Download Exam III Review Note: This is only a sample of the types of exercises

Exam III Review
Note: This is only a sample of the types of exercises that may appear on the exam. Anything covered in class may appear on the
exam.
If Z is a standard normal variable, find the probability.
1) The probability that Z lies between 0 and 3.01
Answer: 0.4987
2) The probability that Z is less than 1.13
Answer: 0.8708
3) The probability that Z is greater than -1.82
Answer: 0.9656
The Precision Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0°C at
the freezing point of water. Tests on a large sample of these thermometers reveal that at the freezing point of water, some
give readings below 0°C (denoted by negative numbers) and some give readings above 0°C (denoted by positive
numbers). Assume that the mean reading is 0°C and the standard deviation of the readings is 1.00°C. Also assume that the
frequency distribution of errors closely resembles the normal distribution. A thermometer is randomly selected and
tested. Find the temperature reading corresponding to the given information.
4) Find P40, the 40th percentile.
Answer: -0.25°
5) If 7% of the thermometers are rejected because they have readings that are too high, but all other thermometers
are acceptable, find the temperature that separates the rejected thermometers from the others.
Answer: 1.48°
6) If 6.3% of the thermometers are rejected because they have readings that are too high and another 6.3% are
rejected because they have readings that are too low, find the two readings that are cutoff values separating the
rejected thermometers from the others.
Answer: -1.53° , 1.53°
Assume that X has a normal distribution, and find the indicated probability.
7) The mean is µ = 15.2 and the standard deviation is = 0.9.
Find the probability that X is greater than 16.1.
Answer: 0.1587
8) The mean is µ = 22.0 and the standard deviation is = 2.4.
Find the probability that X is between 19.7 and 25.3.
Answer: 0.7477
Solve the problem.
9) Scores on a test are normally distributed with a mean of 65.3 and a standard deviation of 10.3. Find P81, which
separates the bottom 81% from the top 19%.
Answer: 74.4
10) Assume that women have heights that are normally distributed with a mean of 63.6 inches and a standard
deviation of 2.5 inches. Find the value of the quartile Q3 .
Answer: 65.3 inches
Find the indicated probability.
11) The incomes of trainees at a local mill are normally distributed with a mean of $1100 and a standard deviation
of $150. What percentage of trainees earn less than $900 a month?
Answer: 9.18%
12) The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a standard
deviation of $45. What is the probability that a randomly selected teacher earns more than $525 a week?
Answer: 0.2177
13) A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a
standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 170
and 220.
Answer: 0.3811
Solve the problem.
14) The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 74 inches,
and a standard deviation of 12 inches. What is the probability that the mean annual snowfall during 36
randomly picked years will exceed 76.8 inches?
Answer: 0.0808
15) The scores on a certain test are normally distributed with a mean score of 51 and a standard deviation of 2.
What is the probability that a sample of 90 students will have a mean score of at least 51.2108?
Answer: 0.1587
16) Find the critical value z /2 that corresponds to a degree of confidence of 98%.
Answer: 2.33
17) Find the value of -z /2 that corresponds to a level of confidence of 98.22 percent.
Answer: -2.37
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
18) n = 58, x = 28; 95 percent
Answer: 0.354 < p < 0.612
^
Find the minimum sample size you should use to assure that your estimate of p will be within the required margin of
error around the population p.
^
^
19) Margin of error: 0.006; confidence level: 96%; p and q are unknown
Answer: 29,185
^
20) Margin of error: 0.04; confidence level: 95%; from a prior study, p is estimated by the decimal equivalent of
92%.
Answer: 177
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
21) Of 346 items tested, 12 are found to be defective. Construct the 98% confidence interval for the proportion of all
such items that are defective.
Answer: 0.0118 < p < 0.0576
22) Of 129 randomly selected adults, 32 were found to have high blood pressure. Construct a 95% confidence
interval for the true percentage of all adults that have high blood pressure.
Answer: 17.4% < p < 32.3%
Use the confidence level and sample data to find a confidence interval for estimating the population µ.
23) Test scores: n = 109, x = 79.1,
= 6.9; 99 percent
Answer: 77.4 < µ < 80.8
24) A random sample of 144 full-grown lobsters had a mean weight of 18 ounces and a standard deviation of 2.9
ounces. Construct a 98 percent confidence interval for the population mean µ.
Answer: 17 < µ < 19
Use the margin of error, confidence level, and standard deviation
an unknown population mean µ.
25) Margin of error: $121, confidence level: 95%, = $528
to find the minimum sample size required to estimate
Answer: 74
26) Margin of error: $126, confidence level: 99%,
= $534
Answer: 120
Do one of the following, as appropriate: (a) Find the critical value z /2, (b) find the critical value t /2, (c) state that
neither the normal nor the t distribution applies.
27) 98%; n = 7; = 27; population appears to be normally distributed.
Answer: z /2 = 2.33
28) 90%; n = 10;
is unknown; population appears to be normally distributed.
Answer: t /2 = 1.833
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume
that the population has a normal distribution.
29) n = 10, x = 12.8, s = 4.9, 95 percent
Answer: 9.29 < µ < 16.31
30) Thirty randomly selected students took the calculus final. If the sample mean was 78 and the standard deviation
was 7.5, construct a 99 percent confidence interval for the mean score of all students.
Answer: 74.23 < µ < 81.77
31) The principal randomly selected six students to take an aptitude test. Their scores were:
77.9 89.1 80.7 78.6 74.4 82.0
Determine a 90 percent confidence interval for the mean score for all students.
Answer: 76.36 < µ < 84.54
Express the null hypothesis H0 and the alternative hypothesis H1 in symbolic form. Use the correct symbol (µ, p,
)for
the indicated parameter.
32) An entomologist writes an article in a scientific journal which claims that fewer than 11 in ten thousand male
fireflies are unable to produce light due to a genetic mutation. Use the parameter p, the true proportion of
fireflies unable to produce light.
Answer: H0 : p = 0.0011
H1 : p < 0.0011
33) A psychologist claims that more than 6.3 percent of the population suffers from professional problems due to
extreme shyness. Use p, the true percentage of the population that suffers from extreme shyness.
Answer: H0 : p = 6.3%
H1 : p > 6.3%
34) A researcher claims that 62% of voters favor gun control.
Answer: H0 : p = 0.62
H1 : p 0.62
Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z
value used to test a null hypothesis.
35) = 0.05 for a two-tailed test.
Answer: ±1.96
36)
= 0.08; H1 is µ 3.24
Answer: ±1.75
^
Find the value of the test statistic z using z =
p -p
.
pq
n
37) A claim is made that the proportion of children who play sports is less than 0.5, and the sample statistics include
n = 1158 subjects with 30% saying that they play a sport.
Answer: -13.61
38) The claim is that the proportion of drowning deaths of children attributable to beaches is more than 0.25, and
the sample statistics include n = 647 drowning deaths of children with 30% of them attributable to beaches.
Answer: 2.94
Use the given information to find the P-value.
39) The test statistic in a right-tailed test is z = 1.43.
Answer: 0.0764
40) The test statistic in a two-tailed test is z = 1.95.
Answer: 0.0512
Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim.
41) An entomologist writes an article in a scientific journal which claims that fewer than 19 in ten thousand male
fireflies are unable to produce light due to a genetic mutation. Assuming that a hypothesis test of the claim has
been conducted and that the conclusion is to reject the null hypothesis, state the conclusion in nontechnical
terms.
Answer: There is sufficient evidence to support the claim that the true proportion is less than 19 in ten thousand.
42) Carter Motor Company claims that its new sedan, the Libra, will average better than 23 miles per gallon in the
city. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is to reject the null
hypothesis, state the conclusion in nontechnical terms.
Answer: There is sufficient evidence to support the claim that the mean is greater than 23 miles per gallon.
43) The owner of a football team claims that the average attendance at games is over 727, and he is therefore
justified in moving the team to a city with a larger stadium. Assuming that a hypothesis test of the claim has
been conducted and that the conclusion is failure to reject the null hypothesis, state the conclusion in
nontechnical terms.
Answer: There is not sufficient evidence to support the claim that the mean attendance is greater than 727.
44) A psychologist claims that more than 75 percent of the population suffers from professional problems due to
extreme shyness. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is
failure to reject the null hypothesis, state the conclusion in nontechnical terms.
Answer: There is not sufficient evidence to support the claim that the true proportion is greater than 75 percent.
Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final
conclusion that addresses the original claim.
45) A manufacturer considers his production process to be out of control when defects exceed 3%. In a random
sample of 85 items, the defect rate is 5.9% but the manager claims that this is only a sample fluctuation and
production is not really out of control. At the 0.01 level of significance, test the manager's claim.
Answer: H0 : p = 0.03. H1 : p > 0.03. Test statistic: z = 1.57. P-value: p = 0.0582.
Critical value: z = 2.33. Fail to reject null hypothesis. There is not sufficient evidence to warrant rejection
of the manager's claim that production is not really out of control.
46) A supplier of 3.5" disks claims that no more than 1% of the disks are defective. In a random sample of 600 disks,
it is found that 3% are defective, but the supplier claims that this is only a sample fluctuation. At the 0.01 level of
significance, test the supplier's claim that no more than 1% are defective.
Answer: H0 : p = 0.01. H1 : p > 0.01. Test statistic: z = 4.92. P-value: p = 0.0001.
Critical value: z = 2.33. Reject null hypothesis. There is sufficient evidence to warrant rejection of the
claim that no more than 1% are defective.
47) A poll of 1,068 adult Americans reveals that 48% of the voters surveyed prefer the Democratic candidate for the
presidency. At the 0.05 level of significance, test the claim that at least half of all voters prefer the Democrat.
Answer: H0 : p = 0.5. H1 : p < 0.5. Test statistic: z = -1.31. P-value: p = 0.0951.
Critical value: z = -1.645. Fail to reject null hypothesis. There is not sufficient evidence to warrant
rejection of the claim that at least half of all voters prefer the Democrat.
48) In a sample of 167 children selected randomly from one town, it is found that 37 of them suffer from asthma. At
the 0.05 significance level, test the claim that the proportion of all children in the town who suffer from asthma
is 11%.
Answer: H0 : p = 0.11. H1 : p
0.11. Test statistic: z = 4.61. P-value: p = 0.0001.
Critical values: z = ±1.96. Reject null hypothesis. There is sufficient evidence to warrant rejection of the
claim that the proportion of all children in the town who suffer from asthma is 11%.
49) The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean
weight of 183.9 lb. Assuming that is known to be 121.2 lb, use a 0.10 significance level to test the claim that
the population mean of all such employees weights is less than 200 lb.
Answer: H0 : µ = 200; H1 : µ < 200; Test statistic: z = -0.98. P-value: 0.1635. Fail to reject H0 . There is not
sufficient evidence to warrrant the rejection of the claim that the mean equals 200.
Test the given claim using the traditional method of hypothesis testing. Assume that the sample has been randomly
selected from a population with a normal distribution.
50) A light-bulb manufacturer advertises that the average life for its light bulbs 900 hours. A random sample of 15
of its light bulbs resulted in the following lives in hours.
995
590
510
539
739
917
571
555
916
728
664
693
708
887
849
At the 10% significance level, do the data provide evidence that the mean life for the company's light bulbs
differs from the advertised mean?
Answer: Test statistic: t = -4.342. Critical values: t = ±1.761. Reject H0 : µ = 900 hours. There is sufficient evidence to
support the claim that the true mean life differs from the advertised mean.