Today we had our introduction to confidence intervals, one of the key inferential tools we will use for the remainder of the year--I love this stuff!
Tonight, practice your interpretations of confidence intervals with the following:
Page 446-448: 5, 9, 21b, 23c
- For 5, determine which statement is correct (if any)! Explain why any incorrect statements are incorrect. (Unfortunately all 5 are incorrect....)
- We will have a stamp like this tomorrow!
- For 21 90% confidence interval is provided (with an inequality)--use your writing template/notes to interpret this interval!
- For 21b and 23c the interval is given below--your job is to interpret the interval:
- 21b): 98% Confidence Interval = (0.18209, 0.21791)
- I used the "One Pop Z Int" feature on my calculator to create this 98% confidence interval....can you do the same?
- What is the value of n? x (x is not given, you must calculate it)? Enter these and your confidence level, then press calculate!
- 23c): 95% Confidence Interval = (0.05147, 0.12552)
- I used the "One Pop Z Int" feature on my calculator to create this 98% confidence interval....can you do the same?
- What is the value of n? x (for this context x is given)? Enter these and your confidence level, then press calculate!
Tomorrow in class we'll discuss interpreting intervals some more, then get into the conditions; finally, we'll move into an exploration of the mathematics behind how a confidence interval is calculated for the end of class tomorrow and into Wednesday. I can't wait!
Here are the textbook problems (answers below):
5.) A catalog sales company promises to deliver orders placed on the Internet within 3 days. Follow-up calls to a few randomly selected customers show that a 95% confidence interval for the proportion of all orders that arrive on time is 88% +/- 6%. What does this mean? Are these conclusions correct? Explain:
a.) Between 82% and 94% of all orders arrive on time.
b.) 95% of all random samples of customers will show that 88% of orders arrive on time.
c.) 95% of all random samples of customers will show that 82% to +5% of orders arrive on time.
d.) We are 95% sure that between 82% and 94% of the orders placed by the customers in this sample arrived on time.
e.) On 95% of the days, between 82% and 94% of the orders will arrive on time.
9.) What fraction of cars are made in Japan? The computer output below summarizes the results of a random sample of 50 autos. Explain carefully what it tells you. (Interpret the interval!)
z-interval for proportion
With 90.00% confidence
0.29938661 < p(japan) < 0.46984416
21b.) Vitamin D, whether ingested as a dietary supplement or produced naturally when sunlight falls upon the skin, is essential for strong, healthy bones. The bone disease rickets was largely eliminated in England during the 1950's, but now there is concern that a generation of children more likely to watch TV or play computer games than spend time outdoors is at increased risk. A recent study of 2700 children randomly selected from all parts of England found 20% of them deficient in Vitamin D.
b.) Explain carefully what your interval means (in context).
98% Confidence Interval = (0.18209, 0.21791)
23c.) In a random survey of 226 college students 20 reported being "only" children (with no siblings). Estimate the proportion of students nationwide who are only children.
c.) Interpret your interval.
95% Confidence Interval = (0.05147, 0.12552)
And here are the textbook problem answers:
5.)
a.) Between 82% and 94% of all orders arrive on time. This statement does not include a level of confidence, and implies that this is definitely true. A confidence interval is not "definitely true" and its interpretation should reference a level of confidence.
b.) 95% of all random samples of customers will show that 88% of orders arrive on time. The confidence level does not tell us what percent of samples will give a certain sample statistic (p-hat). This statement is false.
c.) 95% of all random samples of customers will show that 82% to +5% of orders arrive on time. The confidence level does not tell us what percent of samples will result in this confidence interval. This statement is false.
d.) We are 95% sure that between 82% and 94% of the orders placed by the customers in this sample arrived on time. The problem here is the wording "in this sample." A confidence interval is used to estimate the population percentage, not a sample percent. (We don't need an interval to estimate the sample %, we know it! We know p-hat = 0.88).
e.) On 95% of the days, between 82% and 94% of the orders will arrive on time. This statement is false. 95% is our level of confidence, not the % of days...
9.) What fraction of cars are made in Japan? The computer output below summarizes the results of a random sample of 50 autos. Explain carefully what it tells you. (Interpret the interval!)
z-interval for proportion
With 90.00% confidence
0.29938661 < p(japan) < 0.46984416
We are 90% confident that the true proportion of cars made in Japan falls between 29.938661% and 46.984416% based on this sample of 50 autos.
21b.) Vitamin D, whether ingested as a dietary supplement or produced naturally when sunlight falls upon the skin, is essential for strong, healthy bones. The bone disease rickets was largely eliminated in England during the 1950's, but now there is concern that a generation of children more likely to watch TV or play computer games than spend time outdoors is at increased risk. A recent study of 2700 children randomly selected from all parts of England found 20% of them deficient in Vitamin D.
b.) Explain carefully what your interval means (in context).
98% Confidence Interval = (0.18209, 0.21791)
We are 98% confident that the true proportion of children in England who are deficient in Vitamin D falls between 18.209% and 21.791% based on this sample of 2700 children.
23c.) In a random survey of 226 college students 20 reported being "only" children (with no siblings). Estimate the proportion of students nationwide who are only children.
c.) Interpret your interval. 95% Confidence Interval = (0.05147, 0.12552)
We are 95% confident that the true proportion of students nationwide who are only children falls between 5.147% and 12.552% based on this sample of 226 college students.
