion regarding his future business plans. him obtain a small business loan. Bob also wants to find an d on this type of specialty candy. ook and asks the person that answers how much money obtains the following results (in dollars): 75, 74, 80, 68, 79, 85, 90, 76, 85, 67, 89, 82, 69, 79, 82, 80, 84, 79, 78, 81, 77, d customer base for his new business. Bob also hopes the the loan he needs to start it! the amount citizens spend per year. nt of ALL citizens, should we use a z-score or interval and a 92% confidence interval. ean amount that citizens spend per year. nt per year is? Why? business? Explain. s spend per year. should we use a z-score or a t- confidence interval. =NORMSINV(1-H59/H60) =NORMSINV(1-H64/H65) tizens spend per year. y? . ood customer base tomer base ... View Full Document