1

Day #7: Concluding Confidence Intervals/ Chapter 8 Test Review

I can determine sample statistics from a confidence interval.

EXAMPLES: Confidence Intervals & Margin of Error: Meal Plan: After surveying students at Dartmouth College, a campus organization calculated that a 95% confidence interval for the mean cost of food for one term (of three in Dartmouth trimester calendar) is ($780, $920). Now the organization is trying to write its report, and considering the following interpretations. Determine which statements are true. Comment on each.

a. 95% of all students pay between $780 and $920 for food.

b. 95% of the sampled students paid between $780 and $920.

c. We’re 95% sure that students in this sample averaged between $780 and $920 for food.

d. 95% of samples of students will have average food costs between $780 and $920.

e. We’re 95% sure that the average amount all students pay is between $780 and $920.

2

Cattle: Livestock are given a special feed supplement to see if it will promote weight gain. The researchers report that the 77 cows studied gained an average of 56 pounds, and that a 95% confidence interval for the mean weight gain this supplement produces has a margin of error of 11 pounds. Some students wrote the following conclusions. Did anyone interpret the interval correctly? Explain any misrepresentations. a. 95% of the cows studied gained between 45 and 67 pounds.

b. We’re 95% sure that a cow fed this supplement will gain between 45 and 67 pounds.

c. We’re 95% sure that the average weight gain among the cows in this study was between 45 and 67 pounds.

d. The average weight gain of cows fed this supplement will be between 45 and 67 pounds 95% of the time.

e. If this supplement is tested on another sample of cows, there is a 95% chance that their average weight gain will be between 45 and 67 pounds.

3

Example: How much homework? The principal at a large high school claims that students spend at least 10 hours per week doing homework on average. To investigate this claim, an AP Statistics class selected a random sample of 250 students from their school and asked them how many long they spent doing homework during the last week. The sample mean was 10.2 hours and the sample standard deviation was 4.2 hours. Problem: (a) Construct and interpret a 95% confidence interval for the mean time spent doing homework in the last week for students at this school. (b) Based on your interval in part (a), what can you conclude about the principal’s claim?

4

5

“Does Snickers Really Satisfy?”: Suppose your class is investigating the weights of Snickers 1-ounce fun-size candy bars to see if customers are getting full value for their money. Assume that the weights are normally distributed. Several candy bars are randomly selected and weighed with sensitive balances borrowed from the physics lab. The weights are (in ounces):

.95 1.02 .98 .97 1.05 1.01 .98 1.00

a. What is n?

b. Calculate the sample mean.

c. Calculate the sample standard deviation and the standard error.

d. We want to determine a 90% confidence interval for the true mean, μ. Explain the meaning of the confidence level.

6

e. Determine t*.

f. Find the margin of error.

Using the Snickers data, we want to determine the 90% confidence interval for the mean weight of the candy bars.

7

Day #8: Chapter 8 Review

I can identify a point estimator to help estimate an unknown parameter.

I can correctly interpret the meaning of the margin of error in context.

I can interpret a confidence level in context. I can understand how confidence level or sample size will

affect the margin of error. I can understand that a confidence interval gives a range of

plausible values for the parameter. Consider the basic ‘formula’ for constructing a confidence interval:

Statistic ± (critical value) * (Standard Deviation)

8

I can determine the sample size required to obtain a level C

confidence interval for a population proportion with a specified margin of error.

I can determine the sample size required to obtain a level C confidence interval for a population mean with a specified margin of error.

Some scientists believe that a new drug would benefit about half of all people with a certain blood disorder. To estimate the proportion of patients who would benefit from taking the drug, the scientists will administer it to a random sample of patients who have the blood disorder. What sample size is needed so that the 95% confidence interval will have a margin of error of no more than 3%?

A. 748 B. 1068 C. 1503 D. 2056 E. 2401

9

To assess the accuracy of a laboratory scale, a standard weight that is known to weigh 1 gram is repeatedly weighed a total of n times and the mean ̅ of the weighing’s is computed. Suppose the scale readings are normally distributed with unknown mean μ and standard deviation σ = 0.01 g. How large should n be so

that a 95% confidence interval for μ has a margin of error of ± 0.0001?

A. 100 B. 196 C. 27,061 D. 10,000 E. 38,416

10

I can explain how practical issues like nonresponse, under coverage, and response bias can affect the interpretation of a confidence interval.

The survey in the previous question was conducted by calling land-line telephones, and those conducting the survey are concerned about the possibility of under coverage, since some people do not own a phone or own only a cell phone. Which of the following is the best way for them to correct for this source of bias?

A. Use a lower confidence level, such as 80%. B. Use a higher confidence level, such as 99%. C. Take a larger sample. D. Use a t-interval instead of a z-interval. E. Throw this sample out and start over again with a better sampling method.

I can determine critical values for calculating a confidence interval using a table or your calculator.

You want to calculate a 98% confidence interval for a population mean from a sample of n = 18. What is the appropriate critical t*?

A. 2.110 B. 2.326 C. 2.539 D. 2.552 E. 2.567

11

I can understand why each of the three inference conditions—Random, Normal, and Independent—is important.

A traffic consultant wants to estimate the proportion of cars on a certain street that have more than two occupants. She stands at the side of the road for two hours on a weekday afternoon and flips a coin each time a car approaches. If the coin comes up heads, she counts the number of occupants in the car. After two hours, she has counted 103 cars, 15 of which had more than two occupants. Which condition for constructing a confidence interval for a proportion has she failed to satisfy?

A. n ≥ 30 B. np ≥ 10 C. n(1 – p) ≥ 10 D. The sample is less than 10% of the population. E. The data is an SRS from the population of interest. The college newspaper of a large Midwestern university

periodically conducts a survey of students on campus to determine the attitude on campus concerning issues of interest. Pictures of the students interviewed along with quotes of their responses are printed in the paper. Students are interviewed by a reporter “roaming” the campus selecting students to interview “haphazardly.” On a particular day the reporter interviews five students and asks them if they feel there is adequate student parking on campus. Four of the students say, “no.” Which of the following conditions for inference about a proportion using a confidence interval are violated in this example?

12

I. The data are an SRS from the population of interest. II. The population is at least ten times as large as the sample. III. ̂ 10 and ̂ 10 . A. I only B. II only C. III only D. I and III E. All three conditions are violated I can understand how the margin of error of a confidence

interval changes with the sample size and the level of confidence C.

A nationwide poll of 2,525 adults estimated with 95% confidence that the proportion of Americans who support health care reform is 0.78 ± 0.0162. A member of Congress

thinks that 95% confidence isn't enough. He wants to be 99% confident. How would the margin of error of a 99% confidence interval based on the same sample compare with the 95% interval?

A. It would be smaller, because it omits only 1% of the possible samples instead of 5%. B. It would be the same, because the sample is the same. C. It would be larger, because higher confidence requires a larger margin of error. D. Can't tell, because the margin of error varies from sample to sample. E. Can't tell, because it depends on the size of the population.

13

I can construct and interpret a confidence interval for a population proportion.

I can construct and interpret a confidence interval for a population mean.

I can interpret a confidence interval in context. The heights (in inches) of males in the United States are

believed to be approximately normally distributed with mean µ. The mean height of a random sample of 25 American adult males is found to be ̅ inches and the standard deviation s = 4.15. What is the standard error of ̅?

A. 0.17 B. 0.69 C. 0.83 D. 1.856 E. 2.04 A random sample of 900 individuals has been selected from a

large population. It was found that 180 are regular users of vitamins. Thus, the proportion of the regular users of vitamins in the population is estimated to be 0.20. The standard error of this estimate is approximately

A. 0.1600 B. 0.0002 C. 0.4000 D. 0.0133 E. 0.0267

14

A polling organization announces that the proportion of American voters who favor congressional term limits is 64%, with a 95% confidence margin of error of 3%. Which of the following statements is a correct interpretation of 95% confidence? A. If the poll were conducted again in the same way, there is a

95% chance that the fraction of voters favoring term limits in the second poll would be between 61% and 67%.

B. There is a 95% probability that the true percent of voters favoring term limits is between 61% and 67%.

C. If the poll were conducted again the same way, there is a 95% probability that the percent of voters favoring term limits in the second poll would be within 3% of the percent favoring term limits in the first poll.

D. Among 95% of the voters, between 61% and 67% favor term limits.

E. None of the above.

15

‘t’ Table Practice #1. What critical value from Table C satisfies each of the following conditions?

a. The t distribution with 5 degrees of freedom has probability 0.05 to the right of .

b. The t distribution with 21 degrees of freedom has probability 0.99 to the left of .

c. The one-sample t statistic from an SRS of 20 observations has probability 0.75 to the left of .

d. The one-sample t statistic from a sample of 15 observations has probability 0.025 to the right of .

#2. Using the t tables, estimate:

a. The critical value of t for a 95% confidence interval with df = 7.

b. The critical value of t for a 99% confidence interval with df = 102.

16

#3. What critical value from Table C should be used for a confidence interval for the mean of the population in each of the following situations?

a. A 90% confidence interval based on n = 12 observations.

b. A 95% confidence interval from an SRS of 30 observations.

c. A 80% confidence interval form a sample of size 18.

17

Day #9: Test Review #1. Who Should Get Welfare?”: A news article on a Gallup Poll noted that “28 percent of the 1548 adults questioned felt that those who were able to work should be taken off welfare.” The article also said, “The margin of error for a sample size of 1548 is plus or minus three percentage points.” Opinion polls usually announce margins of error for 95% confidence. Using this fact, explain to someone who knows no statistics what “margin of error plus or minus three percentage points” means. (i.e.: Please correctly interpret this interval in the context of the problem).

18

#2. Doritos: Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. The company states that the population of bags is normally distributed. They carefully weighed the contents of each bag, recording the following weights (in grams):

29.2 28.5 28.7 28.9 29.1 29.5 a. Do these data satisfy the conditions/assumptions for

inference?

b. Find the mean and standard deviation of the observed weights.

c. Create a 95% confidence interval for the mean weight of such bags of chips.

d. Explain in context what your interval means (i.e.: interpret it).

e. Comment on the company’s stated net weight of 28.3 grams.

19

I can carry out the steps in constructing a confidence interval for a population proportion: define the parameter; check conditions; perform calculations; interpret results in context.

I can carry out the steps in constructing a confidence interval for a population mean: define the parameter; check conditions; perform calculations; interpret results in context.

I can determine sample statistics from a confidence interval.

FRAPPY #1: 2011 AP Statistics Exam (B) Question #5: During a flu vaccine shortage in the United States, it was believed that 45 percent of vaccine-eligible people received flu vaccine. The results of a survey given to a random sample of 2,350 vaccine-eligible people indicated that 978 of the 2,350 people had received flu vaccine. Construct and interpret a 99 percent confidence interval for the proportion of vaccine-eligible people who had received flu vaccine. Use your confidence interval to comment on the belief that 45 percent of the vaccine eligible people had received flu vaccine.

20

21

FRAPPY #2. 2002 AP Statistics Exam Question #6 A survey given to a random sample of students at a university included a question about which of two well-known comedy shows, Seinfeld or Friends, students preferred. The students were asked the question, “Do you prefer Seinfeld or Friends?” The responses are show below:

Preference Seinfeld Friends Total

185 139 324

Based on the results of this survey, construct and interpret an 80% confidence interval for the proportion of students in the population who would respond Seinfeld to the question “Do you prefer Seinfeld or Friends?” Be sure to: Check all conditions needed to make a correct inference Show how you would ‘construct’ the interval Correctly interpret the interval in the context of the problem

with the correct ‘tense’.

22

23

FRAPPY #3: Reading Scores: There are many ways to measure the reading ability of children. Research designed to improve reading performance is dependent on good measures of the outcome. One frequently used test is the DRP or Degree of Reading Power. A researcher suspects that the mean score µ of all third graders in Henrico County Schools is different from the national mean, which is 32. To test her suspicion, she administers the DRP to an SRS of 44 Henrico County third-grade students. Their scores were: 40 26 39 14 42 18 25 43 46 27 19 47 19 26 35 34 15 44 40 38 31 46 52 25 35 35 33 29 34 41 49 28 52 47 35 48 22 33 41 51 27 14 54 45 She then asked Minitab to calculate some descriptive statistics from this data set:

MTB > Describe 'DRPscore'. N MEAN MEDIAN STDEV 44 35.09 35.00 11.19 MIN MAX Q1 Q3 14.00 54.00 26.25 44.75

You may assume that DRP scores are approximately normal, and that the standard deviation of scores in Henrico County Schools is unknown. Construct an 80% confidence interval for the mean DRP score in Henrico County Schools. HINT: The box above with the Minitab data, states what the sample mean is. Use this for your interval.

24