Download Comparing Means … Independent Samples

1. __A__ Suppose a 95% confidence interval for μ turns out to be (120, 300).
To make a more useful inference from the data, it is desired to reduce the width of the confidence interval.
Which of the following will result in a reduced interval width?
(A) Decrease the confidence level from 95% to 90%
(C) Decrease the sample size, i.e. use less data
(B) Increase the confidence level from 95% to 99%
(D) Use a statistical software package such as MINITAB
2. __C__ A point estimate is the value of a ____ that estimates the value of a ____.
A) Parameter; Statistic
C) Statistic; Parameter
B) Table Value; Standard Error
D) Standard Error; Table Value
3. Professor Crayfish from the Biology Department is studying the freshwater Bumble Bee shrimp species.
The professor wants to estimate the mean weight of the freshwater Bumble Bee shrimp species.
3.a) __C__ Data analysis for 148 such shrimp resulted in a 95% confidence interval for this mean to be 5.5 grams to 8.5 grams.
Which one of the following is a correct interpretation of this interval?
A.
B.
C.
D.
95% of all Bumble Bee shrimp are between 5.5 and 8.5 grams.
95% of the 148 Bumble Bee shrimp were between 5.5 and 8.5 grams.
5.5 grams to 8.5 grams are reasonable values for the mean weight of all Bumble Bee shrimp at the 95% confidence level.
95% of the time a Bumble Bee shrimp is between 5.5 and 8.5 grams.
3.b) __3/2 = 1.5 grams__ What is the value of the margin of error for the interval listed above in part a)?
3.c) __C__ Suppose Professor Crayfish is unhappy because this margin of error is too large.
What would you advise the Professor to do for his next research study in order to get a smaller margin of error?
A. Take the sample data from a larger population
C. Take a larger sample
B. No need to take a larger sample since it is already above 30
D. Use 99% confidence instead of 95%
4.-6. The U.S. Department of Agriculture (USDA) wants to determine the average yearly consumption of broccoli per person in the United States.
Last year, a survey was conducted of 625 U.S. residents. This sample had a mean of 12.2 pounds of broccoli per person.
The standard deviation of the U.S. population is known to be 15 pounds of broccoli per person.
4. In what parameter is the USDA interested? __C__.
A. The amount of broccoli consumed by the 625 people surveyed last year.
B. The mean amount of broccoli produced per year in the United States.
C. The mean yearly number of pounds of broccoli consumed per person in the United States.
D. Whether or not broccoli is a popular vegetable.
5. What statistic could the USDA use to estimate this parameter? __D__.
A. The mean number of people in the United States that like broccoli.
B. 15 pounds of broccoli per person.
C. 1000 people
D. 12.2 pounds of broccoli per person.
6. What is the value of the standard deviation (i.e. standard error) of the sampling distribution of X ? __A__.
A) 0.6 = 15/sqrt(625)
B) 15
C) 25
D) 0.024
7. (3 pts) A television sports commentator wants to estimate the proportion of citizens who “follow professional football.”
What sample size is needed if he wants to be within 3 percentage points with 98% confidence - he does not use any prior estimates?
n
0.5(1  0.5)( 2.326) 2
p(1  p) z 2
n

 1502.8...  1503
E2
(0.03) 2
8. (4 pts) Crawling Babies
The following data represent the age (in weeks) at which babies first crawl based on a random survey of 12 mothers conducted by
www.essentialbaby.com.au. Construct and Interpret a 95% confidence interval for the mean age at which a baby first crawls.
52
30
44
35
39
26
47
37
56
26
39
28
Descriptive Statistics: AGE
Variable
AGE
N
12
Mean
38.25
StDev
10.00
Minimum
26.00
Maximum
56.00
n = 12
s x  10 / 12  2.887  t-interval for mean
95% confidence … df = 12-1 = 11 t critical-value = 2.201
C.I.  38.25 +- 2.201(2.887)
At the 90% confidence level, plausible values for the mean age at which a baby first crawls (31.9, 44.6) weeks.
9. (4 pts) A survey of 1600 fatal auto accidents showed that 608 involved the use of a cell phone.
Construct and Interpret a 96% confidence interval for the proportion of fatal auto accidents that involve the use of a cell phone.
phat = 608/1600 = 0.38
StdErr of phat = sqrt[0.38(0.62)/1600] = 0.012135
z for 96% confidence = 2.054
0.38 +- 2.054(0.012135)
0.38 +- 0.025
At the 96% confidence level,
plausible values for the proportion of all fatal auto accidents that involve the use of a cell phone  (0.355 0.405)