Assignment:
Q1. 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
Q2. 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
Q3. 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
Q4. 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
| Gestation Period |
Mean Birth Weight |
Standard Deviation |
| Under 28 Weeks |
1.88 |
1.19 |
| 28 to 31 Weeks |
4.07 |
1.87 |
| 32 to 35 Weeks |
5.73 |
1.48 |
| 36 Weeks |
6.46 |
1.2 |
| 37 to 39 Weeks |
7.33 |
1.09 |
| 40 Weeks |
7.72 |
1.05 |
| 41 Weeks |
7.83 |
1.08 |
| 42 Weeks and over |
7.65 |
1.12 |
Q1. 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.
Q2. 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?
Q3. Are the ages of people in the United States normally distributed? Explain your reasoning.
Q4. 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?
Q5. 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?
| Age |
Midpoint |
Relative Frequency |
mid*relfreq |
mid^2*relfreq |
| 0 - 4 |
2 |
6.7% |
0.134 |
0.268 |
| 5 - 9 |
7 |
6.8% |
0.476 |
3.332 |
| 10 - 14 |
12 |
7.4% |
0.888 |
10.656 |
| 15 - 19 |
17 |
7.2% |
1.224 |
20.808 |
| 20 - 24 |
22 |
7.0% |
1.54 |
33.88 |
| 25 - 29 |
27 |
6.2% |
1.674 |
45.198 |
| 30 - 34 |
32 |
6.8% |
2.176 |
69.632 |
| 35 - 39 |
37 |
7.3% |
2.701 |
99.937 |
| 40 - 44 |
42 |
8.1% |
3.402 |
142.884 |
| 45 - 49 |
47 |
7.6% |
3.572 |
167.884 |
| 50 - 54 |
52 |
6.6% |
3.432 |
178.464 |
| 55 - 59 |
57 |
5.5% |
3.135 |
178.695 |
| 60 - 64 |
62 |
4.2% |
2.604 |
161.448 |
| 65 - 69 |
67 |
3.4% |
2.278 |
152.626 |
| 70 - 74 |
72 |
3.0% |
2.16 |
155.52 |
| 75 - 79 |
77 |
2.6% |
2.002 |
154.154 |
| 80 - 84 |
82 |
1.9% |
1.558 |
127.756 |
| 85 - 89 |
87 |
1.0% |
0.87 |
75.69 |
| 90 - 94 |
92 |
0.5% |
0.46 |
42.32 |
| 95 - 99 |
97 |
0.2% |
0.194 |
18.818 |
|
|
|
|
|
|
Population Mean = |
36.48 |
|
|
Population Variance = |
509.18 |
|
Population Standard Deviation = |
22.57 |
| 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 |
| Confidence Level |
Alpha |
Area to Left |
Z-Score |
Z-Score (Rounded) |
| 90% |
0.100 |
0.95 |
1.64485 |
1.645 |
| 95% |
0.050 |
0.975 |
1.95996 |
1.96 |
| 98% |
0.020 |
0.99 |
2.32635 |
2.326 |
| 99% |
0.010 |
0.995 |
2.57583 |
2.576 |
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.
| Confidence Level |
Area in Two Tails |
Sample Size |
t-Score |
t-Score (Rounded) |
| 90% |
0.100 |
25 |
1.71088 |
1.711 |
| 95% |
0.050 |
14 |
2.16037 |
2.16 |
| 98% |
0.020 |
12 |
2.71808 |
2.718 |
| 99% |
0.010 |
27 |
2.77871 |
2.779 |
|
|
|
|
|
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 t-distribution. 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 - 1Questions:
Q1. Using Excel, find the z-score that corresponds to the following Confidence Levels:
a. 80%
b. 85%
c. 92%
d. 97%
Q2. Using Excel, find the t-score that corresponds to the following Confidence Levels and Sample Sizes:
a. 95% with n = 25
b. 96% with n = 15
c. 97% with n = 21
d. 91% with n = 10
Q3. Suppose we wish to estimate the population mean using a confidence interval. When is it appropriate to use a z-score? When is it appropriate to use a t-score?
| Confidence Level |
Alpha |
Area to Left |
Z-Score |
Z-Score (Rounded) |
| 90% |
=(1-N2)/2 |
=1-O2 |
=NORMINV(P2,0,1) |
=ROUND(Q2,3) |
| 95% |
=(1-N3)/2 |
=1-O3 |
=NORMINV(P3,0,1) |
=ROUND(Q3,3) |
| 98% |
=(1-N4)/2 |
=1-O4 |
=NORMINV(P4,0,1) |
=ROUND(Q4,3) |
| 99% |
=(1-N5)/2 |
=1-O5 |
=NORMINV(P5,0,1) |
=ROUND(Q5,3) |
| Confidence Level |
Area in Two Tails |
Sample Size |
t-Score |
t-Score (Rounded) |
| 90% |
=1-N21 |
25 |
=TINV(O21,P21-1) |
=ROUND(Q21,3) |
| 95% |
=1-N22 |
14 |
=TINV(O22,P22-1) |
=ROUND(Q22,3) |
| 98% |
=1-N23 |
12 |
=TINV(O23,P23-1) |
=ROUND(Q23,3) |
| 99% |
=1-N24 |
27 |
=TINV(O24,P24-1) |
=ROUND(Q24,3) |
|
|
|
|
|
Q4. Bob loves making candy, especially varieties of caramel, including plain, chocolate dipped caramels and chocolate dipped caramels with pecans. Bob has received lots of compliments from his friends and neighbors, and several have encouraged him to start his own candy making business.
After several days of research, Bob finds that the national average amount of money spent annually per person on this type of specialty candy is $75. Bob believes that the citizens in his area spend more than that per year. Knowing whether or not this is true could help Bob make a wise decision regarding his future business plans.
Bob wants to use statistics to support his claim, and to help him obtain a small business loan. Bob also wants to find an estimate of the true amount of money local citizens do spend on this type of specialty candy.
Bob randomly selects several people from his local phone book and asks the person that answers how much money they typically spend per year on candy like he will make. He obtains the following results (in dollars): 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.
Based upon these results, Bob is hoping his area has a good customer base for his new business. Bob also hopes the 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 year.
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?
3. Find the z/t-values (as appropriate) for a 95% confidence interval and a 92% confidence interval.
4. Find a 95% and a 92% confidence interval for the true mean amount that citizens spend per year.
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.
| 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 |