Download Stat Test 4 REVIEW

Stat Test 4 REVIEW
Name___________________________________
Solve the problem.
1) The following confidence interval is obtained for a population proportion, p: (0.505, 0.545). Use these confidence
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interval limits to find the point estimate, p.
Answer: 0.525
2) The following confidence interval is obtained for a population proportion, p: 0.537 < p < 0.563. Use these
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confidence interval limits to find the point estimate, p.
Answer: 0.550
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the
given statistics and confidence level. Round the margin of error to four decimal places.
3) 95% confidence; n = 320, x = 60
Answer: 0.0428
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
4) n = 51, x = 27; 95% confidence
Answer: 0.392 < p < 0.666
Use the given data to find the minimum sample size required to estimate the population proportion.
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5) Margin of error: 0.004; confidence level: 95%; p and q unknown
Answer: 60,025
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6) Margin of error: 0.05; confidence level: 99%; from a prior study, p is estimated by 0.15.
Answer: 339
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7) Margin of error: 0.01; confidence level: 95%; from a prior study, p is estimated by the decimal equivalent of 52%.
Answer: 9589
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
8) A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct
the 95% confidence interval for the true proportion of all voters in the state who favor approval.
Answer: 0.438 < p < 0.505
9) A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for
governor. Construct the 98% confidence interval for the true population proportion of all New York State union
members who favor the Republican candidate.
Answer: 0.308 < p < 0.438
Solve the problem.
10) A newspaper article about the results of a poll states: "In theory, the results of such a poll, in 99 cases out of 100
should differ by no more than 5 percentage points in either direction from what would have been obtained by
interviewing all voters in the United States." Find the sample size suggested by this statement.
Answer: 664
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Use the given information to find the minimum sample size required to estimate an unknown population mean µ.
11) Margin of error: $120, confidence level: 95%, = $593
Answer: 94
12) Margin of error: $137, confidence level: 99%,
= $591
Answer: 124
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your
answer to the same number of decimal places as the sample mean.
13) A random sample of 105 light bulbs had a mean life of x = 441 hours with a standard deviation of
Construct a 90% confidence interval for the mean life, µ, of all light bulbs of this type.
= 40 hours.
Answer: 435 hr < µ < 447 hr
14) A random sample of 130 full-grown lobsters had a mean weight of 21 ounces and a standard deviation of 3.0
ounces. Construct a 98% confidence interval for the population mean µ.
Answer: 20 oz < µ < 22 oz
15) 37 packages are randomly selected from packages received by a parcel service. The sample has a mean weight
of 17.0 pounds and a standard deviation of 3.3 pounds. What is the 95% confidence interval for the true mean
weight, µ, of all packages received by the parcel service?
Answer: 15.9 lb < µ < 18.1 lb
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.
16) 99%; n = 17; is unknown; population appears to be normally distributed.
Answer: t /2 = 2.921
17) 90%; n = 10;
is unknown; population appears to be normally distributed.
Answer: t /2 = 1.833
18) 95%; n = 11;
is known; population appears to be very skewed.
Answer: Neither the normal nor the t distribution applies.
Assume that a sample is used to estimate a population mean µ. Use the given confidence level and sample data to find
the margin of error. Assume that the sample is a simple random sample and the population has a normal distribution.
Round your answer to one more decimal place than the sample standard deviation.
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19) 95% confidence; n = 91; x = 16, s = 9.1
Answer: 1.90
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.
20) A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 225 milligrams
with s = 15.7 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such
eggs.
Answer: 215.0 mg < µ < 235.0 mg
21) n = 10, x = 8.7, s = 3.3, 95% confidence
Answer: 6.34 < µ < 11.06
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22) The football coach randomly selected ten players and timed how long each player took to perform a certain
drill. The times (in minutes) were:
7.2 10.5 9.9 8.2 11.0
7.3 6.7 11.0 10.8 12.4
Determine a 95% confidence interval for the mean time for all players.
Answer: 8.15 min < µ < 10.85 min
23) The amounts (in ounces) of juice in eight randomly selected juice bottles are:
15.2 15.5 15.9 15.5
15.0 15.7 15.0 15.7
Construct a 98% confidence interval for the mean amount of juice in all such bottles.
Answer: 15.10 oz < µ < 15.77 oz
Express the null hypothesis and the alternative hypothesis in symbolic form. Use the correct symbol (µ, p, ) for the
indicated parameter.
24) 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.
25) = 0.05 for a left-tailed test.
Answer: -1.645
26)
= 0.09 for a right-tailed test.
Answer: 1.34
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p-p
Find the value of the test statistic z using z =
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pq
n
27) 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 = 681 drowning deaths of children with 30% of them attributable to beaches.
Answer: 3.01
Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null
hypothesis (reject the null hypothesis or fail to reject the null hypothesis).
28) The test statistic in a right-tailed test is z = 0.52.
Answer: 0.3015; fail to reject the null hypothesis
29) The test statistic in a left-tailed test is z = -1.83.
Answer: 0.0336; reject the null hypothesis
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