Course Hero has millions of student submitted documents similar to the one

below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

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

**Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.**

Below is a small sample set of documents:

San Jose State - EE - 271

SAN JOSE STATE UNIVERSITY Charles W. Davidson College of EngineeringDEPARTMENT OF ELECTRICAL ENGINEERING EE271 Homework #1 1. Find ( a b ) + ( a ' b ) + ( a b ' ) + ( a ' b ' ) 2. Show that a = b if and only if (a.b) + (a.b) = 0 3. Find the complement of

San Jose State - EE - 271

SAN JOSE STATE UNIVERSITY Charles W. Davidson College of EngineeringDEPARTMENT OF ELECTRICAL ENGINEERING EE271 Homework #2 1. Write the logic tables and canonical Boolean expressions of the RS-, D-, T-, and JK-flipflops and then minimize these canonical

San Jose State - EE - 271

SAN JOSE STATE UNIVERSITY Charles W. Davidson College of EngineeringDEPARTMENT OF ELECTRICAL ENGINEERINGEE271Homework #3 Solution Problem 1 module tr_latch (q_out, enable, data); output q_out; input enable, data; reg q_out; always @ (enable or data) be

San Jose State - EE - 271

SAN JOSE STATE UNIVERSITY Charles W. Davidson College of EngineeringDEPARTMENT OF ELECTRICAL ENGINEERINGEE271Homework #4 1. For the circuit Y = ( A + D ) ( DBC ) a. Write a gate-level Verilog code and its testbench. b. Write a Verilog code with user de

San Jose State - EE - 271

SAN JOSE STATE UNIVERSITY Charles W. Davidson College of EngineeringDEPARTMENT OF ELECTRICAL ENGINEERINGEE271Homework #5 1. A circuit represented by Boolean equation below: (A AND B) OR (A AND C) OR (B AND C) a. If all gates are 2-input gates: show the

San Jose State - EE - 271

SAN JOSE STATE UNIVERSITY Charles W. Davidson College of EngineeringDEPARTMENT OF ELECTRICAL ENGINEERINGEE271Homework #6 Create a directory named HW6 and a subdirectory src on your UNIX account. Write a Verilog module for a16-bit ripple carry adder (co

San Jose State - EE - 271

SAN JOSE STATE UNIVERSITY Charles W. Davidson College of EngineeringDEPARTMENT OF ELECTRICAL ENGINEERINGEE271Homework #7 1. An addition circuit running at 100MHz with a supply voltage of 3V. Assume that on the average each node change states (from 0 to

San Jose State - EE - 271

SAN JOSE STATE UNIVERSITY Charles W. Davidson College of EngineeringDEPARTMENT OF ELECTRICAL ENGINEERINGEE271Homework #8 1. Derive the formula of mapping explicit values to an implicit value of an integer in one's complementary system: X i = 2n 1 + X e

San Jose State - EE - 271

SAN JOSE STATE UNIVERSITY Charles W. Davidson College of EngineeringDEPARTMENT OF ELECTRICAL ENGINEERINGEE271Homework #9 1. For the 3 memory module below (may not work because they are not yet tested) Develop test benches to test them and correct the m

San Jose State - EE - 271

San Jose State - EE - 271

SOLUTION FOR QUESTION 1 OF HW #9 1. 16 x 8-bit Asynchronous SRAM `define DELAY 1 module Ram (data, address, memw, memr, cs); parameter width = 8, nbytes = 16, addr = 4; input memw, memr, cs; input [addr-1:0] address; inout [width-1:0] data; reg [width-1:0

San Jose State - EE - 210

SOLUTION 001Write analytical expression for the continuous-time waveform f ( t ) , shown in below figure.f (t ) 3-3SOLUTION 1:-157tP = ( x1 = -3, y1 = 0 ) P2 = ( x2 = -1, y1 = 3) 1f (t ) - 0 = 3-0 3 ( t - (-3) ) = ( t + 3) u ( t + 3) 2 -1 - ( -3

San Jose State - EE - 210

PROBLEM 001PROBLEM 1: Write analytical expression for the continuous-time waveform f ( t ) , shown in below figure.f (t ) 3-3-157t1PROBLEM 002PROBLEM 2: Write analytical expression for the continuous-time waveform f ( t ) , shown in below figure

San Jose State - EE - 210

EE 210Problem 1:CT Convolution Sample ProblemSpring 2004Consider the following two continuous time (CT) signals x1 ( t ) and y1 ( t )x1 ( t )2 1 -1 0 1 2 3 4 5 6 7 2ty1 ( t )2 1 -1 0 1 2 3 1 24567tThe signal v1 ( t ) is defined as the convo

San Jose State - EE - 270

EE270 Spring 2008 Homework #1 1) Prove the following relation using only fundamentals of Boolean a. (x+y) (x+z) = x'. ( y z)b. (ab)' = a'+b' 2) Express in DCF and CCF a. ab + a'c + bd b. (x+y+z)(x'+y') 3) Show that the following function is self dual a.

San Jose State - EE - 270

HW#2EE270 Spring 20081. Determine minimum SOP form of the following: a) F(A,B,C,D) = m (4,7,9,15) + d (1,2,3,6) b) F(A,B,C,D) = F(A,B,C,D) = m (0,2,3,4,5) + d (8,9,10,11)2. Determine POS form for the following: a) F(A,B,C,D) = M( 4,7,9,11,12). D(0,1,2,

San Jose State - EE - 270

Homework4 1. DesignafundamentalmodecircuitasdescribedusingNORgatesonly a. Itisatwoinput,oneoutputdevice(x1,x2,z) b. Whenx2=1,z=0 c. Thefirstchangex1from0to1whilex2=0causetheoutputztobe1 d. Theoutputwouldremain1untilx2goesto1,forcingzto0 2. Obtainaprimitiv

Berkeley - CS - 61B

01/21/09 18:40:54CS 61B: Lecture 1 Wednesday, January 21, 2009 Prof. Jonathan Shewchuk, jrs@cory.eecs Email to prof & all TAs at once (preferred): cs61b@cory.eecs Todays reading: Sierra & Bates, pp. 1-9, 18-19, 84. Handout: Course Overview (also availabl

Berkeley - CS - 61B

01/23/09 04:41:09CS 61B: Lecture 2 Friday, January 23, 2009 Todays reading: Sierra & Bates, Chapter 2; pp. 54-58, 154-160, 661, 669.02METHODS = Lets look at some methods that arent constructors. s2 = s1.toUppercase(); -s2 |.+->| YOW! | -String s3 = s2.

Berkeley - CS - 61B

01/26/09 00:58:44CS 61B: Lecture 3 Monday, January 26, 2009 Todays reading: Sierra & Bates, pp. 71-74, 76, 85, 240-249, 273-281, 308-309.031DEFINING CLASSES = An object is a repository of data. _Fields_ are variables that hold the data stored in objec

Berkeley - CS - 61B

01/28/09 03:52:43CS 61B: Lecture 4 Wednesday, January 28, 2009 Todays reading: S&B pp. 10-14, 49-53, 75, 78-79, 86, 117, 286-287, 292, 660.041a | b | a & b | a | b | !a =|= false | false | false | false | true false | true | false | true | true | fals

Berkeley - CS - 61B

01/30/09 03:02:55CS 61B: Lecture 5 Friday, January 30, 2009 Todays reading: LOOPS = "while" Loops -A "while" statement is like an "if" statement, but the body of the statement is executed repeatedly, as long as the condition remains true. The following e

Berkeley - CS - 61B

02/01/09 05:34:22CS 61B: Lecture 6 Monday, February 2, 2009 Todays reading: MORE ARRAYS = Automatic Array Construction -Last lecture, we used a loop to construct all the arrays that the top-level array references. This was necessary to construct a triang

Berkeley - CS - 61B

02/05/09 02:38:20CS 61B: Lecture 7 Wednesday, February 4, 2009 Todays reading: Goodrich & Tamassia, Section 3.2.071public class ListNode cfw_ public int item; public ListNode next; / ListNode is a recursive type / Here were using ListNode before / we

Berkeley - CS - 61B

02/01/09 05:51:29CS 61B: Lecture 8 Friday, February 6, 2009 Todays reading: Goodrich & Tamassia, Section 3.3.081THE "public" AND "private" KEYWORDS = Thus far, weve usually declared fields and methods using the "public" keyword. However, we can also d

Berkeley - CS - 61B

04/24/09 20:07:04CS 61B: Lecture 9 Monday, February 9, 2009 Todays reading: Sierra & Bates pp. 77, 235-239, 258-265, 663.091THE STACK AND THE HEAP = Java stores stuff in two separate pools of memory:Parameter Passing -As in Scheme, Java passes all pa

Berkeley - CS - 61B

02/11/09 23:25:02CS 61B: Lecture 10 Wednesday, February 11, 2009 Todays reading: Sierra & Bates, pp. 95-109, 662.101Modular testing: testing each method and each class separately. Integration testing: testing a set of methods/classes together. Result

Berkeley - CS - 61B

02/16/09 18:36:38CS 61B: Lecture 11 Friday, February 13, 2009 Todays reading: All of Chapter 7, plus pp. 28-33, 250-257.111public TailList(int x) cfw_ super(x); tail = null; The zero-parameter SList() constructor is always called by default, regardle

Berkeley - CS - 61B

02/16/09 18:08:41CS 61B: Lecture 12 Wednesday, February 18, 2009 Todays reading: Sierra & Bates, Chapter 8.121public void listSort(List l) cfw_ . ABSTRACT CLASSES = An abstract class is a class whose sole purpose is to be extended. public abstract cl

Berkeley - CS - 61B

02/19/09 20:33:40CS 61B: Lecture 13 Friday, February 20, 2009 Todays reading: Sierra & Bates, pp. 154-160, 587-591, 667-668.131| | | | | | | | /* list/SListNode.java */ package list; class SListNode cfw_ Object item; SListNode next; JAVA PACKAGES = I

Berkeley - CS - 61B

02/25/09 16:25:41CS 61B: Lecture 15 Wednesday, February 25, 2009 Todays reading: Sierra & Bates, pp. 315-338.151EXCEPTIONS = When a run-time error occurs in Java, the JVM "throws an exception," and the result is an error message. Oddly, an exception i

Berkeley - CS - 61B

02/28/09 19:50:15CS 61B: Lecture 16 Friday, February 27, 2009 Todays reading: Sierra & Bates, pp. 189, 283.161EXCEPTIONS (continued) = The "finally" keyword -A finally clause can also be added to a "try." try cfw_ statementX; return 1; catch (SomeExc

Berkeley - CS - 61B

CS 61B: Lecture 17 Monday, March 2, 2009 Game tree search How could we design a program that plays Tic Tac Toe? The standard technique searches for the best moves by using a game tree, which looks much like a family tree. A game tree is not a data structu

Berkeley - CS - 61B

03/04/09 20:28:41CS 61B: Lecture 18 Wednesday, March 4, 2009 Todays reading: Sierra & Bates, pp. 80-84.181ENCAPSULATION = A _module_ is a set of methods that work together as a whole to perform some task or set of related tasks. A module is _encapsula

Berkeley - CS - 61B

03/06/09 19:03:10CS 61B: Lecture 19 Friday, March 6, 2009 Todays reading: Sierra & Bates, p. 664.19(3) What happens if we invoke l.remove(n), then l.insertAfter(i, n)?1Another way to trash the DList invariants is to treat a node thats been removed fr

Berkeley - CS - 61B

03/03/09 14:57:33CS 61B: Lecture 20 Monday, March 9, 2009 Todays reading: Goodrich & Tamassia, Chapter 4 (especially 4.2 and 4.3).20Next, you must learn how to express this idea rigorously. Here is the central lesson of todays lecture, which will bear

Berkeley - CS - 61B

03/03/09 14:57:33CS 61B: Lecture 21 Wednesday, March 11, 2009 ASYMPTOTIC ANALYSIS (continued): More Formalism = |-| | Omega(f(n) is the set of all functions T(n) that satisfy: | | | | There exist positive constants d and N such that, for all n >= N, | |

Berkeley - CS - 61B

02/25/09 22:50:17CS 61B: Lecture 22 Friday, March 13, 2009 Todays reading: Goodrich & Tamassia, Chapter 5.221BASIC DATA STRUCTURES = Stacks -A _stack_ is a crippled list. You may manipulate only the item at the top of the stack. The main operations: y

Berkeley - CS - 61B

03/16/09 22:42:55CS 61B: Lecture 23 Monday, March 16, 2009 Todays reading: Goodrich & Tamassia, Sections 9.1-9.3.231WARNING: When an object is stored in a hash table, an application should never change the object in a way that will change its hash cod

Berkeley - CS - 61B

03/18/09 16:22:59CS 61B: Lecture 24 Wednesday, March 18, 2009 Todays reading: Goodrich & Tamassia, Chapter 7.241ROOTED TREES = A _tree_ consists of a set of nodes and a set of edges that connect pairs of nodes. A tree has the property that there is ex

Berkeley - CS - 61B

03/18/09 21:12:44CS 61B: Lecture 25 Friday, March 20, 2009 Todays reading: Goodrich & Tamassia, Sections 8.1-8.3.251The entries in a heap satisfy the _heap-order_property_: no child has a key less than its parents key. Observe that any subtree of a bi

Berkeley - CS - 61B

03/27/09 18:38:38CS 61B: Lecture 26 Monday, March 30, 2009 Todays reading: Goodrich & Tamassia, Section 10.1.261Representing Binary Trees -Recall that a binary tree is a rooted tree wherein no node has more than two children. Additionally, every child

Berkeley - CS - 61B

03/26/09 22:43:22CS 61B: Lecture 27 Wednesday, April 1, 2009 2-3-4 TREES = A 2-3-4 tree is a perfectly balanced tree. It has a big advantage over regular binary search trees: because the tree is perfectly balanced, find, insert, and remove operations tak

Berkeley - CS - 61B

04/05/09 01:54:57CS 61B: Lecture 28 Friday, April 3, 2009 GRAPHS = A graph G is a set V of vertices (sometimes called nodes), and a set E of edges (sometimes called arcs) that each connect two vertices together. To confuse you, mathematicians often use t

Berkeley - CS - 61B

04/05/09 22:43:28CS 61B: Lecture 29 Monday, April 6, 2009 GRAPHS (continued) = Breadth-first search (BFS) is a little more complicated than depth-first search, because its not naturally recursive. We use a queue so that vertices are visited in order acco

Berkeley - CS - 61B

04/08/09 03:19:43CS 61B: Lecture 30 Wednesday, April 8, 2009 SORTING = The need to sort numbers, strings, and other records arises frequently. The entries in any modern phone book were sorted by a computer. Databases have features that sort the records r

Berkeley - CS - 61B

04/06/09 00:00:57CS61B: Lecture 31 Friday, April 10, 2009 QUICKSORT = Quicksort is a recursive divide-and-conquer algorithm, like mergesort. Quicksort is in practice the fastest known comparison-based sort for arrays, even though it has a Theta(n^2) wors

Berkeley - CS - 61B

04/15/09 00:32:22CS61B: Lecture 33 Wednesday, April 15, 2009 Todays reading: Goodrich & Tamassia, Section 11.6.33List-Based Disjoint Sets and the Quick-Find Algorithm -The obvious data structure for disjoint sets looks like this. - Each set references

Berkeley - CS - 61B

04/15/09 20:34:08CS61B: Lecture 34 Friday, April 17, 2009 Todays reading: Goodrich & Tamassia, Sections 11.3 & 11.7.34A LOWER BOUND ON COMPARISON-BASED SORTING = Suppose we have a scrambled array of n numbers, with each number from 1.n occurring once.

Berkeley - CS - 61B

04/20/09 20:17:40CS 61B: Lecture 36 Monday, April 20, 2009 Counting Sort -If the items we sort are naked keys, with no associated values, bucket sort can be simplified to become _counting_sort_. In counting sort, we dont need queues at all; we need merel

Berkeley - CS - 61B

04/24/09 16:48:00CS61B: Lecture 37 Friday, April 24, 2009 SPLAY TREES = A splay tree is a type of balanced binary search tree. Structurally, it is identical to an ordinary binary search tree; the only difference is in the algorithms for finding, insertin

Berkeley - CS - 61B

04/27/09 18:30:17CS61B: Lecture 38 Monday, April 27, 2009 AMORTIZED ANALYSIS = Weve seen several data structures for which I claimed that the average time for certain operations is always better than the worst-case time. These data structures include has

Berkeley - CS - 61B

04/29/09 23:28:06CS61B: Lecture 39 Wednesday, April 29, 2009 RANDOMIZED ANALYSIS = Randomized analysis, like amortized analysis, is a mathematically rigorous way of saying, "The worst-case running time of this operation is slow, but nobody cares, because

Berkeley - CS - 61B

04/30/09 00:05:23CS61B: Lecture 40 Friday, May 1, 2009 Expression Parsing (an application of stacks) -Arithmetic expressions can be written in prefix, infix, or postfix. Infix is the usual way of writing expressions that were all familiar with: 3 + 4 * 7

Berkeley - CS - 61B

05/04/09 03:52:42CS61B: Lecture 41 Monday, May 4, 2009 GARBAGE COLLECTION = Objects take up space in memory. If your program creates lots of objects, throws them away, and creates more, you might eventually run out of memory. To reduce the chance that th

Berkeley - CS - 61B

05/06/09 18:40:24CS61B: Lecture 42 Wednesday, May 6, 2009 Generational Garbage Collection -Studies of memory allocation have shown that most objects allocated by most programs have short lifetimes, while a few go on to survive through many garbage collec

Berkeley - CS - 61B

01/23/09 04:29:03CS 61B Homework 1 Due 5pm Wednesday, January 28, 2009 This homework assignment is meant to make sure you can write, compile, and run simple Java programs. This is an individual assignment; you may not share code with other students. You

Berkeley - CS - 61B

01/29/09 03:22:24CS 61B Homework 2 Due 5pm Wednesday, February 4, 2009 This homework assignment is designed to help you learn about building Java classes and to observe the decomposition of a complicated task into simple subtasks. This is an individual a

Berkeley - CS - 61B

02/07/09 19:53:10CS 61B Homework 3 Due 5pm Wednesday, February 11, 2006 This homework assignment is designed to give you practice working with arrays, linked lists, and nested loops. It will also give you practice for the similar but harder run-length en

Al Akhawayn University - FINANCE - fin 3301

FIN 3301 Homework # 1 Due: Monday, February 22Name_Show all your work. Time lines, formulas, and inputs to your financial calculator as needed. Problem # 1 Suppose a government bond will pay $1,000 three years from now. If the going interest rate on 3-y

Al Akhawayn University - SBA - fin3301

CHAPTER 5Time Value of MoneyTopics Covered Time lines Future Values and Compound Interest Present Values Multiple Cash Flows Level Cash Flows Perpetuities and Annuities Effective Annual Interest Rates Inflation & Time ValueTime lines0I%1 CF12 CF2