Deck 8: Confidence Intervals
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Deck 8: Confidence Intervals
1
Find the confidence level for an interval which has a critical value of 1.05.
A) 85.31%
B) 14.69%
C) 70.63%
D) 29.37%
A) 85.31%
B) 14.69%
C) 70.63%
D) 29.37%
70.63%
2
In a small-overlap front crash test, a car is crashed into a simulated telephone pole and the maximum intrusion of debris into the passenger compartment of a specific model of car is
Measured. This intrusion is normally distributed with a standard deviation of 2.5 cm.
How many cars must be crashed to establish that a 98% confidence interval for the mean intrusion of
Debris into the passenger compartment will have a margin of error of 0.3 cm?
A) 49
B) 376
C) 1
D) 20
Measured. This intrusion is normally distributed with a standard deviation of 2.5 cm.
How many cars must be crashed to establish that a 98% confidence interval for the mean intrusion of
Debris into the passenger compartment will have a margin of error of 0.3 cm?
A) 49
B) 376
C) 1
D) 20
376
3
The three confidence intervals below were constructed from the same sample. One of them was computed at a confidence level of 90%, another at a confidence level of 95%, and another at a confidence level of 98%.
Which is the confidence level at 98%?
A)
B)
C) cannot be determined
D)
Which is the confidence level at 98%?
A)

B)

C) cannot be determined
D)


4
A college admissions officer takes a simple random sample of 60 entering freshmen and computes their mean mathematics SAT score to be 439. Assume the population standard deviation is
Based on a 98% confidence interval for the mean mathematics SAT score, is it likely that the mean
Mathematics SAT score for entering freshmen class is greater than 456? (Hint: you should first
Construct the 98% confidence interval for the mean mathematics SAT score.)
A) No
B) The likelihood cannot be determined.
C) Yes

Mathematics SAT score for entering freshmen class is greater than 456? (Hint: you should first
Construct the 98% confidence interval for the mean mathematics SAT score.)
A) No
B) The likelihood cannot be determined.
C) Yes
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5
A random sample of a specific brand of snack bar is tested for calorie count, with the following results:
Assume the population standard deviation is
and that the population is approximately normal. Construct a 95% confidence interval for the calorie count of the snack bars.
A) (139.8, 152.2)
B) (99.0, 193.0)
C) (140.1, 151.9)
D) (130.3, 161.7)


A) (139.8, 152.2)
B) (99.0, 193.0)
C) (140.1, 151.9)
D) (130.3, 161.7)
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6
A college admissions officer takes a simple random sample of 110 entering freshmen and computes their mean mathematics SAT score to be 465. Assume the population standard deviation is
Construct a 90% confidence interval for the mean mathematics SAT score for the entering freshmen
Class.
A) (463, 467)
B) (451, 479)
C) (373, 557)
D) (446, 484)

Class.
A) (463, 467)
B) (451, 479)
C) (373, 557)
D) (446, 484)
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7
Find the critical value
needed to construct a(n) 99.3% confidence interval.
A) 2.46
B) 2.70
C) 3.59
D) 2.35

A) 2.46
B) 2.70
C) 3.59
D) 2.35
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8
A sample of size n = 42 is drawn from a population whose standard deviation is
Find the margin of error for a 95% confidence interval for 
A) 1.49
B) 1.31
C) 0.88
D) 2.57


A) 1.49
B) 1.31
C) 0.88
D) 2.57
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9
A sample of size n = 16 is drawn from an approximately normal population whose standard deviation is
The sample mean is
Construct a 99% confidence interval for 
A) (48.91, 53.09)
B) (38.28, 63.72)
C) 44.24, 57.76
D) (51.00, 57.76)



A) (48.91, 53.09)
B) (38.28, 63.72)
C) 44.24, 57.76
D) (51.00, 57.76)
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10
Scientists want to estimate the mean weight gain of mice after they have been fed a special diet. From previous studies, it is known that the weight gain is normally distributed with standard
Deviation 2 grams.
How many mice must be weighed so that a 99% confidence interval for mean weight will have a margin
Of error of 0.4 grams?
A) 26
B) 1
C) 13
D) 166
Deviation 2 grams.
How many mice must be weighed so that a 99% confidence interval for mean weight will have a margin
Of error of 0.4 grams?
A) 26
B) 1
C) 13
D) 166
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11
Find the critical value
needed to construct a(n) 97% confidence interval.
A) 2.17
B) 1.88
C) 2.75
D) 1.92

A) 2.17
B) 1.88
C) 2.75
D) 1.92
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12
A population has a standard deviation
How large a sample must be drawn so that a 98% confidence interval for μ will have a margin of error equal to 4.1?
A) 192
B) 5
C) 109
D) 10

A) 192
B) 5
C) 109
D) 10
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13
The following MINITAB output presents a 95% confidence interval. 
Fill in the blanks: We are______ confident that the population mean is between ________and________
A) 5%, 46.270,51.416
B) 95%, 0,48.843
C) 95%, 46.270,51.416
D) 5%, 0,48.843

Fill in the blanks: We are______ confident that the population mean is between ________and________
A) 5%, 46.270,51.416
B) 95%, 0,48.843
C) 95%, 46.270,51.416
D) 5%, 0,48.843
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14
A random sample of electronic components had the following operational times before failure, in hours.
Assume the population standard deviation is
and that the population is approximately normal. Construct a 90% confidence interval for the operational time before failure.
A) (318.9, 358.4)
B) (331.3, 346.1)
C) (279.4, 397.9)
D) (332.1, 345.3)


A) (318.9, 358.4)
B) (331.3, 346.1)
C) (279.4, 397.9)
D) (332.1, 345.3)
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15
The following display from a TI-84 Plus calculator presents a 95% confidence interval.
Fill in the blanks: We are ________ confident that the population mean is between _______ and _______.
A) 5%, 0, 46.695
B) 95%, 0, 46.695
C) 95%, 42.447, 50.943
D) 5%, 42.447, 50.943

A) 5%, 0, 46.695
B) 95%, 0, 46.695
C) 95%, 42.447, 50.943
D) 5%, 42.447, 50.943
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16
A random sample of 60 adults is chosen and their mean serum cholesterol level is found to be 196 milligrams per deciliter. Assuming that the population standard deviation is
, compute a 99%
Confidence interval for the mean serum cholesterol level for adults.
A) (153, 239)
B) (172, 220)
C) (182, 210)
D) (194, 198)

Confidence interval for the mean serum cholesterol level for adults.
A) (153, 239)
B) (172, 220)
C) (182, 210)
D) (194, 198)
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17
A sample of 36 light bulbs had a mean lifetime of 508 hours. A 95% confidence interval for the population mean was
Which one of the following statements is the correct interpretation of the results?
A) 95% of the light bulbs in the sample had lifetimes between 500.8 hours and 515.2 hours
B) None of these are true.
C) We are 95% confident that the mean lifetime of all the bulbs in the population is between 500.8 hours and 515.2 hours.
D) The probability that the population mean is between 500.8 hours and 515.2 hours is 0.95.

A) 95% of the light bulbs in the sample had lifetimes between 500.8 hours and 515.2 hours
B) None of these are true.
C) We are 95% confident that the mean lifetime of all the bulbs in the population is between 500.8 hours and 515.2 hours.
D) The probability that the population mean is between 500.8 hours and 515.2 hours is 0.95.
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18
A random sample of 9 TI-89 Titanium calculators being sold over the internet had the following prices, in dollars.
Assume the population standard deviation is
and that the population is approximately normal.
Construct a 90% confidence interval for the mean price for all the TI-89's being sold over the
Internet.
A) (134.6, 147.9)
B) (122.6, 159.9)
C) (85.3, 197.2)
D) (134.2, 148.2)


Construct a 90% confidence interval for the mean price for all the TI-89's being sold over the
Internet.
A) (134.6, 147.9)
B) (122.6, 159.9)
C) (85.3, 197.2)
D) (134.2, 148.2)
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19
A random sample of 80 adults is chosen and their mean serum cholesterol level is found to be 202 milligrams per deciliter. Assume that the population standard deviation is
Based on a 95% confidence interval for the mean serum cholesterol, is it likely that the mean serum
Cholesterol is greater than 219? (Hint: you should first construct the 95% confidence interval.)
A) No
B) The likelihood cannot be determined.
C) Yes

Cholesterol is greater than 219? (Hint: you should first construct the 95% confidence interval.)
A) No
B) The likelihood cannot be determined.
C) Yes
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20
A simple random sample of kitchen toasters is to be taken to determine the mean operational lifetime in hours. Assume that the lifetimes are normally distributed with population standard
Deviation
hours.
Find the sample size needed so that a 98% confidence interval for the mean lifetime will have a margin
Of error of 4.
A) 13
B) 150
C) 3
D) 257
Deviation

Find the sample size needed so that a 98% confidence interval for the mean lifetime will have a margin
Of error of 4.
A) 13
B) 150
C) 3
D) 257
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21
Six measurements were made of the magnesium ion concentration (in parts per million, or ppm) in a city's municipal water supply, with the following results. It is reasonable to assume that the population is
Approximately normal.
Construct a 99% confidence interval for the mean magnesium ion concentration.
A) (151.4, 156.9)
B) (137.0, 171.3)
C) (138.5, 169.8)
D) (151.6, 156.7)
Approximately normal.

A) (151.4, 156.9)
B) (137.0, 171.3)
C) (138.5, 169.8)
D) (151.6, 156.7)
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22
A sample of size n = 14 has a sample mean
and sample standard deviation s = 2.1. It is reasonable to assume that the population is approximately normal. Construct a 99% confidence
Interval for the population mean
A) (10.7, 13.1)
B) (11.4, 12.4)
C) (10.2, 13.6)
D) (10.4, 13.4)

Interval for the population mean

A) (10.7, 13.1)
B) (11.4, 12.4)
C) (10.2, 13.6)
D) (10.4, 13.4)
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23
In a sample of 8 children, the mean age at which they first began to combine words was 16.41 months, with a standard deviation of 5.11 months. It is reasonable to assume that the population is
Approximately normal. Construct a 95% confidence interval for the mean age at which children first
Begin to combine words.
A) (15.7, 17.1)
B) (-17.8, 50.6)
C) (14.9, 17.9)
D) (12.1, 20.7)
Approximately normal. Construct a 95% confidence interval for the mean age at which children first
Begin to combine words.
A) (15.7, 17.1)
B) (-17.8, 50.6)
C) (14.9, 17.9)
D) (12.1, 20.7)
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24
The following MINITAB output presents a confidence interval for a population mean.
Find the critical value
for a 98% confidence interval.
A) 2.462
B) 2.457
C) 2.750
D) 2.760


A) 2.462
B) 2.457
C) 2.750
D) 2.760
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25
A sample of 69 tobacco smokers who recently completed a new smoking-cessation program were asked to rate the effectiveness of the program on a scale of 1 to 10, with 10 corresponding to
"completely effective" and 1 corresponding to "completely ineffective". The average rating was 4.5
And the standard deviation was 4.3.
Construct a 95% confidence interval for the mean score.
A) (4.0, 5.0)
B) (0, 4.5)
C) (3.5, 5.5)
D) (4.1, 4.9)
"completely effective" and 1 corresponding to "completely ineffective". The average rating was 4.5
And the standard deviation was 4.3.
Construct a 95% confidence interval for the mean score.
A) (4.0, 5.0)
B) (0, 4.5)
C) (3.5, 5.5)
D) (4.1, 4.9)
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26
The following MINITAB output presents a 95% confidence interval. 
Use the appropriate critical value along with the information in the computer output to construct a 99 confidence interval.
A) (44.845,51.835)
B) (45.185,51.495)
C) (47.206,49.474)
D) (45.681,50.999)

Use the appropriate critical value along with the information in the computer output to construct a 99 confidence interval.
A) (44.845,51.835)
B) (45.185,51.495)
C) (47.206,49.474)
D) (45.681,50.999)
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27
The following MINITAB output presents a 95% confidence interval. 
Find the sample size needed so that the 95% confidence interval will have a margin of error of 1.1.
A) 472
B) 579
C) 19
D) 336

Find the sample size needed so that the 95% confidence interval will have a margin of error of 1.1.
A) 472
B) 579
C) 19
D) 336
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28
The following MINITAB output presents a confidence interval for a population mean.
Use the information in the output to construct a 98% confidence interval.
A) (92.161,124.093)
B) (105.546,110.707)
C) (93.878,122.376)
D) (105.235,111.018)

A) (92.161,124.093)
B) (105.546,110.707)
C) (93.878,122.376)
D) (105.235,111.018)
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29
Find the point estimate for the given values of x and n. 
A) 106
B) 0.4564
C) 0.5436
D) 0.03567

A) 106
B) 0.4564
C) 0.5436
D) 0.03567
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30
Use the given data to construct a confidence interval of the requested level. x = 78, n = 124, confidence level 98%
A) (0.586, 0.672)
B) (0.394, 0.864)
C) (0.544, 0.714)
D) (0.528, 0.730)
A) (0.586, 0.672)
B) (0.394, 0.864)
C) (0.544, 0.714)
D) (0.528, 0.730)
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31
The following MINITAB output presents a confidence interval for a population mean.
How many degrees of freedom are there?
A) 66
B) 3.2236
C) 68
D) 67

A) 66
B) 3.2236
C) 68
D) 67
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32
A sample of size n = 16 is drawn from a normal population. Find the critical value
needed to construct a 90% confidence interval.
A) 1.645
B) 1.341
C) 1.746
D) 1.753

A) 1.645
B) 1.341
C) 1.746
D) 1.753
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33
Find the critical value
needed to construct a confidence interval of the given level with the given sample size.
Level 98%, sample size 5
A) 3.365
B) 2.999
C) 2.326
D) 3.747

Level 98%, sample size 5
A) 3.365
B) 2.999
C) 2.326
D) 3.747
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34
The following display from a TI-84 Plus calculator presents a 95% confidence interval.
Fill in the blanks: We are ________ confident that the population mean is between _______ and _______.
A) 95%, 52.45,54.30
B) 95%, 0,53.375
C) 5%, 0,53.375
D) 5%, 52.45,54.30

A) 95%, 52.45,54.30
B) 95%, 0,53.375
C) 5%, 0,53.375
D) 5%, 52.45,54.30
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35
Find the standard error for the given values of x and n. x = 95, n = 158
A) 0.6013
B) 95
C) 0.03895
D) 0.3987
A) 0.6013
B) 95
C) 0.03895
D) 0.3987
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36
A survey asked 26 adults how many years of education they had. The sample mean was 13.03 with a standard deviation of 3.22. It is reasonable to assume that the population is approximately normal.
Construct a 90% confidence interval for the mean number of years of education.
A) (-14.1, 14.1)
B) (12.8, 13.2)
C) (-15.0, 41.1)
D) (12.0, 14.1)
Construct a 90% confidence interval for the mean number of years of education.
A) (-14.1, 14.1)
B) (12.8, 13.2)
C) (-15.0, 41.1)
D) (12.0, 14.1)
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37
Six measurements were made of the magnesium ion concentration (in parts per million, or ppm) in a city's municipal water supply, with the following results. It is reasonable to assume that the population is
Approximately normal.
Based on a 95% confidence interval for the mean magnesium ion concentration, is it reasonable to
Believe that the mean magnesium ion concentration may be greater than 199.5? (Hint: you should
First calculate the 95% confidence interval for the mean magnesium ion concentration.)
A) No
B) The likelihood cannot be determined.
C) Yes
Approximately normal.

Believe that the mean magnesium ion concentration may be greater than 199.5? (Hint: you should
First calculate the 95% confidence interval for the mean magnesium ion concentration.)
A) No
B) The likelihood cannot be determined.
C) Yes
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38
The following MINITAB output presents a 95% confidence interval. 
Find the sample size needed so that the 99% confidence interval will have a margin of error of 1.3.
A) 301
B) 14
C) 175
D) 246

Find the sample size needed so that the 99% confidence interval will have a margin of error of 1.3.
A) 301
B) 14
C) 175
D) 246
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39
Find the margin of error for the given confidence level and values of x and n. x = 118, n = 252, confidence level 98%
A) 0.03143
B) 0.07311
C) 0.4683
D) 0.5317
A) 0.03143
B) 0.07311
C) 0.4683
D) 0.5317
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40
Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximately normal.
Construct a 95% confidence interval for the mean weight.
A) (21.847, 22.126)
B) (21.782, 22.192)
C) (21.853, 22.120)
D) (21.790, 22.183)

A) (21.847, 22.126)
B) (21.782, 22.192)
C) (21.853, 22.120)
D) (21.790, 22.183)
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41
In a survey of 318 registered voters, 176 of them wished to see Mayor Waffleskate lose her next election. Construct a 95% confidence interval for the proportion of registered voter who want to see
Mayor Waffleskate defeated.
A) (0.526, 0.581)
B) (0.446, 0.661)
C) 0.508, 0.599)
D) (0.499, 0.608)
Mayor Waffleskate defeated.
A) (0.526, 0.581)
B) (0.446, 0.661)
C) 0.508, 0.599)
D) (0.499, 0.608)
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42
Following are the heights in inches of 12 two-year-old apple trees. Assume that the population is normally distributed.
Construct a 99% confidence interval for the population standard deviation σ.
A) (4.39, 13.31)
B) (4.20, 12.75)
C) (4.32, 13.85)
D) (4.49, 12.79)

A) (4.39, 13.31)
B) (4.20, 12.75)
C) (4.32, 13.85)
D) (4.49, 12.79)
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43
The following MINITAB output presents a confidence interval for a proportion.
Fill in the blanks: We are ________ confident that the population mean is between _______ and _______.
A) 5%, 0.467457,0.631135
B) 95%, 0.480610,0.617982
C) 95%, 0.467457,0.631135
D) 97.5%, 0.480610,0.617982

A) 5%, 0.467457,0.631135
B) 95%, 0.480610,0.617982
C) 95%, 0.467457,0.631135
D) 97.5%, 0.480610,0.617982
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44
Scores on the math SAT are normally distributed. A sample of 15 SAT scores had a standard deviation s = 80. Construct a 95% confidence interval for the population standard deviation 
A) (61.51, 116.78)
B) (57.09, 119.62)
C) (59.10, 123.82)
D) (58.57, 126.17)

A) (61.51, 116.78)
B) (57.09, 119.62)
C) (59.10, 123.82)
D) (58.57, 126.17)
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45
Measurements were made of the milk fat content (in percent) in six brands of feta cheese (a variety of goat cheese), with the following results. Assume that the population is normally distributed.
Construct a 90% confidence interval for the population standard deviation σ.
A) (1.66, 5.16)
B) (1.56, 4.32)
C) (1.70, 4.73)
D) (1.82, 4.35)

A) (1.66, 5.16)
B) (1.56, 4.32)
C) (1.70, 4.73)
D) (1.82, 4.35)
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46
In a survey of 464 registered voters, 133 of them wished to see Mayor Waffleskate lose her next election. The Waffleskate campaign claims that no more than 29% of registered voters wish to see
Her defeated. Does the 98% confidence interval for the proportion support this claim? (Hint: you
Should first construct the 98% confidence interval for the proportion of registered voters who wish
To see Waffleskate defeated.)
A) Yes
B) No
C) The reasonableness of the claim cannot be determined.
Her defeated. Does the 98% confidence interval for the proportion support this claim? (Hint: you
Should first construct the 98% confidence interval for the proportion of registered voters who wish
To see Waffleskate defeated.)
A) Yes
B) No
C) The reasonableness of the claim cannot be determined.
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47
An Internet service provider sampled 590 customers and found that 85 of them experienced an interruption in their service during the previous month. Find a point estimate for the population
Proportion of all customers who experienced an interruption.
A) 85
B) 0.1441
C) 0.8559
D) 0.0145
Proportion of all customers who experienced an interruption.
A) 85
B) 0.1441
C) 0.8559
D) 0.0145
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48
A simple random sample of size 41 has mean
and standard deviation s = 15.73. The population distribution is unknown. Determine the correct method of finding a 90% confidence
Interval for the population mean and compute it.
A) t-method: (66.41, 74.69)
B) z-method: (66.41, 74.69)
C) z-method: (66.51, 74.59)
D) Cannot compute: the population size is too small.

Interval for the population mean and compute it.
A) t-method: (66.41, 74.69)
B) z-method: (66.41, 74.69)
C) z-method: (66.51, 74.59)
D) Cannot compute: the population size is too small.
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49
An Internet service provider sampled 540 customers and found that 55 of them experienced an interruption in their service during the previous month. Construct a 95% confidence interval for the
Proportion of all customers who have experienced a service interruption.
A) (0.076, 0.127)
B) (0.873, 0.924)
C) (0.102, 0.898)
D) (0.080, 0.123)
Proportion of all customers who have experienced a service interruption.
A) (0.076, 0.127)
B) (0.873, 0.924)
C) (0.102, 0.898)
D) (0.080, 0.123)
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50
Find the critical values for a 98% confidence interval using the chi-square distribution with 20 degrees of freedom.
A) 7.633, 36.191
B) 7.434, 39.997
C) 8.260, 37.566
D) 9.237, 35.020
A) 7.633, 36.191
B) 7.434, 39.997
C) 8.260, 37.566
D) 9.237, 35.020
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51
An Internet service provider sampled 570 customers and found that 70 of them experienced an interruption in their service during the previous month. The company claims that no more than 10%
Of customers experienced an interruption in the past month. Does the 95% confidence interval for
The proportion support this claim? (Hint: you should first construct the 95% confidence interval for
The proportion of customers who experienced an interruption.)
A) Yes
B) No
C) The reasonableness of the claim cannot be determined.
Of customers experienced an interruption in the past month. Does the 95% confidence interval for
The proportion support this claim? (Hint: you should first construct the 95% confidence interval for
The proportion of customers who experienced an interruption.)
A) Yes
B) No
C) The reasonableness of the claim cannot be determined.
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52
Twenty-two concrete blocks were sampled and tested for crushing strength in order to estimate the proportion that were sufficiently strong for a certain application. Eighteen of the 22 blocks were
Sufficiently strong. Use the small-sample method to construct a 95% confidence interval for the
Proportion of blocks that are sufficiently strong.
A) (0.593, 0.945)
B) (0.657, 0.979)
C) (0.633, 0.905)
D) (0.607, 0.931)
Sufficiently strong. Use the small-sample method to construct a 95% confidence interval for the
Proportion of blocks that are sufficiently strong.
A) (0.593, 0.945)
B) (0.657, 0.979)
C) (0.633, 0.905)
D) (0.607, 0.931)
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53
The following display from a TI-84 Plus calculator presents a 99% confidence interval for a proportion.
Fill in the blanks: We are ________ confident that the population mean is between _______ and _______.
A) 99%, 0,0.633097
B) 1%, 0,0.633097
C) 99%, 0.473812,0.792382
D) 1%, 0.473812,0.792382

A) 99%, 0,0.633097
B) 1%, 0,0.633097
C) 99%, 0.473812,0.792382
D) 1%, 0.473812,0.792382
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54
The following MINITAB output presents a confidence interval for a proportion.
Use the information in the display to construct a 98% confidence interval for p.
A) (0.592, 0.733)
B) (0.603, 0.722)
C) (0.579, 0.747)
D) (0.570, 0.756)

A) (0.592, 0.733)
B) (0.603, 0.722)
C) (0.579, 0.747)
D) (0.570, 0.756)
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55
Find the critical values for a 99% confidence interval using the chi-square distribution with 20 degrees of freedom.
A) 8.260, 37.566
B) 6.844, 38.582
C) 7.434, 39.997
D) 7.633, 36.191
A) 8.260, 37.566
B) 6.844, 38.582
C) 7.434, 39.997
D) 7.633, 36.191
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56
Construct a 90% confidence interval for the population standard deviation
if a sample of size 8 has standard deviation s = 20.
A) (14.11, 35.94)
B) (14.37, 34.22)
C) (15.26, 31.44)
D) (13.44, 32.01)

A) (14.11, 35.94)
B) (14.37, 34.22)
C) (15.26, 31.44)
D) (13.44, 32.01)
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57
In a survey of 331 registered voters, 158 of them wished to see Mayor Waffleskate lose her next election. Find a point estimate for the proportion of registered voters who wish to see Mayor
Waffleskate defeated.
A) 0.4773
B) 158
C) 0.5227
D) 0.02745
Waffleskate defeated.
A) 0.4773
B) 158
C) 0.5227
D) 0.02745
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58
A researcher wants to construct a 98% confidence interval for the proportion of elementary school students in Seward County who receive free or reduced-price school lunches. What sample size is
Needed so that the confidence interval will have a margin of error of 0.09?
A) 119
B) 167
C) 7
D) 16
Needed so that the confidence interval will have a margin of error of 0.09?
A) 119
B) 167
C) 7
D) 16
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59
A researcher wants to construct a 98% confidence interval for the proportion of elementary school students in Seward County who receive free or reduced-price school lunches. A state-wide survey
Indicates that the proportion is 0.60. Using this estimate, what sample size is needed so that the
Confidence interval will have a margin of error of 0.07?
A) 265
B) 19
C) 8
D) 189
Indicates that the proportion is 0.60. Using this estimate, what sample size is needed so that the
Confidence interval will have a margin of error of 0.07?
A) 265
B) 19
C) 8
D) 189
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60
The following display from a TI-84 Plus calculator presents a 99% confidence interval for a proportion.
Use the information in the display to construct a 95% confidence interval for p.
A) (0.348, 0.652)
B) (0.285, 0.715)
C) (0.363, 0.637)
D) (0.337, 0.663)

A) (0.348, 0.652)
B) (0.285, 0.715)
C) (0.363, 0.637)
D) (0.337, 0.663)
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61
A simple random sample of size 51 has mean
. The population distribution is approximately normal, with standard deviation
. Determine the correct method of finding
A 95% confidence interval for the population mean and compute it.
A) z-method: (66.15, 75.29)
B) Cannot compute: the population size is too small.
C) t-method: (66.15, 75.29)
D) z-method: (66.26, 75.18)


A 95% confidence interval for the population mean and compute it.
A) z-method: (66.15, 75.29)
B) Cannot compute: the population size is too small.
C) t-method: (66.15, 75.29)
D) z-method: (66.26, 75.18)
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62
A simple random sample of size 20 has mean
and standard deviation s = 16.77. The population distribution is unknown. Determine the correct method of finding a 90% confidence
Interval for the population mean and compute it.
A) Cannot compute: the population size is too small.
B) z-method: (64.72, 77.06)
C) t-method: (64.41, 77.37)
D) z-method: (64.41, 77.37)

Interval for the population mean and compute it.
A) Cannot compute: the population size is too small.
B) z-method: (64.72, 77.06)
C) t-method: (64.41, 77.37)
D) z-method: (64.41, 77.37)
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