Name ______________Solutions_________________________________ Date ________________________
AP Statistics
1.
Jonathon is the manager of a used video game store. He wants to estimate the average amount a
customer spends per visit. A random sample of 80 customers’ receipts gave a mean of $6.90 and a
standard deviation of $2.45.
a. What conditions are necessary to calculate a 90% confidence interval for the average amount spent
by all customers?
Sample size is greater than 30
b.
2.
Chapter 8 Review
For a day when the video game store had 80 customers, estimate the total income on that day. (as an
interval)
90% C.I. (6.441, 7.3559)
Allen is an appliance salesman who works on commission. A random sample of 39 days showed that the
sample standard deviation value of sales was $215. How many more days should be included in a sample
to be 95% sure the population mean µ is within $50 of the sample mean?
n = 39 C.I. = 95% E = 50 𝑛 = (
𝑧𝑐 𝜎 2
)
𝐸
n= 72
33 More Days
3.
What percentage of male athletes wear contact lenses during performances? If you have no preliminary
estimate for p, how many male athletes should you include in a random sample to be 90% sure the point
estimate will be within a distance of 0.05 from p.
1
𝑧
4
𝐸
n = ( 𝑐 )2
n = 271
Studies show that approximately 19% of male athletes wear contact lenses during performances. How
many male athletes should you include in a random sample to be 90% sure the point estimate will be
within a distance of 0.05 from p.
𝑧
n = pq( 𝑐)2
𝐸
n = 167
4.
At a large office supply store, the daily sales of two similar brand-name laser printers are being compared.
A random sample of 16 days showed that Brand I had mean daily sales $2464 and a standard deviation
$529. A random sample of 19 days showed that Brand II had mean daily sales $2285 and a standard
deviation $440. Assume sales follow an approximately normal distribution.
a. Find a 90% confidence interval for the population mean difference in sales.
µ1 = mean daily sales of laser printers brand 1
µ2 = mean daily sales of brand 2
90% C.I. µ1 - µ2 (-101.9, 459.86)
b. What does the interval tell you about the mean sales of one printer compared to that of the other?
At the 90% C.L. my interval contains both positive and negative values; it does not appear that the
population mean daily sales of printers differ.
5.
A production manager is studying the effect of overtime on different shifts. On shift I at least half the
workers were on overtime. A random sample of 245 items assembly line showed that 24 were defective.
Shift II had no overtime workers. A random sample of 258 items from the assembly line showed that 11
had defects.
a. Find and interpret a 90% confidence interval of the proportion of defective items for Shift I versus
Shift II.
P1 = shift 1 population proportion of defective items 𝑝̂ 1 =
P2 = shift 2 population proportion of defective items 𝑝̂ 2 =
24
245
11
258
2 Prop z-interval p1- p2 90% (.018, .093)
I am 90% confident that p1 p2 . or I am 90% confident that the proportion of defects from shift 1 minus
the proportion of defects from shift 2 is contained in(.018, .093).
b.
6.
What assumptions are required for your calculations from a. Do you think these assumptions are
satisfied?
n1p1  5
n1q1 5 and n2p2  5
n2q2 5
The table below shows data from Canadian and U.S. ninth-grader test scores.
U.S.
Canadians
a.
Sample Size
Sample Mean
300
250
95
90.5
Write the formula for the 95% confidence interval in terms of the values shown above.
(µ1 - µ2) ± (zc 1.96 or tc 1.98) (E) because n is so large could use zc
(95 − 90.5) ± (1.96)√
b.
Sample Standard
Deviation
4.5
4.1
Determine the confidence interval.
95% C.I. µ1 - µ2 (3.78, 5.22)
4.52 4.12
+
300 250
c.
Provide an interpretation of the confidence interval. Include in your interpretation a conclusion that
can be drawn about which country has a higher average test score.
µ1 – U.S. ninth-grader test score population mean
µ2 - Canadian. ninth-grader test score population mean
I am 95% confident that µ1 - µ2 is between 3.78 and 5.22. or I am 95% confident the U.S. ninth grade
test scores are higher than Canadian test scores.
7.
You want to create a 95% C.I. with a margin of error of no more than 0.05 for a population proportion.
The historical data indicate that the population has remained constant at about 0.55. What is the
minimum size random sample you need to construct this interval?
𝑧
n = pq( 𝑐)2
n = 381
𝐸
8.
A random sample of 30 households was selected as part of a study on electricity usage, and the number
of kilowatt-hours (kWh) was recorded for each household in the sample for the March quarter 2011. The
average usage was found to be 450 kWh. In a very large study in the March quarter of the previous year it
was found that the standard deviation of the usage was 81 kWh. Assuming the standard deviation is
unchanged and that the usage is normally distributed provide an expression for calculating a 99% C.I. for
the mean usage in March quarter of 2011.
A. 450 ± 2.756 ×
B.
450 ± 2.575 ×
C.
450 ± 2.33 ×
81
√30
9
√30
81
√30
81
D. 450 ± 2.575 ×
√30
σ used from previous year study
E.
None of the above
9.
If our sample size from problem 8 were tripled, how would this change the confidence interval size?
A. Divides the interval size by 3.
B. Multiples the interval by 1.732
C. Divides the interval size by 1.732 increasing sample size by mult. of d divides the interval est. by √𝑑
D. Triples the interval size.
E. None of the above.
10. Suppose you construct a 95% C.I. from a random sample of size n=20 with a sample mean 100 taken from
a population with unknown µ and a known σ = 10. And the interval is fairly wide. Which of the following
conditions would not lead to a narrower C.I.?
A.
B.
C.
D.
E.
If you decrease your C.I.
If you increase your sample size
If the sample size was smaller
If the population standard deviation was smaller
All of the above would lead to a narrower confidence interval.
11. Consider the following data:
1, 7, 3, 3, 6, 4
Assuming these data are drawn from a normal population with mean = µ and variance = 6, the 95% C.I. for
the appropriate z-score is ____ standard units long.
A. 5.14
B. 4.9
C. 4.04
D. 3.92 zc = 1.96 2(1.96) = 3.92
E. None of the above
12. In a middle school with 500 students, it was found that the true proportion of brunettes is 17.8%. If you
were to construct a C.I. estimate of the proportion of brunettes with sample size n=50, which of the
following statements is true?
I.
II.
III.
A.
B.
C.
D.
E.
The interval contains 17.8%
A 95% C.I. estimate contains 17.8%
The center of the interval is 17.8%
I and II
II and III
I and III
All of the above
None of the above
13. That is the critical t-value of a 95% C.I. estimate for a sample size of 25?
A. 2.06
B. 1.711
C. 2.064
D. 1.708
E. None of the above
14. In a recent online survey of 1,700 adults it was revealed that only 17% felt the internet was secure. With
what degree of confidence can you say that 17% ± 2% of adults believe that online security is secure?
A.
B.
C.
D.
E.
72.9%
90%
95%
97.2%
98.6%
15. A C.I. estimate is determined from the SAT scores of a SRS of n students. All other things being equal,
which of the following will result in a small margin of error?
I.
II.
III.
A.
B.
C.
D.
E.
Smaller standard deviation
Smaller sample size
Smaller C.I.
I and II
I and III ME varies by zc and standard deviation
II and III
I, II, and III
None of the above
16. If the C.I. for the population proportion changes from 90% to 98% with all else being equal, then
____________
A. The size of the interval increases by 41% ±1.645 𝑡𝑜 ± 2.326 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑠𝑖𝑧𝑒
B.
C.
D.
E.
The size of the interval decreases by 9%
The size of the interval increases by 9%
The size of the interval decreases by 41%
None of the above
17. Which of the following margins of error in a C.I. results using z-scores?
I.
II.
III.
A.
B.
C.
D.
E.
In obtaining the sample surveys error occurred due to no response
Random sample surveys
Errors occurred due to using sample standard deviation as estimates for σ
I only
II only ME due to variation of chance, nothing to do with defective survey
III only
I and II
I and III
2.326
1.645
𝑏𝑦 41%