Question 1

Data collected by child development scientists produced the following  90%90 \%90%  confidence interval for the average age (in months)  at which children say their first word:  10.6<μ(10.6 < \mu (10.6<μ(  age  ) <13.7)  < 13.7) <13.7 .

Accepted Answer

A)   Based on this sample, we can say, with  90%90 \%90%  confidence, that the mean age at which children say their first word is between  10.610.610.6  and  13.713.713.7  months.
B)    90%90 \%90%  of the children in this sample said their first word when they were between  10.610.610.6  and  13.713.713.7  months old.
C)   We are  90%90 \%90%  sure that a child will say his first word when he is between  10.610.610.6  and  13.713.713.7  months old.
D)   We are  90%90 \%90%  sure that the average age at which children in this sample said their first word was between  10.610.610.6  and  13.713.713.7  months.
E)   If we took many random samples of children, about  90%90 \%90%  of them would produce this confidence interval. F) B) and E)
G) B) and D)

Question 2

Use the t-tables, software, or a calculator to estimate the indicated P-value: P-value for  t≥\mathrm { t } \geqt≥   1.761.761.76  with 24 degrees of freedom.

Accepted Answer

A)    0.05920.05920.0592 
B)    0.95440.95440.9544 
C)    0.04560.04560.0456 
D)    0.09120.09120.0912 
E)    0.02280.02280.0228  F) A) and E)
G) C) and D)

Question 3

Thirty randomly selected students took the calculus final.If the sample mean was 95 and the standard deviation was 6.4, construct a 99% confidence interval for the mean score of all students.

Accepted Answer

A)  (92.12, 97.88) 
B)  (91.78, 98.22) 
C)  (93.01, 98.21) 
D)  (93.01, 96.99) 
E)  (91.79, 98.21) F) A) and B)
G) A) and C)

Question 4

A random sample of clients at a weight loss center were given a dietary supplement to see if it would promote weight loss.The center reported that the 100 clients lost an average of 48 pounds, and that a 95% confidence interval for the mean weight loss this supplement produced has a margin of error of ±7 pounds.

Accepted Answer

A)  95% of the people who use this supplement will lose between 41 and 55 pounds.
B)  The average weight loss of clients who take this supplement will be between 41 and 55 pounds.
C)  95% of the clients in the study lost between 41 and 55 pounds.
D)  We are 95% sure that the average weight loss among the clients in this study was between 41 and 55 pounds.
E)  We are 95% confident that the mean weight loss produced by the supplement in weight loss center clients is between 41 and 55 pounds.F) C) and E)
G) B) and D)

Question 5

Use the  ttt -tables, software, or a calculator to estimate the indicated  PPP -value:
 PPP -value for  t≤1.76t \leq 1.76t≤1.76  with 24 degrees of freedom

Accepted Answer

A)    0.02280.02280.0228 
B)    0.90880.90880.9088 
C)    0.95440.95440.9544 
D)    0.97720.97720.9772 
E)    0.04560.04560.0456  F) A) and B)
G) B) and C)

Question 6

The principal randomly selected six students to take an aptitude test.Their scores were:  71.681.088.980.478.172.0\begin{array} { l l l l l l } 71.6 & 81.0 & 88.9 & 80.4 & 78.1 & 72.0\end{array}71.6​81.0​88.9​80.4​78.1​72.0​  Determine a 90% confidence interval for the mean score for all students.

Accepted Answer

A)  (73.26, 84.07) 
B)  (84.07, 73.26) 
C)  (83.97, 83.97) 
D)  (73.36, 83.97)  E) None of the above
F) A) and C) D

Question 7

A certain population is bimodal.We want to estimate its mean, so we will collect a sample.
Which should be true if we use a large sample rather than a small one?
I.The distribution of our sample data will be more clearly bimodal.
II.The sampling distribution of the sample means will be approximately normal.
III.The variability of the sample means will be smaller.

Accepted Answer

A)  I only
B)  II only
C)  III only
D)  II and III
E)  I, II, and IIIF) B) and C)
G) A) and B)

Question 8

The amounts (in ounces)  of juice in eight randomly selected juice bottles are:  15.015.915.815.715.415.215.215.315.0\quad15 .9\quad15 .8\quad15 .7\quad15 .4\quad15 .2\quad15 .2\quad15 .315.015.915.815.715.415.215.215.3  Construct a 98% confidence interval for the mean amount of juice in all such bottles.

Accepted Answer

A)  (15.04, 15.84) 
B)  (15.74, 15.14) 
C)  (15.14, 15.74) 
D)  (15.04, 15.74)  E) None of the above
F) C) and D)

Question 9

Using the t-tables, software, or a calculator, estimate the critical value of t for the given
Confidence interval and degrees of freedom: 80% confidence interval with df = 11

Accepted Answer

A)  1.372
B)  2.718
C)  1.280
D)  1.356
E)  1.363F) C) and E)
G) A) and C)

Question 10

Using the t-tables, software, or a calculator, estimate the critical value of t for the given confidence interval and degrees of freedom: 90% confidence interval with df = 4.

Accepted Answer

A)  1.645
B)  1.533
C)  2.353
D)  2.132
E)  4.604F) B) and C)
G) A) and E)

Question 11

A credit union took a random sample of 40 accounts and yielded the following  90%90 \%90%  confidence interval for the mean checking account balance at the institution:
 $2183<μ\$ 2183 < \mu$2183<μ  (balance)   <$3828< \$ 3828<$3828 .

Accepted Answer

A)   If we took random samples of checking accounts at this credit union, about nine out of ten of them would produce this confidence interval.
B)   We are  90%90 \%90%  confident that the mean checking account balance at this credit union is between  $2183\$ 2183$2183  and  $3828\$ 3828$3828 , based on this sample.
C)   We are  90%90 \%90%  confident that the mean checking account balance in the U.S. is between  $2183\$ 2183$2183  and \$3828.
D)   About 9 out of 10 people have a checking account balance between  $2183\$ 2183$2183  and  $3828\$ 3828$3828 .
E)   We are  90%90 \%90%  sure that the mean balance for checking accounts in the sample was between  $2183\$ 2183$2183  and  $3828\$ 3828$3828  F) A) and B)
G) B) and E)

Question 12

Which is true about a 95% confidence interval based on a given sample?
I.The interval contains 95% of the population.
II.Results from 95% of all samples will lie in the interval.
III.The interval is narrower than a 98% confidence interval would be.

Accepted Answer

A)  I only
B)  II only
C)  III only
D)  II and III onlyE) B) and D)
F) None of the above C

Question 13

Based on a sample of size 49 , a  95%95 \%95%  confidence interval for the mean score of all students,  μ\muμ , on an aptitude test is from  59.259.259.2  to  64.864.864.8 . Find the margin of error.

Accepted Answer

A)    5.65.65.6 
B)    0.780.780.78 
C)    0.050.050.05 
D)    2.82.82.8 
E)   There is not enough information to find the margin of error. F) A) and B)
G) B) and C)

Question 14

A certain population is strongly skewed to the right.We want to estimate its mean, so we will collect a sample.Which should be true if we use a large sample rather than a small one?
I.The distribution of our sample data will be closer to normal.
II.The sampling model of the sample means will be closer to normal.
III.The variability of the sample means will be greater.

Accepted Answer

A)  I only
B)  II only
C)  III only
D)  I and III only
E)  II and III onlyF) B) and D)
G) B) and C)

Question 15

Based on a sample of 30 randomly selected years, a 90% confidence interval for the meanannual precipitation in one city is from 48.7 inches to 51.3 inches.Find the margin of error.

Accepted Answer

A)  0.39 inches
B)  0.10 inches
C)  2.6 inches
D)  1.3 inches
E)  There is not enough information to find the margin of error.F) D) and E)
G) All of the above

Question 16

Analysis of a random sample of 250 Illinois nurses produced a  95%95 \%95%  confidence interval for the mean annual salary of  $42,846<μ\$ 42,846 < \mu$42,846<μ  (Nurse Salary)   <$49,686< \$ 49,686<$49,686

Accepted Answer

A)   About  95%95 \%95%  of Illinois nurses earn between  $42,846\$ 42,846$42,846  and  $49,686\$ 49,686$49,686 .
B)   We are  95%95 \%95%  confident that the average nurse salary in the U.S. is between  $42,846\$ 42,846$42,846  and  $49,686\$ 49,686$49,686 .
C)   If we took many random samples of Illinois nurses, about  95%95 \%95%  of them would produce this confidence interval.
D)   We are  95%95 \%95%  confident that the interval from  $42,846\$ 42,846$42,846  to  $49,686\$ 49,686$49,686  contains the true mean of all Illinois nurses.
E)   About  95%95 \%95%  of the nurses surveyed earn between  $42,846\$ 42,846$42,846  and  $49,686\$ 49,686$49,686 . F) A) and E)
G) A) and B)

Question 17

n=12,xˉ=28.0,s=5.7n = 12 , \bar { x } = 28.0 , s = 5.7n=12,xˉ=28.0,s=5.7 . Find a  99%99 \%99%  confidence interval for the mean.

Accepted Answer

A)    (22.79,33.21) ( 22.79,33.21 ) (22.79,33.21)  
B)    (22.89,33.11) ( 22.89,33.11 ) (22.89,33.11)  
C)    (22.79,33.09) ( 22.79,33.09 ) (22.79,33.09)  
D)    (22.91,33.09) ( 22.91,33.09 ) (22.91,33.09)  
E)    (23.53,32.47) ( 23.53,32.47 ) (23.53,32.47)   F) A) and E)
G) A) and C)

Question 18

A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test.Their mean score is 76.2 and their standarddeviation is 21.4.Construct the 95% confidence interval for the mean score of all such subjects.

Accepted Answer

A)  (64.2, 83.2) 
B)  (64.2, 88.2) 
C)  (74.6, 77.8) 
D)  (69.2, 83.2) 
E)  (67.7, 84.7) F) A) and E)
G) C) and D)

Question 19

n=10,xˉ=13.0,s=4.5n = 10 , \bar { x } = 13.0 , s = 4.5n=10,xˉ=13.0,s=4.5 . Find a  95%95 \%95%  confidence interval for the mean.

Accepted Answer

A)    (9.80,16.20) ( 9.80,16.20 ) (9.80,16.20)  
B)    (9.78,16.22) ( 9.78,16.22 ) (9.78,16.22)  
C)    (10.39,15.61) ( 10.39,15.61 ) (10.39,15.61)  
D)    (9.83,16.20) ( 9.83,16.20 ) (9.83,16.20)  
E)    (9.83,16.17) ( 9.83,16.17 ) (9.83,16.17)   F) A) and D)
G) D) and E) B

Exam 14: Confidence Intervals for Means

Data collected by child development scientists produced the following $90 \%$ confidence interval for the average age (in months) at which children say their first word: $10.6 < \mu ($ age $) < 13.7$ .

Correct Answer
verified

Use the t-tables, software, or a calculator to estimate the indicated P-value: P-value for $\mathrm { t } \geq$ $1.76$ with 24 degrees of freedom.

Correct Answer
verified

Thirty randomly selected students took the calculus final.If the sample mean was 95 and the standard deviation was 6.4, construct a 99% confidence interval for the mean score of all students.

Correct Answer
verified

Correct Answer
verified

Use the $t$ -tables, software, or a calculator to estimate the indicated $P$ -value: $P$ -value for $t \leq 1.76$ with 24 degrees of freedom

Correct Answer
verified

The principal randomly selected six students to take an aptitude test.Their scores were: $\begin{array} { l l l l l l } 71.6 & 81.0 & 88.9 & 80.4 & 78.1 & 72.0\end{array}$ Determine a 90% confidence interval for the mean score for all students.

Correct Answer
verified
D

Correct Answer
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The amounts (in ounces) of juice in eight randomly selected juice bottles are: $15.0\quad15 .9\quad15 .8\quad15 .7\quad15 .4\quad15 .2\quad15 .2\quad15 .3$ Construct a 98% confidence interval for the mean amount of juice in all such bottles.

Correct Answer
verified

Using the t-tables, software, or a calculator, estimate the critical value of t for the given Confidence interval and degrees of freedom: 80% confidence interval with df = 11

Correct Answer
verified

Using the t-tables, software, or a calculator, estimate the critical value of t for the given confidence interval and degrees of freedom: 90% confidence interval with df = 4.

Correct Answer
verified

A credit union took a random sample of 40 accounts and yielded the following $90 \%$ confidence interval for the mean checking account balance at the institution: $\$ 2183 < \mu$ (balance) $< \$ 3828$ .

Correct Answer
verified

Which is true about a 95% confidence interval based on a given sample? I.The interval contains 95% of the population. II.Results from 95% of all samples will lie in the interval. III.The interval is narrower than a 98% confidence interval would be.

Correct Answer
verified
C

Based on a sample of size 49 , a $95 \%$ confidence interval for the mean score of all students, $\mu$ , on an aptitude test is from $59.2$ to $64.8$ . Find the margin of error.

Correct Answer
verified

Correct Answer
verified

Based on a sample of 30 randomly selected years, a 90% confidence interval for the meanannual precipitation in one city is from 48.7 inches to 51.3 inches.Find the margin of error.

Correct Answer
verified

Analysis of a random sample of 250 Illinois nurses produced a $95 \%$ confidence interval for the mean annual salary of $\$ 42,846 < \mu$ (Nurse Salary) $< \$ 49,686$

Correct Answer
verified

$n = 12 , \bar { x } = 28.0 , s = 5.7$ . Find a $99 \%$ confidence interval for the mean.

Correct Answer
verified

A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test.Their mean score is 76.2 and their standarddeviation is 21.4.Construct the 95% confidence interval for the mean score of all such subjects.

Correct Answer
verified

$n = 10 , \bar { x } = 13.0 , s = 4.5$ . Find a $95 \%$ confidence interval for the mean.

Correct Answer
verified
B

Exam 14: Confidence Intervals for Means

Data collected by child development scientists produced the following 90%90 \%90% confidence interval for the average age (in months) at which children say their first word: 10.6<μ(10.6 < \mu (10.6<μ( age ) <13.7) < 13.7) <13.7 .

Correct AnswerverifiedShow Answer

Use the t-tables, software, or a calculator to estimate the indicated P-value: P-value for t≥\mathrm { t } \geqt≥ 1.761.761.76 with 24 degrees of freedom.

Correct AnswerverifiedShow Answer

Thirty randomly selected students took the calculus final.If the sample mean was 95 and the standard deviation was 6.4, construct a 99% confidence interval for the mean score of all students.

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Use the ttt -tables, software, or a calculator to estimate the indicated PPP -value: PPP -value for t≤1.76t \leq 1.76t≤1.76 with 24 degrees of freedom

Correct AnswerverifiedShow Answer

Correct AnswerverifiedD

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Using the t-tables, software, or a calculator, estimate the critical value of t for the given Confidence interval and degrees of freedom: 80% confidence interval with df = 11

Correct AnswerverifiedShow Answer

Using the t-tables, software, or a calculator, estimate the critical value of t for the given confidence interval and degrees of freedom: 90% confidence interval with df = 4.

Correct AnswerverifiedShow Answer

A credit union took a random sample of 40 accounts and yielded the following 90%90 \%90% confidence interval for the mean checking account balance at the institution: $2183<μ\$ 2183 < \mu$2183<μ (balance) <$3828< \$ 3828<$3828 .

Correct AnswerverifiedShow Answer

Which is true about a 95% confidence interval based on a given sample? I.The interval contains 95% of the population. II.Results from 95% of all samples will lie in the interval. III.The interval is narrower than a 98% confidence interval would be.

Correct AnswerverifiedC

Based on a sample of size 49 , a 95%95 \%95% confidence interval for the mean score of all students, μ\muμ , on an aptitude test is from 59.259.259.2 to 64.864.864.8 . Find the margin of error.

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Based on a sample of 30 randomly selected years, a 90% confidence interval for the meanannual precipitation in one city is from 48.7 inches to 51.3 inches.Find the margin of error.

Correct AnswerverifiedShow Answer

Analysis of a random sample of 250 Illinois nurses produced a 95%95 \%95% confidence interval for the mean annual salary of $42,846<μ\$ 42,846 < \mu$42,846<μ (Nurse Salary) <$49,686< \$ 49,686<$49,686

Correct AnswerverifiedShow Answer

n=12,xˉ=28.0,s=5.7n = 12 , \bar { x } = 28.0 , s = 5.7n=12,xˉ=28.0,s=5.7 . Find a 99%99 \%99% confidence interval for the mean.

Correct AnswerverifiedShow Answer

A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test.Their mean score is 76.2 and their standarddeviation is 21.4.Construct the 95% confidence interval for the mean score of all such subjects.

Correct AnswerverifiedShow Answer

n=10,xˉ=13.0,s=4.5n = 10 , \bar { x } = 13.0 , s = 4.5n=10,xˉ=13.0,s=4.5 . Find a 95%95 \%95% confidence interval for the mean.

Correct AnswerverifiedB

Data collected by child development scientists produced the following $90 \%$ confidence interval for the average age (in months) at which children say their first word: $10.6 < \mu ($ age $) < 13.7$ .

Correct Answer
verified

Use the t-tables, software, or a calculator to estimate the indicated P-value: P-value for $\mathrm { t } \geq$ $1.76$ with 24 degrees of freedom.

Correct Answer
verified

Correct Answer
verified

Correct Answer
verified

Use the $t$ -tables, software, or a calculator to estimate the indicated $P$ -value: $P$ -value for $t \leq 1.76$ with 24 degrees of freedom

Correct Answer
verified

Correct Answer
verified
D

Correct Answer
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Correct Answer
verified

Correct Answer
verified

Correct Answer
verified

A credit union took a random sample of 40 accounts and yielded the following $90 \%$ confidence interval for the mean checking account balance at the institution: $\$ 2183 < \mu$ (balance) $< \$ 3828$ .

Correct Answer
verified

Correct Answer
verified
C

Based on a sample of size 49 , a $95 \%$ confidence interval for the mean score of all students, $\mu$ , on an aptitude test is from $59.2$ to $64.8$ . Find the margin of error.

Correct Answer
verified

Correct Answer
verified

Correct Answer
verified

Analysis of a random sample of 250 Illinois nurses produced a $95 \%$ confidence interval for the mean annual salary of $\$ 42,846 < \mu$ (Nurse Salary) $< \$ 49,686$

Correct Answer
verified

$n = 12 , \bar { x } = 28.0 , s = 5.7$ . Find a $99 \%$ confidence interval for the mean.

Correct Answer
verified

Correct Answer
verified

$n = 10 , \bar { x } = 13.0 , s = 4.5$ . Find a $95 \%$ confidence interval for the mean.

Correct Answer
verified
B