Test 10B
AP Statistics
Name:
Directions: Work on these sheets. Answer completely, but be concise. A normal probability table is
attached.
Part 1: Multiple Choice. Circle the letter corresponding to the best answer.
1. The value of z* required for a 70% confidence interval is
(a) -0.5244
(b) 1.036
(c) 0.5244
(d) 0.6179
(e) The answer can’t be determined from the information given.
(f) None of the above. The answer is _____________________.
맞습니다.
2. A significance test allows you to reject a hypothesis H0 in favor of an alternative Ha at the 5%
level of significance. What can you say about significance at the 1% level?
(a) H0 can be rejected at the 1% level of significance.
(b) There is insufficient evidence to reject H0 at the 1% level of significance.
(c) There is sufficient evidence to accept H0 at the 1% level of significance.
(d) Ha can be rejected at the 1% level of significance.
(e) The answer can’t be determined from the information given.
맞습니다.
3. A 95% confidence interval for the mean  of a population is computed from a random sample
and found to be 9 ± 3. We may conclude that
(a) There is a 95% probability that  is between 6 and 12.
(b) There is a 95% probability that the true mean is 9 and a 95% chance the true margin of error is
3.
(c) If we took many, many additional random samples and from each computed a 95%
confidence interval for  , approximately 95% of these intervals would contain  .
(d) If we took many, many additional random samples and from each computed a 95% confidence
interval for  , 95% of them would cover the values from 6 to 12.
(e) All of the above.
맞습니다.
4. A 95% confidence interval for the mean reading achievement score for a population of third grade
students is (44.2, 54.2). Suppose you compute a 99% confidence interval using the same
information. Which of the following statements is correct?
(a) The intervals have the same width.
(b) The 99% interval is shorter.
(c) The 99% interval is longer.
(d) The answer can’t be determined from the information given.
(e) None of the above. The answer is ______________________.
맞습니다.
Chapter 10
1
Test 10B
5. Which of the following are correct?
I. The power of a significance test depends on the alternative value of the parameter.
II. The probability of a Type II error is equal to the significance level of the test.
III. Type I and Type II errors only make sense when a significance level has been chosen in
advance.
(a) I and II only
(b) I and III only
(c) II and III only
(d) I, II, and III
(e) None of the above gives the complete set of true responses.
맞습니다.
6. In a test of H0: µ = 100 against Ha: µ  100, a sample of size 80 produces z = 0.8 for the value of
the test statistic. The P-value of the test is thus equal to:
(a)
(b)
(c)
(d)
(e)
0.20
0.40
0.29
0.42
0.21
two-tailed test 이므로 0.21*2 =0.42 (d) 입니다.
7. 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 x of the weighings is computed. Suppose the
scale readings are normally distributed with unknown mean  and standard deviation
 = 0.01 g. How large should n be so that a 95% confidence interval for  has a margin of error
of ± 0.0001?
(a) 100
(b) 196
(c) 27061
(d) 10000
(e) 38416
맞습니다.
8. 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 reject H0 at level  = 0.05.
(e) We would accept H0 at level  = 0.05.
Chapter 10
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Test 10B
95% confidence interval이 (1.7, 3.5) 이므로
X
1.7  3.5
 2.6
2
이고
hypothesis test 에서
m argin of error= Z / 2
test statistic Z =

n
 1.96

n
 0.9
X   2.6  0

 5.65
0.46
/ n
P-value = P(Z>5.65) * 2 = 0.000 (너무작아서표시안됨, two-tailed test)
P-value < 0.05 이므로 H0 reject 답은 (d) 입니다.
(위와 같은 과정을 거쳐 confidence interval을 이용해 hypothesis test 결과를 알 수 있습니다.)
Part 2: Free Response
Communicate your thinking clearly and completely.
9. A steel mill’s milling machine produces steel rods that are supposed to be 5 cm in diameter. When
the machine is in statistical control, the rod diameters vary according to a normal distribution with
mean µ = 5 cm and standard deviation  = 0.02 cm. A large sample of 150 produced by the
machine yields a sample mean diameter of 5.005 cm.
(a) Construct a 99% confidence interval for the true mean diameter of the rods produced by the
milling machine. Follow the inference toolbox.
I assume that the sample is random and that the distribution is approximately normal
The pop of interest is the steel rods produced by the machines. Mu =5,l std.=.02
I will now estimate mu.
5.005+/- 2.576* (.02/root150)= (3.95335, 5.00921
Im 95% →99% confident that the true mean diameter of the rods are between (3.95335, 5.00921
계산을 실수 한 것 같습니다.
.02
5.005  2.576
 5.005  0.0042 이므로
150
(5.0008, 5.0092)
(b) Does the interval in (a) give you reason to suspect that the machine is not producing rods of the
correct diameter? State appropriate hypotheses and a significance level. Then explain your
conclusion.
H null: mu =5
H alternative: mu <5
→ H alternative: mu ≠ 5 (correct 인지 물어봤으므로 크거나 작은 것 모두 불량으로 보는 것이
더 타당해 보입니다.)
I will test at .05 significance level
→ (confidence level 99%를 사용했으므로 significance level도 비슷하게 0.01을 사용하는게 좋아
보입니다.)
I divine H alternative as mu < 5 because the interval suggests that the machines may be producing
smaller rods
(5.005-5)/(.02/root150)= 3.06
P at 3.06 = .9989
Chapter 10
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Test 10B
1-.9989= .0011
→ P-value = 2*0.0011 = 0.0022 (two-tailed test)
.0011< .05
because the p value is smaller, I reject H null and accept that the company is making
smaller rods
P –value = 0.0022 < 0.01 이 H0 reject
(c) Describe a Type II error in the context of this problem. How could the manufacturer decrease
the probability of a Type II error.
Type two error: if the company believes that they make 5 cm diameter but aer actually producing
smaller rods.
→ Type II error 는 Ha가 true 일 때, false라고 판단할 오류 입니다.
즉, rod 가 5cm diameter가 아님에도 정상(5cm)으로 판단하는 경우입니다.
(소비자 위험 – 불량제품을 정상으로 판단하는 경우이므로 소비자에게 손해)
10. A pharmaceutical manufacturer does a chemical analysis to check the potency of products. The
standard release potency for cephalothin crystals is 910. An assay of 16 lots gives the following
potency data:
897
918
914
906
913
895
906
893
916
908
918
906
905
907
921
901
Assume a population standard deviation  = 8.2.
(a) Construct a 99% confidence interval for the population mean. Follow the Inference Toolbox.
I am finding the mu interval for all the cephalothin crystals. Mu = 910 std, =8.2
Assuming this is an SRS and that the sample distribution is approximately normal,
911.5 + / - 8.2/root 16 *2.576
= 906.2 ~ 916.8
I am 99% confident that the true potency of cephalothin is between (906.2 and 916.8)
→ sample mean = 907.75 나옵니다. (911.5는 계산기 입력이 틀린 듯)
confidence interval= sam ple statstics  m argin of error
m argin of error = criticalvalue * standard deviation
criticalvalue = Z /2  Z.005  2.58
따라서, interval은
907.75  2.58 
8.2
 907.75  5.29
16
You want to test hypotheses about the mean population potency,
H 0 : µ = 910
H a : µ < 910


at the 1% level of significance. The z test statistic is z = x  910 8.2
Chapter 10
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
16 .
Test 10B
(b) What is the rule for rejecting H 0 in terms of z?
If 1 – the corresponding z-score is less than the levl of significance, we reject H null
→decision rule : p-value > significance level 이면 H0 accept
p-value < significance level 이면 H0 reject

x  910 
907.75  910 

P  z 
  P  z 
  P ( z  1.096)  0.1365
8.2
/
16
8.2 / 16 



p-value = 0.1365 (one-tailed test)이므로 significance level 0.01이면 H0 accept
(c) What values of x would lead you to reject H 0 ?
(X -910) / (8.2/ root 16) < .2.325
X< 905.234
→ significance level 0.01 에서 고려하면
P( Z<-2.326)=0.01 이므로
x  910
 2.326 에서 x  905.232
8.2 / 16
(sign 주의!)
(d) Describe a Type I error in the context of this problem. What is the probability of a Type I error?
When I accept H alternative, the statement that potency, is < 910 when H null, the statement that mu =
910, is actually true
Significance level alpha is always the probability of type 1 error
Probability of type 1 error =.01
→ Type I error : H0가 true 이지만 reject하는 오류 이므로 맞습니다.
significance = probability of Type I error
Chapter 10
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Test 10B