Download 5th six weeks review-test next week File

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
Transcript
AP Statistics
Review for 5th six weeks test
Name _________________
Part 1: Multiple Choice: Circle the letter corresponding to the best answer.
1. Suppose that the population of the scores of all high school seniors who took the SAT Math test this year
follows a normal distribution with mean  and standard deviation  = 100. You read a report that says,
“on the basis of a simple random sample of 100 high school seniors that took the SAT-M test this year, a
confidence interval for  is 512.00 ± 25.76.” The confidence level for this interval is
(a) 90%.
(b) 95%.
(c) 99.5%.
(d) 99%.
(e) over 99.9%.
2. Is the proportion of marshmallows in Mr. Miller’s favorite breakfast cereal lower than it used to be? To
determine this, you test the hypotheses Ho: p old = p new, Ha: p old > p new at the α = 0.05 level. You calculate a
test statistic of 1.980. Which of the following is the appropriate P-value and conclusion for your test?
(a) P-value = 0.047; fail to reject Ho; we do not have convincing evidence that the proportion of
marshmallows has been reduced.
(b) P-value = 0.047; accept Ho; there is convincing evidence that the proportion of marshmallows has been
reduced.
(c) P-value = 0.024; fail to reject Ho; we do not have convincing evidence that the proportion of
marshmallows has been reduced.
(d) P-value = 0.024; reject Ho; we have convincing evidence that the proportion of marshmallows has been
reduced.
(e) P-value = 0.024; fail to reject Ho; we have convincing evidence that the proportion of marshmallows has
not changed.
3. The government claims that students earn an average of $4500 during their summer break from studies. A
random sample of students gave a sample average of $3975 and a 95% confidence interval was found to be
$3525 < µ < $4425. This interval is interpreted to mean that:
(a) If the study were to be repeated many times, there is a 95% probability that the true average summer
earnings is not $4500 as the government claims.
(b) Because our specific confidence interval does not contain the value $4500 there is a 95% probability that
the true average summer earnings is not $4500.
(c) If we repeat our survey many times, then about 95% of our confidence intervals will contain the
true value of the average earnings of students.
(d) If we were to repeat our survey many times, then about 95% of all the confidence intervals will contain
the value $4500.
(e) There is a 95% probability that the true average earnings are between $3525 and $4425 for all students.
4. The heights (in inches) of males in the United States are believed to be normally distributed with mean µ. The
average height of a random sample of 25 American adult males is found to be x = 69.72 inches and the
standard deviation of the 25 heights is found to be s = 4.15. The standard error of x is
(a) 0.17
(b) 0.83
(c) 0.69
(d) 1.856
(e) 2.04
5. Suppose we want a 90% confidence interval for the average amount spent on books by freshmen in their
first year at a major university. The interval is to have a margin of error of $2, and the amount spent has
a normal distribution with a standard deviation  = $30. The number of observations required is closest to
(a) 25.
(b) 30.
(c) 609.
(d) 608.
(e) 865.
1
6. I collect a random sample of size n from a population and from the data collected compute a 95%
confidence interval for the mean of the population. Which of the following would produce a new
confidence interval with larger width (larger margin of error) based on these same data?
(a) Use a smaller confidence level.
(b) Use a larger confidence level.
(c) Use the same confidence level, but compute the interval n times. Approximately 5% of these intervals
will be larger.
(d) Increase the sample size.
(e) Nothing can guarantee absolutely that you will get a larger interval. One can only say the chance of
obtaining a larger interval is 0.05.
7. A 95% confidence interval for µ is calculated to be (1.7, 3.5). It is now decided to test the hypothesis H0: µ
= 0 vs. Ha: µ  0 at the  = 0.05 level, using the same data as was used to construct the confidence interval.
(a) We cannot test the hypothesis without the original data.
(b) We cannot test the hypothesis at the  = 0.05 level since the  = 0.05 test is connected to the 97.5%
confidence interval.
(c) We can only make the connection between hypothesis tests and confidence intervals if the sample sizes
are large.
(d) We would accept H0 at level  = 0.05.
(e) We would reject H0 at level  = 0.05.
Part 2: Free Response: Answer completely, but be concise. Write sequentially and show all steps.
8. Researchers studying the learning of speech often compare measurements made on the recorded speech of
adults and children. One variable of interest is called the voice onset time (VOT). Here are the results for
6-year-old children and adults asked to pronounce the word “bees.” The VOT is measured in milliseconds
and can be either positive or negative.
(a) What is the mean and standard error of the difference between the mean VOT for children and adults?
(b) The researchers were investigating whether VOT distinguishes adults from children. Construct a 95%
confidence interval for the difference in mean VOTs when pronouncing the word “bees.”
Mars Inc., makers of M&M candies, claims that they produce plain M&Ms with the following distribution:
Brown: 30%
Red:
20%
Yellow: 20%
Orange: 10%
Green:
10%
Blue:
10%
A bag of plain M&Ms was selected randomly from the grocery store shelf, and the color counts were as
follows:
Brown: 16
Red:
11
Yellow: 19
Orange: 5
Green:
7
Blue:
3
9. You want to conduct an appropriate test of the manufacturer’s claim for the proportion of yellow
M&Ms. Identify the population a parameter of interest. Then state hypotheses. State and verify the
conditions for performing the significance test. Calculate the test statistic and the P-value. What do you
conclude about the manufacturer’s claim? Explain.
2