Which of the following statistics are unbiased estimators of population​ parameters?
 
Choose the correct answer below.  Select all that apply.
A. Sample variance used to estimate a population variance.
B. Sample range used to estimate a population range.
C. Sample proportion used to estimate a population proportion.
D. Sample mean used to estimate a population mean.
E. Sample median used to estimate a population median.
F. Sample standard deviation used to estimate a population standard deviation.
 
Fill in the blank.
_____________ is the distribution of all values of the statistic when all possible samples of the same
size n are taken from the same population.

The sampling distribution of statistics  
Which of the following is NOT a property of the sampling distribution of the sample​ mean? Choose the correct answer below.
 
A. The distribution of the sample mean tends to be skewed to the right or left.
B. The expected value of the sample mean is equal to the population mean.
C. The sample means target the value of the population mean.
D. The mean of the sample means is the population mean.  
Which of the following is NOT a property of the sampling distribution of the​ variance? Choose the correct answer below.
 
A. The distribution of sample variances tends to be a normal distribution.
B. The mean of the sample variances is the population variance. 
C. The sample variances target the value of the population variance.
D. The expected value of the sample variance is equal to the population variance.  
_________is the distribution of all values of the statistic when all possible samples of the same size n are taken
from the same population

the sampling distribution of a statistic  
Which of the following is a biased​ estimator? That​ is, which of the following does not target the population​ parameter? Choose the correct answer below.
 
Proportion
Median
Variance
Mean  
Homework 6.4
 
Assume that females have pulse rates that are normally distributed with a mean of u  = 73.0   beats per minute and a standard deviation of sigma equals 12.5   beats per minute. Complete parts​ (a) through​ (c) below.
 
If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 80 beats per minute.
 
The probability is 0.7122 (Round to four decimal places as​ needed.)
 
80 – 73 / 12.5 = 0.56

TI84 Plus Instructions

2nd     VARS     Normalcdf
lower: -9999
upper: 0.56
u: 0
σ: 1
=   0.7122
 
b. If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than
80 beats per minute.
 
The probability is 0.9974
​(Round to four decimal places as​ needed.)

25 / 2 = 12.5
12.5 / √25 = 2.5
(80 – 73) / 2.5 = 2.8
 
TI84 Plus Instructions

2ND        VARS       Normalcdf
lower: -9999
upper: 2.8
u: 0
σ: 1
= 0.9974
 
c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?
 
A. Since the distribution is of sample​ means, not​ individuals, the distribution is a normal distribution for any sample size.
B. Since the original population has a normal​ distribution, the distribution of sample means is a normal
distribution for any sample size.
C. Since the distribution is of​ individuals, not sample​ means, the distribution is a normal distribution for any sample size.
D. Since the mean pulse rate exceeds​ 30, the distribution of sample means is a normal distribution for any sample size.  
Assume that females have pulse rates that are normally distributed with a mean of mu = 75.0 beats per minute
and a standard deviation of sigma equals 12.5 beats per minute. Complete parts​ (a) through​ (c) below.

a. If 1 adult female is randomly​ selected,
find the probability that her pulse rate is between 68 beats per minute and 82 beats per minute.
 
The probability is 0.4545
​(Round to four decimal places as​ needed.)
 
68 – 75 / 12.5 = -.56
82 – 75 / 12.5 = .56
 
TI84 Plus Instructions

2ND       VARS       Normalcdf: -9999       -0.56       0       1      =    0.2877 (4 decimal places)
2ND       VARS       Normalcdf: -9999        0.56        0       1      =    0.7122 (4 decimal places)
0.7122 - 0.2877 = .4245
 
b. If 4 adult females are randomly​ selected, find the probability that they have pulse rates with a mean between
68 beats per minute and 82 beats per minute.

The probability is 0.7372
​(Round to four decimal places as​ needed.)
 
12.5 / √4 = 6.25
68 – 75 / 6.25 =    -1.12
82 – 75 / 6.25 =     1.12
 
TI84 Plus Instructions

2ND         VARS
Normalcdf: -9999      -1.12       0      1       = 0.1314 (4 decimal places)
Normalcdf: -9999      1.12        0       1      = 0.8686 (4 decimal places)
0.8686 - 0.1314 = 0.7372
 
c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?
 
A. Since the original population has a normal​ distribution, the distribution of sample means is a normal
distribution for any sample size.
B. Since the distribution is of​ individuals, not sample​ means, the distribution is a normal distribution for any sample size.
C. Since the mean pulse rate exceeds​ 30, the distribution of sample means is a normal distribution for any sample size.
D. Since the distribution is of sample​ means, not​ individuals, the distribution is a normal distribution for any sample size.
 
An elevator has a placard stating that the maximum capacity is 2370 lb --- 15 passengers.​
So, 15 adult male passengers can have a mean weight of up to 2370 / 15 equals 158 pounds. 
If the elevator is loaded with 15   adult male​ passengers, find the probability that it is overloaded because
they have a mean weight greater than 158   lb.​ (Assume that weights of males are normally distributed with
a mean of 165 lb and a standard deviation of 34 lb ​.)  Does this elevator appear to be​ safe?
 
The probability the elevator is overloaded is 0.7881
(Round to four decimal places as​ needed.)

(158 – 165) / (34 / √15) = -0.7973789242 (-0.80) 2 decimal places
 
TI84 Plus Instructions

2ND       VARS      Normalcdf: -9999     -0.80      0      1      =      0.2118553337 (round to 4 decimals) 0.2119
1 - 0.2119 = 0.7881
 
Does this elevator appear to be​ safe?
 
A. ​Yes, there is a good chance that 15 randomly selected people will not exceed the elevator capacity.
B. No, there is a good chance that 15 randomly selected people will exceed the elevator capacity.
C. Yes, 15 randomly selected people will always be under the weight limit.
D. ​No, 15 randomly selected people will never be under the weight limit.
 
An engineer is going to redesign an ejection seat for an airplane.
The seat was designed for pilots weighing between 120 lb and 161 lb.
The new population of pilots has normally distributed weights with a mean of 130 lb. and a standard deviation of 28.2 lb.
 
If a pilot is randomly​ selected, find the probability that his weight is between 120 lb. and 161 lb.
 
The probability is approximately 0.5012  (Round to four decimal places as​ needed.)
 
120 – 130 / 28.2 =   -0.35
161 – 130 / 28.2 =    1.10
 
TI84 Plus Instructions

2ND   VARS
Normalcdf:    -9999     -0.35      0     1    = 0.3631
Normalcdf:    -9999      1.10       0     1   = 0.8643
0.8643 - 0.3632 = 0.5012
 
If 35 different pilots are randomly​ selected, find the probability that their mean weight is between 120 lb and 161 lb
The probability is approximately 0.9820
​(Round to four decimal places as​ needed.)
 
28.2 / √35 = 4.766669997
120 – 130 / 4.766669997 = -2.10
161 – 130 / 4.766669997 = 6.50
 
TI84 Plus Instructions

2ND       VARS    Normalcdf:   -9999       -0.2.10       0      1    = 0.0179
2ND       VARS     Normalcdf:   -9999        6.50        0        1     = 1 (we need to use 0.9999 when results are = to 1)
0.9999 - 0.0179 = 0.9820
 
c. When redesigning the ejection​ seat, which probability is more​ relevant?
 
A. Part​ (a) because the seat performance for a sample of pilots is more important.
B. Part​ (b) because the seat performance for a sample of pilots is more important.
C. Part​ (b) because the seat performance for a single pilot is more important.
D. Part​ (a) because the seat performance for a single pilot is more important.  
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between
150 lb and 201 lb. The new population of pilots has normally distributed weights with a mean of 158 lb.
and a standard deviation of 29.9 lb.
 
If a pilot is randomly​ selected, find the probability that his weight is between 120 lb. and 161 lb.
 
The probability is approximately 0.5316
​(Round to four decimal places as​ needed.)
 
150 – 158 / 29.9 = -0.27
201 – 158 / 29.9 = 1.44
 
TI84 Plus Instructions

Press 2ND        VARS   Normalcdf:      -9999       -0.35      0       1 = 0.3935
Press 2ND        VARS   Normalcdf:      -9999        1.10       0      1 = 0.9251
0.9251 - 0.3935 = 0.5316
 
If 31 different pilots are randomly​ selected, find the probability that their mean weight is between 150 lb and 201 lb
 
The probability is approximately 0.9319
​(Round to four decimal places as​ needed.)
 
29.9 / √31 = 5.370198531
150 – 158 / 5.370198531 =   -1.49
201 - 158 / 5.370198531 =    8.01
 
TI84 Plus Instructions

2ND      VARS      Normalcdf:      -9999      -1.49      0       1 = 0.0681
2ND      VARS      Normalcdf:      -9999        8.01      0      1 =  1 (we need to use 0.9999 when results are = to 1)
.9999 - 0.0681 = 0.9318

 
c. When redesigning the ejection​ seat, which probability is more​ relevant?
 
A. Part​ (a) because the seat performance for a sample of pilots is more important.
B. Part​ (b) because the seat performance for a sample of pilots is more important.
C. Part​ (b) because the seat performance for a single pilot is more important.
D. Part​ (a) because the seat performance for a single pilot is more important
 
The​ _______ tells us that for a population with any​ distribution, the distribution of the sample means approaches
a normal distribution as the sample size increases.
Central Limit Theorem  
The standard deviation of the distribution of sample means is​ _______.
 
   σ / √n  
Which of the following is NOT a conclusion of the Central Limit​ Theorem?  Choose the correct answer below.
 
A. The distribution of the sample means x overbar   ​will, as the sample size​ increases, approach a normal distribution.
B. The standard deviation of all sample means is the population standard deviation divided by the square root of the sample size.
C. The mean of all sample means is the population mean mu
D. The distribution of the sample data will approach a normal distribution as the sample size increases.  
Homework 7. 1
 
A newspaper provided a​ "snapshot" illustrating poll results from 1910 professionals who interview job applicants.
The illustration showed that​ 26% of them said the biggest interview turnoff is that the applicant did not make an effort t
o learn about the job or the company.
The margin of error was given as plus or minus 3   percentage points. What important feature of the poll was​ omitted?
Choose the correct answer.
 
The sample size
The confidence interval
The confidence level
The point estimate  
A magazine provided results from a poll of 1500 adults who were asked to identify their favorite pie.
Among the 1500 ​respondents, 14% chose chocolate​ pie, and the margin of error was given as plus or minus
5 percentage points. What values do p, q​,  ​n, E, and p​ represent? If the confidence level is 99 ​% what is the value of a?
 
The value of p̂ is
the sample size.
the margin of error.
the sample proportion
the population proportion.
 
The value of q is
 
found from evaluating 1 - p̂
the sample proportion.
the sample size.
the population proportion.
the margin of error.
 
The value of n is
the population proportion.
the sample size
the margin of error.
the sample proportion.
found from evaluating 1 - p̂
 
The value of E is
 
the population proportion
the margin of error.
the sample proportion
the sample size
 
The value of p is
 
the population proportion.
the margin of error.
the sample proportion.
the sample size.
 
If the confidence level is 99%, what is the value of alpha?
a = 0.01
​(Type an integer or a decimal. Do not​ round.)
 
100% - 99% = 1% (0.01)
 
A magazine provided results from a poll of 1000   adults who were asked to identify their favorite pie. Among the 1000 ​respondents, 14 ​% chose chocolate​ pie, and the margin of error was given as +/- 3% points. Given specific sample​ data, which confidence interval is​ wider: the 95 ​% confidence interval or the 80% confidence​ interval? Why is it​ wider? Choose the correct answer below.
 
A. An   80% confidence interval must be wider than a 95% confidence interval because it contains ​100% - 80​% = 20​% of the true population​ parameters, while the 95​% confidence interval only contains ​100% - 95​% = 5 ​%  of the true population parameters.
B. A 95​%  confidence interval must be wider than an   80 ​%  confidence interval because it contains 95%  of the true population​ parameters, while the 80 ​%  confidence interval only contains 80 ​%  of the true population parameters.
C. A 95​% confidence interval must be wider than an 80%  confidence interval in order to be more confident that it captures the true value of the population proportion.
D. An 80%  confidence interval must be wider than a 95 ​%  confidence interval in order to be more confident that it captures the true value of the population proportion. Find the critical value ^za/2 that corresponds to the given confidence level.
82 ​%
^za/2 = 1.34
(Round to two decimal places as​ needed.)
 
100% – 82% = 1 - 0.82 = 0.18
^za/2     =     0.18/2 = 0.09

TI84 Plus Instructions

2nd     VARS     invNorm:
area: 0.09
u: 0
σ: 1
= -1.340755035 = 1.34
   
Express the confidence interval (0.069, 0.135) in the form of p^ - E < p < p^ + E
 
0.069 < p < 0.135
​(Type integers or​ decimals.)
 
Use the sample data and confidence level given below to complete parts​ (a) through​ (d). 
A drug is used to help prevent blood clots in certain patients. In clinical​ trials, among 4160   patients treated with the​ drug,
106 developed the adverse reaction of nausea. Construct a 99​% confidence interval for the proportion of adverse reactions. ​
 
Find the best point estimate of the population proportion 0.025
(Round to three decimal places as​ needed.)
 
106 / 4160 = 0.025
 
TI84 Plus Instructions

stat       test        A: 1 - PropZint:
x: 106
n: 4,160
C-Level: 0.99
 
​b) Identify the value of the margin of error E.
E = 0.006   (Round to three decimal places as​ needed.)
 
Margin of Error
 
TI84 Plus Instructions

stat       test        A:1 - PropZint:
x: 106
n: 4,160
C-Level: 0.99
0.01919,.03177 = 0.03177 - 0.01919 / 2 = .00629
 
Construct the confidence interval.
0.019 < p < 0.031   (Round to three decimal places as​ needed.)
0.025 - 0.006 = 0.019
0.025 + 0.006 = 0.031
 
Write a statement that correctly interprets the confidence interval. Choose the correct answer below.
 
A. One has 99​% confidence that the interval from the lower bound to the upper bound actually does
contain the true value of the population proportion.
B. One has 99% confidence that the sample proportion is equal to the population proportion.
C. There is a 99​% chance that the true value of the population proportion will fall between the lower bound and the upper bound.
D. 99​% of sample proportions will fall between the lower bound and the upper bound.  
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 560 babies were​ born, and 308  
of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born.
Based on the​ result, does the method appear to be​ effective?
 
.496 < p < .604
(Round to three decimal places as​ needed.)
 
TI84 Plus Instructions

stat       test        A: 1 - PropZint:
x: 308
n: 560
C-Level: 0.99
Press Calculate

(0.49585, 0.60415)
0.55
0.49585 - .60415 / 2 = .05415
0.55 - 0.05415 = 0.49585
0.55 + .1083 = 0.60415
 
Does the method appear to be​ effective?
 
No, the proportion of girls is not significantly different from 0.5.
Yes, the proportion of girls is significantly different from 0.5.  
A genetic experiment with peas resulted in one sample of offspring that consisted of 429 green peas and 160 yellow peas.
a. Construct a 90% confidence interval to estimate of the percentage of yellow peas.
b. It was expected that​ 25% of the offspring peas would be yellow.
Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations?

A. Construct a 90% confidence interval. Express the percentages in decimal form.
 
.2416 < p < .3016 (Round to three decimal places as​ needed.)
 
160 + 429 = 589
160 ÷ 589 = 0.2716 (rounded to 4)
1 - 0.90 ÷ 2 = 0.05
 
TI84 Plus Instructions

2nd       VARS      invNorm: 0.05      0      1 = 1.644 (rounded to 3)
q = 1 - .2716 = .7284
1.64√.2716 (.7284 ÷ 589) = 0.030 (rounded to 4)
Lower: 0.2716 - 0.0300 = 0.2416
Upper: 0.2716 + 0.0300 = 0.3016

Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations?
 
​No, the confidence interval includes​ 0.25, so the true percentage could easily equal​ 25%
​Yes, the confidence interval does not include​ 0.25, so the true percentage could not equal​ 25%
 
In a survey of 3042 adults aged 57 through 85​ years, it was found that 85.1% of them used at least one prescription medication. Complete parts​ (a) through​ (c) below.
 
How many of the 3042 subjects used at least one prescription​ medication?
2589 (Round to the nearest integer as​ needed.)
 
3042 x 0.851 = 2589
 
Construct a​ 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who
use at least one prescription medication.

84.0% < p < 86.2%
​(Round to one decimal place as​ needed.)
 
^Za/2 = 100% - 90% / 2 = 1 - .90 / 2 = .05
 
TI84 Plus Instructions

2nd       VARS      invNorm:    0.5     0      1 = 1.64 (rounded to 2)
1.64√.851 x (1 - 0.851) / 3042 = 1.64√.851 x 0.149 / 3042 = 0.0106
0.851 - 0.0106 = 0.8404 = 84.0%
0.851 + 0.0106 = 0.8616 = 86.2%
 
What do the results tell us about the proportion of college students who use at least one prescription​ medication?
 
A. The results tell us​ that, with​ 90% confidence, the probability that a college student uses at least one prescription
medication is in the interval found in part​ (b).
B. The results tell us nothing about the proportion of college students who use at least one prescription medication.
C. The results tell us that there is a​ 90% probability that the true proportion of college students who use at least one prescription medication is in the interval found in part​ (b).
D. The results tell us​ that, with​ 90% confidence, the true proportion of college students who use at least one prescription
medication is in the interval found in part​ (b).  
A study of 420004 cell phone users found that 133 of them developed cancer of the brain or nervous system.
Prior to this study of cell phone​ use, the rate of such cancer was found to be 0.0204% for those not using cell phones.
Complete parts​ (a) and​ (b).
 
Use the sample data to construct a 90% confidence interval estimate of the percentage of cell phone users
who develop cancer of the brain or nervous system.
 
​.027% < p < .036%
​(Round to three decimal places as​ needed.)
 
133 / 420004 = 0.00031666 (round to 8 decimals)
100% - 90% / 2 = 1 - .90 / 2 = .05
 
TI84 Plus Instructions
2nd | VARS
invNorm: .5 / 0 / 1 = 1.64 (rounded to 2)
1.64√.00031666 x 0.99968334 / 420004 = 0.00004502409
0.00031666 - 0.00004502 = 0.00027164 x 100 = 0.027%
0.00031666 + 0.00004502 = 0.00036168 x 100 = 0.036%
 
Do cell phone users appear to have a rate of cancer of the brain or nervous system that is
different from the rate of such cancer among those not using cell​ phones?
Why or why​ not?
 
A. No ​, because 0.0204% is included in the confidence interval.
B. Yes ​, because 0.0204% is not included in the confidence interval.
C. Yes ​, because 0.0204% is included in the confidence interval.
D. No ​, because 0.0204​% is not included in the confidence interval.
 
In a study of government financial aid for college​ students, it becomes necessary to estimate the percentage of​
full-time college students who earn a​ bachelor's degree in four years or less. Find the sample size needed to
estimate that percentage. Use a 0.01   margin of error and use a confidence level of 95%. 
Complete parts​ (a) through​ (c) below.
 
Assume that nothing is known about the percentage to be estimated.
N = 9604
(Round up to the nearest​ integer.)
 
100% - 95% / 2 = 1 - 0.95 / 2 = 0.025
 
TI84 Plus Instructions

2nd | VARS
invNorm: .025 / 0 / 1 = 1.960 (rounded to 3)
1.960² x 0.25 / 0.01² = 9604
 
Assume prior studies have shown that about 60​% of​ full-time students earn​ bachelor's degrees in four years or less.
n = 9220
 
1.960² x 0.60 (1 - 0.60) / .01² = 9219.84 (9220)
 
Does the added knowledge in part​ (b) have much of an effect on the sample​ size?
 
A. ​No, using the additional survey information from part​ (b) does not change the sample size.
B. ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.
C. ​Yes, using the additional survey information from part​ (b) only slightly increases the sample size.
D. ​No, using the additional survey information from part​ (b) only slightly reduces the sample size.
 
You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats.
How many randomly selected air passengers must you​ survey?
Assume that you want to be 99​% confident that the sample percentage is within 5.5 percentage points of the
true population percentage. Complete parts​ (a) and​ (b) below. a. Assume that nothing is known about the
percentage of passengers who prefer aisle seats.
 
N = 548
​(Round up to the nearest​ integer.)
 
100% - 99% / 2 = 1 - 0.99 / 2 = .005
 
TI84 Plus Instructions

2nd       VARS     invNorm: 0.005       0      1 = 2.576 (rounded to 3)
2.576² x 0.25 / 0.055² = 548.411
 
Assume that a prior survey suggests that about 38​% of air passengers prefer an aisle seat.
 
N = 517
​(Round up to the nearest​ integer.)
 
2.576² x 0.38 x 1 - 0.38 / .055² = 516.8227 (517)
 
You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats.
How many randomly selected air passengers must you​ survey? Assume that you want to be 90​% confident that the
sample percentage is within 2.5   percentage points of the true population percentage.
Complete parts​ (a) and​ (b) below.
a. Assume that nothing is known about the percentage of passengers who prefer aisle seats.

N= 1082
​(Round up to the nearest​ integer.)
 
100% - 90% / 2 = 1 - 0.90 / 2 = 0.05
 
TI84 Plus Instructions

2nd | VARS
invNorm: 0.05 / 0 / 1 = 1.645 (rounded to 3)
1.645² x 0.25 / 0.025² = 1082.41
 
Assume that a prior survey suggests that about
35%
of air passengers prefer an aisle seat.
N = 985
​(Round up to the nearest​ integer.)
 
 
1.645² x .35 (1 - 0.35) / 0.025² = 984.9931  
Fill in the blank.
A critical​ value, ^za​, denotes the​ _______.

z-score with an area of a to its right  
Which of the following groups has terms that can be used interchangeably with the​ others? Choose the correct answer below.
 
Critical​ Value, Probability, and Proportion
Critical​ Value, Percentage, and Proportion
Critical​ Value, Percentage, and Probability ​
Percentage, Probability, and Proportion  
A​ _______ is a single value used to approximate a population parameter.

point estimate  
Which of the following is NOT needed to determine the minimum sample size required to estimate a population​ proportion?
Choose the correct answer below.
 
^Za/2
Margin of Error
Standard Deviation
a  
Homework 7.2
 
Refer to the accompanying data display that results from a sample of airport data speeds in Mbps.
Complete parts​ (a) through​ (c) below.            
 
TInterval ​(13.046, 22.15)
x =17.598
Sx=16.01712719
n=50
 
Express the confidence interval in the format that uses the​ "less than" symbol.
Given that the original listed data use one decimal​ place, round the confidence interval limits accordingly.
 
13.05 Mbps < μ < 22.15 Mbps ​(Round to two decimal places as​ needed.)
13.046 (rounded to 13.5)
 
b. Identify the best point estimate of mμ and the margin of error.
The point estimate of mμ is 17.60 Mbps. (Round to two decimal places as​ needed.)
x = 17.598 rounded to 17.60
 
The margin of error is = 4.55 Mbps.
​(Round to two decimal places as​ needed.)
 
17.598 – 13.046 = 4.552
 
In constructing the confidence interval estimate of mμ​, why is it not necessary to confirm that the sample data appear to be from a population with a normal​ distribution?
 
A. Because the sample is a random​ sample, the distribution of sample means can be treated as a normal distribution.
B. Because the sample size of 50 is greater than​ 30, the distribution of sample means can be treated as a normal distribution.
C. Because the sample standard deviation is​ known, the normal distribution can be used to construct the confidence interval.
D. Because the population standard deviation is​ known, the normal distribution can be used to construct the confidence interval
 
Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. The results in the screen display are based on a​ 95% confidence level. Write a statement that correctly interprets the confidence interval.                
 
TInterval ​(13.046,22.15)
x = 17.598
Sx = 16.01712719
n = 50
 
Choose the correct answer below.
 
A. The limits of 13.05 Mbps and 22.15 Mbps contain the true value of the mean of the population of all data speeds at the airports.
B. The limits of 13.05 Mbps and 22.15 Mbps contain​ 95% of all of the data speeds at the airports.
C. We have​ 95% confidence that the limits of 13.05 Mbps and 22.15 Mbps contain the sample mean of the data speeds at the airports.
D. We have​ 95% confidence that the limits of 13.05 Mbps and 22.15 Mbps contain the true value of the mean of the population of all data speeds at the airports.
 
Assume that we want to construct a confidence interval. Do one of the​ following, as​ appropriate:
(a) find the critical value ^tα/2​, ​
(b) find the critical value ^zα/2​, or​
(c) state that neither the normal distribution nor the t distribution applies.
 
Here are summary statistics for randomly selected weights of newborn​ girls: n = 156, x = 32.1 ​hg, s = 6.1 hg.
The confidence level is 90%.
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
 
A. ^tα/2 = 1.65
​(Round to two decimal places as​ needed.)
 
A = 1.00 - 0.90 = .10 = 0.10 / 2 = .05
156 – 1 = 155
 
TI84 Plus Instructions

2^nd      VARS      invT: .05      155 = 1.65
 
B. ^zα/2 =
​(Round to two decimal places as​ needed.)
 
C. Neither the normal distribution nor the t distribution applies.
 
The sample size of 300 is greater than 30 one of the conditions for using the normal distribution
or the t distribution is met.  
Here are summary statistics for randomly selected weights of newborn​ girls: n = 226​, x =26.2 ​hg, s = 7.2 hg.
Construct a confidence interval estimate of the mean. Use a 98​% confidence level. Are these results very
different from the confidence interval 24.8 hg < μ <28.4 hg with only 18 sample​ values, x=26.6 hg, and s = 2.9 hg?
 
What is the confidence interval for the population mean mu​?
 
 25.1 hg < u < 27.3 hg
​(Round to one decimal place as​ needed.)
 
226 – 1 = 225
1 - 0.98 / 2 = 0.01
 
TI84 Plus Instructions

2^nd      VARS      invT: 0.01     225      =    2.343
26.2 – (2.343 x 7.2 / √226) = 25.0778509
26.2 + (2.343 x 7.2 / √226) = 27.3221491

 
Are the results between the two confidence intervals very​ different?
A. ​Yes, because one confidence interval does not contain the mean of the other confidence interval.
B. ​No, because each confidence interval contains the mean of the other confidence interval.
C. No, because the confidence interval limits are similar.
D. ​Yes, because the confidence interval limits are not similar
 
Here are summary statistics for randomly selected weights of newborn​ girls: n = 151​, x = 31.9 ​hg, s = 6.4 hg.
Construct a confidence interval estimate of the mean. Use a 99% confidence level.
Are these results very different from the confidence interval 31.0 hg < mμ < 33.8 hg with only
18 sample​ values, x = 32.4 hg, and s = 2.1 hg?
 
What is the confidence interval for the population mean mu​?
 
30.5 hg < u < 33.3 hg
​(Round to one decimal place as​ needed.)
 
151 – 1 = 150
1 - .99 / 2 = 0.005
 
TI84 Plus Instructions

2^nd       VARS      invT:     0.05      150 = 2.609
31.9 – (2.609 x 6.4 / √151) = 30.5
31.9 + (2.609 x 6.4 / √151) = 33.3
 
Are the results between the two confidence intervals very​ different?
A. ​Yes, because one confidence interval does not contain the mean of the other confidence interval.
B. ​No, because each confidence interval contains the mean of the other confidence interval.
C. No, because the confidence interval limits are similar.
D. ​Yes, because the confidence interval limits are not similar  
In a test of the effectiveness of garlic for lowering​ cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes ​(before - after)  in their levels of LDL cholesterol​ (in mg/dL) have a mean of 4.8 and a standard deviation of 19.6  Construct a 90 % confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL​ cholesterol?
 
What is the confidence interval estimate of the population mean mu?
 
​-0.29 mg/dL < μ < 9.89 ​mg/dL
​(Round to two decimal places as​ needed.)
 
T a/2
1.0       - .09 / 2 = 0.05
42 – 1 = 41
 
TI84 Plus Instructions

2^nd     VARS     invT: 0.05      41    = 1.683 (round to 3 decimals)
1.683(19.6 / √42) = 5.090 (round to 3 decimals)
4.8 – 5.090 = -0.29
4.8 + 5.090 = 9.89

 
What does the confidence interval suggest about the effectiveness of the​ treatment?
 
A. The confidence interval limits contain 0, suggesting that the garlic treatment did   affect the LDL cholesterol levels.
B. The confidence interval limits contain   ​0, suggesting that the garlic treatment did not   affect the LDL cholesterol levels.
C. The confidence interval limits do not contain   ​0, suggesting that the garlic treatment did not   affect the LDL cholesterol levels.
D. The confidence interval limits do not contain   ​0, suggesting that the garlic treatment did   affect the LDL cholesterol levels.
 
An IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 99% confidence that the sample mean is within 6 IQ points of the true mean. Assume that sd = 15 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real-world calculation.
 
The required sample size is 42
​(Round up to the nearest​ integer.)
 
TI84 Plus Instructions

1.0       - 0.99 / 2 = 0.05
2^nd      VARS      invNorm
area: 0.05
u: 0
σ: 1
= 2.58 (round to 2 decimals)
(2.58 x 15 / 6)² = 41.6025
 
Would it be reasonable to sample this number of​ students?
 
No, This number of IQ test scores is a fairly small number.
Yes, This number of IQ test scores is a fairly small number.
Yes, This number of IQ test scores is a fairly large number.
No, This number of IQ test scores is a fairly large number.
 
Assume that all​ grade-point averages are to be standardized on a scale between 0 and 5 .  How many​ grade-point averages must be obtained so that the sample mean is within 0.007 of the population​ mean? Assume that a 95​% confidence level is desired. If using the range rule of​ thumb, sd can be estimated as range / 4 = 5 – 0 / 4 = 1.25. Does the sample size seem​ practical?
 
The required sample size is 122500
​(Round up to the nearest whole number as​ needed.)
 
1.0       - .95 / 2 = .025

TI84 Plus Instructions

2^nd      VARS    invNorm
0.025
0
1
= 1.96 (round to 2 decimals)
(1.96 x 1.25 / .007)² = 122500
 
Does the sample size seem​ practical?
 
A. ​No, because the required sample size is a fairly small number.
B. No, because the required sample size is a fairly large number.
C. ​Yes, because the required sample size is a fairly large number.
D. ​Yes, because the required sample size is a fairly small number.  
The ages of a group of 139 randomly selected adult females have a standard deviation of 17.6   years. Assume that the ages of female statistics students have less variation than ages of females in the general​ population, so let sd  = 17.6  years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics​ students? Assume that we want 90% confidence that the sample mean is within​ one-half year of the population mean. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general​ population?
 
The required sample size is 3333
​(Round up to the nearest whole number as​ needed.)
 
 
1 - .90 / 2 = 0.05
1 - .90 = 0.10
Margin of error = ½ year = 0.5
 
TI84 Plus Instructions

2^nd VARS invNorm:
0.05
0
1
= 1.64 (round to 2 decimals)
(1.64 x 17.6 / .5)² = 3332.521
 
Does it seem reasonable to assume that the ages of female statistics students have less variation than
ages of females in the general​ population?
 
A. ​No, because there is no age difference between the population of statistics students and the general population.
B. ​No, because statistics students are typically older than people in the general population.
C. ​Yes, because statistics students are typically younger than people in the general population.
D. ​Yes, because statistics students are typically older than people in the general population.
 
Salaries of 34 college graduates who took a statistics course in college have a​ mean, x ​ of $68,800 .  Assuming a standard​ deviation, sd ​of ​$17928, construct a 95% confidence interval for estimating the population mu .
 
$62774 < u < $74826
​(Round to the nearest integer as​ needed.)
 
1 - .95 / 2 = .025
 
TI84 Plus Instructions

2^nd VARS
invNorm: 0.025      0        1 = 1.96 (round to 2 decimals)
1.960 x 17928 / √34 = 6026.268 (round to 3 decimals)
68800 – 6026.268 = 62773.752
68800 + 6026.268 = 74826.268
 
Fill in the blank.
The number of​ _______ for a collection of sample data is the number of sample values that can vary after
certain restrictions have been imposed on all data values.

Degrees of freedom  
Which of the following is NOT a property of the Student t​ distribution?
Choose the correct answer below.
 
A. The standard deviation of the Student t distribution is s = 1.
B. The Student t distribution has a mean of t = 0.
C. The Student t distribution has the same general symmetric bell shape as the standard normal​ distribution, but it reflects the greater variability that is expected with small samples.
D. The Student t distribution is different for different sample sizes.  
Fill in the blank.
The​ _______ is the best point estimate of the population mean

Sample mean Which of the following would be a correct interpretation of a​ 99% confidence interval such as 4.1 < u <  ​5.6? Choose the correct answer below.
 
A. There is a​ 99% chance that mu   will fall between 4.1 and 5.6.
B. It means that​ 99% of all data values are between 4.1 and 5.6.
C. We are​ 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of u
D. It means that​ 99% of sample means fall between 4.1 and 5.6.  
Which of the following would be a correct interpretation of a​ 99% confidence interval such as 4.1 < u <  ​5.6? Choose the correct answer below.
 
A. There is a​ 99% chance that mu   will fall between 4.1 and 5.6.
B. It means that​ 99% of all data values are between 4.1 and 5.6.
C. We are​ 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of u
D. It means that​ 99% of sample means fall between 4.1 and 5.6.  
Which of the following is NOT an equivalent expression for the confidence interval given by 161.7 < u < 189.5? Choose the correct answer below.
 
A. 161.7 +- 27.8
B. 175.6 - 13.9 < u < 175.6 + 13.9
C. 175.6 +- 13.9
D. ​(161.7,189.5)
 
TI84 Plus Instructions

Press
STAT, EDIT
Input your data points in L1 and hit ENTER after each point.
Press STAT again move to the “TEST” menu.
Scroll down to 8: “TInterval” option and hit ENTER
Input: select the “data” option by highlighting it and hit ENTER.
Scroll down to the “List” item and hit 2nd and then digit 1, press ENTER. This tells the calculator to use the data values in list as the input.
5. Scroll past “Freq” item to the “C-Level” row and enter the level of confidence 12)
Scroll down to the “Calculate” item and hit ENTER.
the output will be displayed that has the desired CI  
Listed below are the amounts of net worth​ (in millions of​ dollars) of the ten wealthiest celebrities in a country. Construct a 99​% confidence interval. What does the result tell us about the population of all​ celebrities? Do the data appear to be from a normally distributed population as​ required?
 
261   208   187   171   166   162   155   155   155   150
 
What is the confidence interval estimate of the population mean mu​?
​$148.4 million < u < $194 million
​(Round to one decimal place as​ needed.)
 
Mean = 1712 / 10 = 171.2
1 - 0.99 / 2 =
N = 10 – 1 = 9
 
TI84 Plus Instructions

2^nd VARS    invT:    0.005      3.250 (round to 3 decimals)
Standard Deviation = 39.30451

https://www.socscistatistics.com/descriptive/variance/default.aspx
 
Listed below are the amounts of net worth​ (in millions of​ dollars) of the ten wealthiest celebrities in a country.
Construct a 90​% confidence interval. What does the result tell us about the population of all​ celebrities?
Do the data appear to be from a normally distributed population as​ required?
 
242      210      176      160      156      154      150      150      150      150
 
What is the confidence interval estimate of the population mean mu​?
​$151.5 million < u < $188.1 million
​(Round to one decimal place as​ needed.)
 
Explanation:
1 - .90 / 2 = .05
n = 10 – 1 = 9
 
TI84 Plus Instructions


2^nd VARS    invT: 0.05       9       =    -1.833112923 (this will need to be positive) 1.833
 
Now we find the standard deviation:
 
This format and SAMPLE Option.
242, 210, 176, 160, 156, 154, 150, 150, 150, 150

https://www.calculator.net/standard-deviation-calculator.html
 
TI84 Plus Instructions

stat | edit | enter data     (242      210      176      160      156      154      150      150      150      150)
stat | calc | 1-var | List: L1 | FreqList: blank | calculate.
sample standard deviation (smaller one) Sx = 31.57
 
1.833 · 31.57 / √10 =
169.8 – 18.3 = 151.5
169.8 + 18.3 = 188.1
 
What does the result tell us about the population of all​ celebrities? Select the correct choice below​ and, if​ necessary, fill in the answer​ box(es) to complete your choice.
 
A. Because the ten wealthiest celebrities are not a representative​ sample, this​ doesn't provide any information about the population of all celebrities.
B. We are 90 ​%  confident that the interval from ​$nothing   million to ​$nothing   million actually contains the true mean net worth of all celebrities. ​(Round to one decimal place as​ needed.)
C. We are confident that 90 ​%  of all celebrities have a net worth between ​$nothing   million and ​$nothing   million. ​(Round to one decimal place as​ needed.)
 
Do the data appear to be from a normally distributed population as​ required?
 
A. ​Yes, but the points in the normal quantile plot do not lie reasonably close to a straight line or show a systematic pattern that is a straight line pattern.
B. ​Yes, because the pattern of the points in the normal quantile plot is reasonably close to a straight line.
C. ​No, because the points lie reasonable close to a straight​ line, but there is a systematic pattern that is not a straight line pattern.
D. No, but the points in the normal quantile plot lie reasonably close to a straight line and show some systematic pattern that is a straight line pattern.
 
Salaries of 34 college graduates who took a statistics course in college have a​ mean, x​ of $60500 Assuming a standard​ deviation, sd of ​$15003 construct a 90% confidence interval for estimating the population mean mu .
 
$56267 < mu < $64733
​(Round to the nearest integer as​ needed.)
 
1 - .9 / 2 = .05 = 1.645 (round to 3 decimals)
 
1.645 x 15003 / √34 = 4232.574 (round to 3 decimals)
60500 - 4232.574 = 56267.426
60500 + 4232.574 = 64732.574  
Review 6.2 – 7
 
The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. How many adults must he survey in order to be 90% confident that his estimate is within five   percentage points of the true population​ percentage? Complete parts​ (a) through​ (c) below.
 
Assume that nothing is known about the percentage of adults who have heard of the brand.
N = 271
(Round up to the nearest​ integer.)
 
1 - .9 / 2 = .1
 
TI84 Plus Instructions

2^nd      VARS      invNorm: .1 / 0 / 1 = 1.6449 (round to 4 decimals)
0.50 is the estimated proportion (when nothing is known, 0.5 is assumed)
E = .05
1.6449² x 0.5(1-.5) / 0.05² = 270.569601
 
Assume that a recent survey suggests that about 85% of adults have heard of the brand.
n = 138  (Round up to the nearest​ integer.)
 
1.6449² x 0.85(1 - 0.85) / 0.06² = 137.9904
 
Use the sample data and confidence level given below to complete parts​ (a) through​ (d). A research institute poll asked respondents if they felt vulnerable to identity theft. In the​ poll, n = 925 and x = 562   who said​ "yes." Use a 95 %   confidence level.
 
Find the best point estimate of the population proportion p.
 
Find the best point estimate of the population proportion p.
(Round to three decimal places as​ needed.)

    0.608
 
562 / 925 = 0.608
 
Identify the value of the margin of error E.
    0.031  (Round to three decimal places as​ needed.)
 
1 - .95 / 2 = 0.025
 
TI84 Plus Instructions

2^nd VARS  invNorm:        0.025      0       1     =   -1.96
1.96 √0.608 (1 - 0.608) / 925 = 0.03146
 
Construct the confidence interval.
.577 < p < .639
(Round to three decimal places as​ needed.)
 
.608 - .031 = .577
.608 + .031 = .639
 
Write a statement that correctly interprets the confidence interval. Choose the correct answer below.
 
A. One has 95​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
B. 95% of sample proportions will fall between the lower bound and the upper bound.
C. There is a 95​% chance that the true value of the population proportion will fall between the lower bound and the upper bound.
D. One has 95​% confidence that the sample proportion is equal to the population proportion.  
A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 98% confidence interval estimate of the mean amount of mercury in the population.
 
Mercury (ppm)
0.56      0.81     0.11     0.88     1.23     0.53     0.91
 
What is the confidence interval estimate of the population mean mu​?
 0.295 ppm < mu < 1.142 ppm
​(Round to three decimal places as​ needed.)
 
TI84 Plus Instructions

STAT
EDIT
*Enter List Data (0.56      0.81     0.11     0.88     1.23     0.53     0.91 )
STAT
TESTS
Tintevals
Data
Lists: L1
Freq: 1
C-Level: 0.98
Calculate: ans = (0.2948, 1.1423)

https://youtu.be/GAK799Ts_jY
 
Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts​ (a) through​ (b) below
 
TInterval
(13.046,22.15)
x = 17.598
Sx = 16.01712719
n = 50
 
What is the number of degrees of freedom that should be used for finding the critical value ^ta/2?
df = 49
​(Type a whole​ number.)
 
50 – 1 = 49
 
b. Find the critical value ta/2 corresponding to a​ 95% confidence level.
 
^ta/2 = 2.01
​(Round to two decimal places as​ needed.)
 
TI84 Plus Instructions

2^nd        VARS      invT:    0.025     49    =   2.00957
 
A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 16 subjects had a mean wake time of 101.0 min. After​ treatment, the 16 subjects had a mean wake time of 71.7 min and a standard deviation of 24.4   min. Assume that the 16 sample values appear to be from a normally distributed population and construct a 99 ​%  confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 101.0   min before the​ treatment? Does the drug appear to be​ effective?
 
Construct the 99% confidence interval estimate of the mean wake time for a population with the treatment.
 
53.7 min < u < 89.7 min ​
(Round to one decimal place as​ needed.)
 
1 - 0.99 / 2 = .005
16 - 1 = 15
 
TI84 Plus Instructions

2^nd VARS     invT:     0.005      15    = 2.947 (round to 3 decimals)
2.947 (24.4 / √16) = 17.9767 (round 3 decimal places to 17.977)
 
71.7 – 17.977 = 53.723
71.7 + 17.977 = 89.677
 
The confidence interval does not include includes the mean wake time of 105.0 min before the​ treatment, so the means before and after the treatment are different. This result suggests that the drug treatment has a significant effect.