Download ch 7 practice test Section 7.2 Solve the problem

ch 7 practice test
Section 7.2
Solve the problem.
1) Find the critical value z /2 that corresponds to a degree of confidence of 91%.
A) 1.34
B) 1.645
C) 1.75
1)
D) 1.70
^
Express the confidence interval in the form of p ± E.
2) 0.02 < p < 0.48
^
A) p = 0.25 - 0.23
2)
^
^
B) p = 0.25 ± 0.23
C) p = 0.25 ± 0.5
^
D) p = 0.23 ± 0.5
Solve the problem.
3) The following confidence interval is obtained for a population proportion, p:
(0.298, 0.338)
3)
^
Use these confidence interval limits to find the point estimate, p .
A) 0.318
B) 0.321
C) 0.298
D) 0.338
4) The following confidence interval is obtained for a population proportion, p:
0.855 < p < 0.897
Use these confidence interval limits to find the margin of error, E.
A) 0.042
B) 0.022
C) 0.876
4)
D) 0.021
Find the margin of error for the 95% confidence interval used to estimate the population proportion.
5) In a survey of 4100 T.V. viewers, 20% said they watch network news programs.
A) 0.0122
B) 0.00915
C) 0.0160
D) 0.0140
6) In a clinical test with 2353 subjects, 1136 showed improvement from the treatment.
A) 0.0273
B) 0.0202
C) 0.0226
D) 0.0172
5)
6)
^
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.
^
^
7) Margin of error: 0.011; confidence level: 92%; p and q unknown
A) 6327
B) 1
C) 40
D) 6328
^
8) Margin of error: 0.01; confidence level: 95%; from a prior study, p is estimated by the decimal
equivalent of 69%.
A) 7396
B) 14,184
C) 8218
D) 26,507
Solve the problem.
9) 464 randomly selected light bulbs were tested in a laboratory, 424 lasted more than 500 hours. Find
a point estimate of the true proportion of all light bulbs that last more than 500 hours.
A) 0.086
B) 0.914
C) 0.477
D) 0.912
1
7)
8)
9)
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
10) A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. 10)
Construct the 95% confidence interval for the true proportion of all voters in the state who favor
approval.
A) 0.444 < p < 0.500
B) 0.438 < p < 0.505
C) 0.435 < p < 0.508
D) 0.471 < p < 0.472
11) Of 132 adults selected randomly from one town, 33 of them smoke. Construct a 99% confidence
interval for the true percentage of all adults in the town that smoke.
A) 17.6% < p < 32.4%
C) 18.8% < p < 31.2%
11)
B) 15.3% < p < 34.7%
D) 16.2% < p < 33.8%
Section 7.3
Solve the problem.
12) Find the critical value z /2 that corresponds to a degree of confidence of 98%.
A) 2.575
B) 2.05
C) 1.75
12)
D) 2.33
Use the confidence level and sample data to find the margin of error E.
13) Replacement times for washing machines: 90% confidence; n = 36, x = 10.0 years, = 2.1 years
A) 0.4 years
B) 6.0 years
C) 0.1 years
D) 0.6 years
14) College students' annual earnings: 99% confidence; n = 71, x = $3660,
A) $243
B) $8
C) $269
= $879
D) $1118
Use the confidence level and sample data to find a confidence interval for estimating the population µ.
15) A random sample of 79 light bulbs had a mean life of x = 400 hours with a standard deviation of
= 28 hours. Construct a 90 percent confidence interval for the mean life, µ, of all light bulbs of this
type.
A) 392 < µ < 408
B) 393 < µ < 407
C) 395 < µ < 405
D) 394 < µ < 406
16) A laboratory tested 90 chicken eggs and found that the mean amount of cholesterol was 230
milligrams with = 16.0 milligrams. Construct a 95 percent confidence interval for the true mean
cholesterol content, µ, of all such eggs.
A) 227 < µ < 233
B) 226 < µ < 232
C) 228 < µ < 234
D) 226 < µ < 233
13)
14)
15)
16)
Use the margin of error, confidence level, and standard deviation to find the minimum sample size required to estimate
an unknown population mean µ.
17) Margin of error: $126, confidence level: 99%, = $534
17)
A) 61
B) 120
C) 69
D) 105,268
Section 7.4
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.
18) 98%; n = 7; = 27; population appears to be normally distributed.
18)
A) t /2 = 2.575
B) z /2 = 2.33
C) t /2 = 1.96
D) z /2 = 2.05
19) 91%; n = 45; is known; population appears to be very skewed.
A) t /2 = 1.34
B) t /2 = 1.645
C) z /2 = 1.75
2
D) z /2 = 1.70
19)
20) 99%; n = 17; is unknown; population appears to be normally distributed.
A) t /2 = 2.921
B) t /2 = 2.898
C) z /2 = 2.567
D) z /2 = 2.583
21) 90%; n = 10; is unknown; population appears to be normally distributed.
A) z /2 = 1.383
B) z /2 = 2.262
C) t /2 = 1.833
D) t /2 = 1.812
20)
21)
22) 95%; n = 11; is known; population appears to be very skewed.
A) Neither the normal nor the t distribution applies.
B) z /2 = 1.812
C) z /2 = 1.96
22)
23) 90%; n =9; = 4.2; population appears to be very skewed.
A) Neither the normal nor the t distribution applies.
B) z /2 = 2.306
23)
D) t /2 = 2.228
C) z /2 = 2.365
D) z /2 = 2.896
Find the margin of error.
_
24) 99% confidence interval; n = 201; x = 217; s = 34
A) 5.6
B) 6.2
C) 8.4
D) 4.7
_
25) 95% confidence interval; n = 51; x = 388; s = 204
A) 51.7
B) 57.4
C) 74.6
D) 120.5
24)
25)
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.
26) A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 193
26)
milligrams with s = 15.4 milligrams. Construct a 95 percent confidence interval for the true mean
cholesterol content of all such eggs.
A) 185.0 < µ < 201.0
B) 183.1 < µ < 202.9
C) 183.3 < µ < 202.7
D) 183.2 < µ < 202.8
27) A savings and loan association needs information concerning the checking account balances of its
local customers. A random sample of 14 accounts was checked and yielded a mean balance of
$664.14 and a standard deviation of $297.29. Find a 98% confidence interval for the true mean
checking account balance for local customers.
A) $492.52 < µ < $835.76
B) $455.65 < µ < $872.63
C) $453.59 < µ < $874.69
D) $493.71 < µ < $834.57
27)
28) The football coach randomly selected ten players and timed how long each player took to perform
a certain drill. The times (in minutes) were:
5.5 5.7 5.0 5.9 5.3
5.9 5.5 5.9 5.0 5.5
Determine a 95 percent confidence interval for the mean time for all players.
A) 5.19 < µ < 5.85
B) 5.75 < µ < 5.29
C) 5.85 < µ < 5.19
D) 5.29 < µ < 5.75
28)
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Section 7.5
Solve the problem.
29) Find the critical value
2
R corresponding to a sample size of 11 and a confidence level of 90
percent.
A) 23.209
B) 18.307
C) 25.188
29)
D) 3.94
Solve the problem.
30) Find the chi-square value
percent.
A) 23.337
2
L corresponding to a sample size of 13 and a confidence level of 98
B) 26.217
C) 3.571
30)
D) 4.404
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation .
Assume that the population has a normal distribution.
31) Weights of men: 90% confidence; n = 14, x = 160.9 lb, s = 12.6 lb
A) 9.9 lb < < 16.3 lb
B) 9.6 lb < < 18.7 lb
C) 9.3 lb < < 17.7 lb
D) 10.2 lb < < 2.7 lb
31)
32) To find the standard deviation of the diameter of wooden dowels, the manufacturer measures 19
randomly selected dowels and finds the standard deviation of the sample to be s = 0.16. Find the
95% confidence interval for the population standard deviation .
A) 0.15 < < 0.21
B) 0.12 < < 0.24
C) 0.13 < < 0.22
D) 0.11 < < 0.25
32)
Find the appropriate minimum sample size.
33) You want to be 95% confident that the sample variance is within 30% of the population variance.
A) 723
B) 97
C) 346
D) 130
33)
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation .
Assume that the population has a normal distribution.
34) The amounts (in ounces) of juice in eight randomly selected juice bottles are:
34)
15.7 15.6 15.3 15.3
15.1 15.6 15.3 15.4
Find a 98 percent confidence interval for the population standard deviation .
A) (0.13, 0.42)
B) (0.13, 0.48)
C) (0.17, 0.66)
D) (0.12, 0.42)
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