Appendix A: Answers to Your Turn Exercises

Chapter 13

1. The variable type takes the following values: adventure, platform, sports, shooter, action, role-playing, and racing.
2. The observation for Spiderman 2 for PS4 is as follows:
   t

   Game	Platform	Studio	Type	Sales for Week	Sales Total	Weeks on List
   Spiderman 2 for PS4	PS4	Activision	Action	6510	49,292	3

1. Since Sirius XM is a stock, it is an element.
2. Another name for the variable Exchange would be Market.

1. Because the number of medals won is finite, the variable number of medals won is discrete.
2. Because the time to run a 100-meter dash can take on an infinite number of values, the variable racing time in the 100-meter dash is continuous.

1. Exchange represents nominal data, because the data cannot be ordered in a natural or obvious way. Also, no arithmetic can be performed on exchange.
2. Last price represents ratio data. Here division makes sense. A last price of $20 is twice a last price of $10.

1. Since Florida has more than 3 counties, this represents a sample.
2. Since we are talking about all of the counties in Florida, this represents a population.

1. Since this represents a sample, the most expensive hotel in the 3 counties represents a statistic.
2. Since this represents a population, the most expensive hotel in Florida represents a parameter.

1. The average grade on the first quiz is a descriptive statistic, since the average grade is only for your class. However, no inference is made regarding a larger population.
2. Jessica won 2 out of 10 games of ping pong to her friend Lu Li. Therefore she won $\left(2/10\right)·100\%=20\%$ of her ping pong games to Lu Li. This represents a statistic. She then used this statistic to perform statistical inference to conclude that she will only win 20% of her games of ping pong to Lu Li.

1. Systematic sample: Bill Gates, Charles Koch, Jim Walton, Michael Bloomberg, Larry Page, Jacqueline Mars, George Soros
2. Systematic sample: Larry Ellison, Alice Walton, Larry Page, Carl Icahn

1. Since you used a random sample from your school, this is not convenience sampling.
2. Since you obtain data from your 5 closest friends at school, this is convenience sampling. You are choosing a sample that is convenient for you.

1. Convenience sampling: you are choosing a sample convenient for you.
2. Systematic Sampling: where every kth student is taken, with $k=10$.
3. Cluster sampling: (a) the population was divided into clusters (classes), (b) a random sample of one cluster (class) is taken, and (c) every student in that cluster (class) is selected.
4. Stratified sampling: (a) the population was divided into subgroups (nursing majors and all others), and (b) a random sample from each of the groups was drawn.
5. Random sampling: the two names were selected randomly.

Chapter 2

1. Borough	Frequency
   Brooklyn	5
   Manhattan	7

2. Violation type	Frequency
   Cell phone	4
   Safety belt	3
   Speeding	2
   Disobey sign	3

1. Borough	Relative frequency
   Brooklyn	$5/12=0.42$
   Manhattan	$7/12=0.58$

2. Violation type	Relative frequency
   Cell phone	$4/12=0.33$
   Safety belt	$3/12=0.25$
   Speeding	$2/12=0.17$
   Disobey sign	$3/12=0.25$

1. 
2. 
3. 
4. 

• (a)

  SAT Writing score	Frequency
  200–390	347,920
  400–590	1,015,121
  600–800	297,006
  Total	1,660,047

• (b)
  

1. 
2. 

1. 
2. 

1. $2+5=7$
2. No. The class boundaries are from 6 to 8.5.
3. 3

Chapter 3

$\frac{10+15+16+9+19+19+19+14}{8}=15.125$

• (a) The sample mean would be lower than $337.50 without the Sony Xperia Z2 since the price of this phone is $600, which is higher than the prices of the other phones.
• $\bar{x}=\frac{\sum x}{n}=\frac{200+250+300}{3}=\$250$. Yes, the sample mean of $250 is lower than $337.50.

1. 
2. Darts has the larger range.
3. $\begin{array}{l}{\text{range}}_{\text{Darts}}=\hspace{0.17em}\text{largest value}- \text{smallest value}=72.9-11.2=61.7\\ {\text{range}}_{\text{DJIA}}=\hspace{0.17em}\text{largest value}- \text{smallest value}=17.7-15.8=1.9\end{array}$

1. $\mu =\frac{\sum x}{N}=\frac{7.8+1.9+14.9+1.5+2.7+1.6}{6}=\$5.07\hspace{0.17em}\text{billion}$

2. $x$	$x-\mu$	${\left(x-\mu \right)}^2$
   7.8	$7.8-5.07=2.73$	7.4529
   1.9	$1.9-5.07=3.17$	10.0489
   14.9	$14.9-5.07=9.83$	96.6289
   1.5	$1.5-5.07=-3.57$	12.7449
   2.7	$2.7-5.07=-2.37$	5.6169
   1.6	$1.6-5.07=-3.47$	12.0409
   		$\sum {\left(x-\mu \right)}^2=144.5334$

${\sigma }^2=\frac{\sum {\left(x-\mu \right)}^2}{6}=\frac{144.5334}{6}=24.0889$

$\sigma =\sqrt{{\sigma }^2}=\sqrt{24.0889}=4.9080$

1. Smaller. The sample contains the 3 smallest numbers in the population.

2. First we need to find the sample mean.

   $\bar{x}=\frac{\sum x}{n}=\frac{1.5+1.9+1.6}{3}=\$1.67\hspace{0.17em}\text{million}$

   $x$	$x-\bar{x}$	${\left(x-\bar{x}\right)}^2$
   1.5	$1.5-1.67=-0.17$	0.0289
   1.9	$1.9-1.67=0.23$	0.0529
   1.6	$1.6-1.67=-0.07$	0.0049
   		$\sum {\left(x-\bar{x}\right)}^2=0.0867$

   From the table, $\sum {\left(x-\bar{x}\right)}^2=0.0867$. Therefore the sample variance is $s=\frac{\sum {\left(x-x\right)}^2}{n-1}=\frac{0.0867}{3-1}=0.04335$ billion of dollars squared.

3. From Problem 2 the sample variance is $s^2=0.04335$ billion of dollars squared. Therefore the sample standard deviation is $s=\sqrt{s^2}=\sqrt{0.04335}=\$0.2082\hspace{0.17em}\text{billion}$.
4. For this sample of CDC funding to fight HIV/AIDS in the northeastern United States, the typical difference between a state's funding and the mean funding is $0.2082 billion.

1. $\mu -1\sigma =70-1·5=65$ mph and m $\mu +1\sigma =70+1·5=75$ mph Therefore the percentage of vehicle speeds that lie between 65 mph and 75 mph is the percentage of vehicle speeds that lie within 1 standard deviation of the mean. Thus, from the Empirical Rule, approximately 68% of vehicle speeds lie between 65 mph and 75 mph.
2. From Problem 1, 65 mph lies 1 standard deviation below the mean. Therefore the Empirical Rule tells us that $\frac{1}{2}\left(100\%-68\%\right)=16\%$ of all vehicle speeds are at most 65 mph.

$\bar{x}=\frac{\sum \left(w·x\right)}{\sum w}=\frac{\left(50\right)\left(90\right)+\left(20\right)\left(70\right)+\left(30\right)\left(85\right)}{50+20+30}=\frac{8450}{100}=84.5$

1. The amount Austin spent on video games is $x=\$136$.

   $\text{The}\hspace{0.17em}z-\text{score}=\frac{\text{data}\hspace{0.17em}\text{value}-\hspace{0.17em}\text{mean}}{\text{standard deviation}}=\frac{x-\mu }{\sigma }=\frac{136-96}{40}=1.$

2. The amount Brian spent on music downloads is $x=\$16$.

   $\text{The}\hspace{0.17em}z-\text{score}=\frac{\text{data}\hspace{0.17em}\text{value}-\hspace{0.17em}\text{mean}}{\text{standard deviation}}=\frac{x-\mu }{\sigma }=\frac{16-96}{40}=-2.$

3. The amount Courtney spent on gifts for her friends is $x=\$256$.

   $\text{The}\hspace{0.17em}z-\text{score}=\frac{\text{data}\hspace{0.17em}\text{value}-\hspace{0.17em}\text{mean}}{\text{standard deviation}}=\frac{x-\mu }{\sigma }=\frac{256-96}{40}=4.$

1. David's $z$-score was −1.5. Therefore he spent

   $x=z-\text{score}·\sigma +\mu =\left(-1.5\right)·\left(\$40\right)+\$96=\$36$

2. Emily's $z$-score was 2.5. Therefore she spent

   $x=z-\text{score}·\sigma +\mu =\left(2.5\right)·\left(\$40\right)+\$96=\$196$

3. Frances's $z$-score was 0. Therefore she spent

   $x=z-\text{score}·\sigma +\mu =\left(0\right)·\left(\$40\right)+\$96=\$96$

$\text{The}\hspace{0.17em}z-\text{score}=\frac{\text{data}\hspace{0.17em}\text{value}-\text{mean}}{\text{standard}\hspace{0.17em}\text{deviation}}=\frac{x-\mu }{\sigma }=\frac{120-96}{40}=\mathrm{0.6.}$

$\text{The}\hspace{0.17em}z-\text{score}=\frac{\text{data}\hspace{0.17em}\text{value}-\text{mean}}{\text{standard}\hspace{0.17em}\text{deviation}}=\frac{x-\mu }{\sigma }=\frac{120-85}{40}=\mathrm{0.875.}$

1. Austin's $z$-score is 1, which lies in the range, −2 , $z$-score < 2. Therefore the amount Austin spent on video games is not considered unusual.
2. Brian's $z$-score is −2, which lies in the range, −3 , $z$-score ≤ −2. Therefore the amount Brian spent on music downloads may be considered moderately unusual.
3. Courtney's $z$-score is 4, which is ≥3. Therefore the amount Courtney spent on gifts for her friends may be considered an outlier.

$\frac{\text{number}\hspace{0.17em}\text{of values in data set}\hspace{0.17em}\le 9.0}{\text{total number of values is data set}}·100\%=\frac{8}{9}·100\%=89\%.$

1. Minimum = 1.9
2. First quartile, Q1 = 3.6
3. Median = Q2 = 5.4
4. Third quartile, Q3 = 7.1
5. Maximum = 9.3

1. Minimum = 1.9
2. First quartile, Q1 = 3.6
3. Median = Q2 = 5.4
4. Third quartile, Q3 = 7.1
5. Maximum = 9.3

Chapter 4

1. Because the response variable depends on the predictor variable, and because the weight of a person depends in part on the height of the person, height is the predictor ($x$) variable and weight is the response ($y$) variable.
2. 

• Step 1

  $\begin{array}{c}\bar{x}=\frac{\sum x}{n}=\frac{63.5+65.9+62.8+61.8+61.3+66.9+62.6+65.4}{8}\\ =63.775\hspace{0.17em}\text{inches}\\ \bar{y}=\frac{\sum y}{n}=\frac{113.8+130.1+108.5+138.9+118.2+130.1+104.9+153.9}{8}\\ =124.8\hspace{0.17em}\text{pounds}\end{array}$

• Step 2

  $x$	$y$	${\left(x-\bar{x}\right)}^2$	$\left(y-\bar{y}\right)$	${\left(y-\bar{y}\right)}^2$	$\left(x-\bar{x}\right)\left(y-\bar{y}\right)$
  63.5	113.8	−0.275	0.075625	−11	121	3.025
  65.9	130.1	2.125	4.515625	5.3	28.09	11.2625
  62.8	108.5	-0.975	0.950625	-16.3	265.69	15.8925
  61.8	138.9	-1.975	3.900625	14.1	198.81	-27.8475
  61.3	118.2	-2.475	6.125625	-6.6	43.56	16.335
  66.9	130.1	3.125	9.765625	5.3	28.09	16.5625
  62.6	104.9	-1.175	1.380625	-19.9	396.01	23.3825
  65.4	153.9	1.625	2.640625	29.1	846.81	47.2875
  			$\begin{array}{l}\sum {\left(x-\bar{x}\right)}^2=\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}29.355\end{array}$		$\begin{array}{l}\sum {\left(y-\bar{y}\right)}^2=\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1928.06\end{array}$	$\begin{array}{l}\sum \left(x-\bar{x}\right)\left(y-\bar{y}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=105.9\end{array}$

• Step 3

  $\begin{array}{c}{s}_x=\sqrt{\frac{{\sum \left(x-\bar{x}\right)}^2}{n-1}}=\sqrt{\frac{29.355}{8-1}}\approx 2.047821142\\ s_y=\sqrt{\frac{\sum {\left(y-\bar{y}\right)}^2}{n-1}}=\sqrt{\frac{1928.06}{8-1}}\approx 16.59629907\end{array}$

• Step 4

  $\begin{array}{l}r=\frac{\sum \left(x-\bar{x}\right)\left(y-\bar{y}\right)}{\left(n-1\right)s_x{s}_y}=\frac{105.9}{\left(8-1\right)\left(2.047821142\right)\left(16.59629907\right)}\\ \hspace{1em}\approx 0.4451379745\approx 0.4451\end{array}$

• (a) From Your Turn #4, page 194:
  • Step 1 Calculate the respective sample means $\bar{x}$ and $\bar{y}$. We have already done this in Your Turn #4 (page 194): $\bar{x}=63.775$ inches and $\bar{y}=124.8$ pounds.
  • Step 2 Calculate the respective sample standard deviations $s_x$ and $s_y$. We have already done this in Your Turn #4 (page 194): $s_x\approx 2.047821142$ $s_y\approx 16.59629907$.
  • Step 3 Find the correlation coefficient $r$. We have already done this in Your Turn #4 (page 194): $r\approx 0.4451379745$.

  • Step 4 Combine the statistics from Steps 2 and 3 to calculate $b_1$:

    $b_1=r·\frac{s_y}{s_x}=0.4451379745·\frac{16.59629907}{2.047821142}\approx 3.607562595\approx \mathrm{3.6076.}$

• (b) $\hat{y}=3.6076x-105.2723$
• (c) Because $y$ and $x$ represent weight and height, respectively, this regression equation is read as follows: “The estimated weight of a woman is 3.6076 times her height minus 105.2723 pounds.”

• (a) For a woman with height 0 inches, her estimated weight is −105.2723 pounds.
• (b) For each increase of 1 inch to a woman's height, her estimated weight increases by 3.6076 pounds.

• (a) For $x=65$ inches, $\hat{y}=3.6076x-105.2723=3.6076\left(65\right)-105.2722=129.2217$ pounds. Because $x=65$ inches lies between 61.3 inches and 66.9 inches, inclusive, this estimate does not represent extrapolation.
• (b) For $x=70$ inches, $\hat{y}=3.6076x-105.2722=3.6076\left(70\right)-105.2723=147.2597$ pounds. Because $x=70$ inches does not lie between 61.3 inches and 66.9 inches, this estimate represents extrapolation.

Chapter 5

• (a) The chance of a recession near zero means that the probability of a recession is near zero. Therefore it is very unlikely that a recession will occur.
• (b) The chance of satisfaction with the purchase of a new car is 100% means the probability that you will be satisfied with the purchase of a new car is 1. Therefore it is certain that you will be satisfied with the purchase of a new car.

• (a) A total of 52 outcomes is included in the sample space, so $N\left(S\right)=52$ Let $E$ be the event a heart is drawn. Then $E$ consists of the 13 hearts so $N\left(E\right)=13$. Therefore the probability of drawing a heart is $P\left(E\right)=\frac{N\left(E\right)}{N\left(S\right)}=\frac{13}{52}=\frac{1}{4}$.
• (b) A total of 52 outcomes is included in the sample space, so $N\left(S\right)=52$. Let $F$ be the event a black card is drawn. Then $F$ consists of the 26 black cards so $n\left(F\right)=26$. Therefore the probability of drawing a black card is $P\left(F\right)=\frac{N\left(F\right)}{N\left(S\right)}=\frac{26}{52}=\frac{1}{2}$.

• (a) Since only a roll of 6 wins, the other 5 possible rolls {1, 2, 3, 4, 5} don't win. Therefore the probability of not winning is $\frac{5}{6}$.
• (b) There are 6 possible rolls. Only one of them is a 5. Therefore the probability of rolling a 5 is $\frac{1}{6}$.

• (a)
  
• (b) {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

• (a) Let $E$ be the event toss two heads. Then $E=\left\{\text{HH}\right\}$, so $N\left(E\right)=1$. Therefore $P\left(E\right)=\frac{N\left(E\right)}{N\left(S\right)}=\frac{1}{4}$.
• (b) Let $F$ be the event toss two tails. Then $F=\left\{\text{TT}\right\}$, so $N\left(F\right)=1$. Therefore $P\left(F\right)=\frac{N\left(F\right)}{N\left(S\right)}=\frac{1}{4}$.

• (a) Let $E$ denote the event roll a sum of 7. Then $E$ consists of the outcomes {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}, so $N\left(E\right)=6$. Therefore $P\left(E\right)=\frac{N\left(E\right)}{N\left(S\right)}=\frac{6}{36}=\frac{1}{6}$.
• (b) Let $F$ denote the event roll a sum of 12. Then F consists of the outcome {(6, 6)}, so $N\left(F\right)=1$. Therefore $P\left(E\right)=\frac{N\left(E\right)}{N\left(S\right)}=\frac{1}{36}$.
• (c) Let $G$ denote the event roll a sum of 2. Then $G$ consists of the outcome {(1,1)}, so $N\left(G\right)=1$. Therefore $P\left(G\right)=\frac{N\left(G\right)}{N\left(S\right)}=\frac{1}{36}$.

• (a) Define $D$: Patient did not complete treatment.
• (b) Relative frequency of $D=\frac{\text{frequency of}D}{\text{number of trials in the experiment}}=\frac{944}{1413}\approx 0.6681$.
• (c) $P\left(D\right)\approx \hspace{0.17em}\text{relative frequency of}D\approx 0.6681$.

• (a) $B$ consists of the outcomes {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}, so $N\left(B\right)=5$. Therefore $P\left(B\right)=\frac{N\left(B\right)}{N\left(S\right)}=\frac{5}{36}$.
• (b) $B^C$ consists of all of the outcomes in the sample space that are not in $B$. Therefore $B$ consists of the outcomes {(1,1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 4), (5, 5), (5, 6), (6, 1), (6, 3), (6, 4), (6, 5), (6, 6)}, so $N\left(B^C\right)=31$. Therefore $P\left(B^C\right)=\frac{N\left(B^C\right)}{N\left(S\right)}=\frac{31}{36}$.

• (a) The set ($M$ and $N$) is the set of outcomes that are common to both $M$ and $N$. The green cell belongs to both the Male column and the Person did not survive row. The green cell represents (M and Nd, which includes the 1364 males who did not survive.

  	Female	Male	Total
  Did not survive	126	1364	1490
  Survived	344	367	711
  Total	470	1731	2201

• (b) The set ($M$ or $N$) is the set of outcomes that are in $M$ or $N$ or both. The green cells are the people that were either male or did not survive or both. Therefore ($M$ or $N$) is represented by the green cells.

  	Female	Male	Total
  Did not survive	126	1364	1490
  Survived	344	367	711
  Total	470	1731	2201

• (a) Here we seek $P\left(M\right)$. There were 1731 males out of a total of 2201 passengers aboard the Titanic. Therefore $P\left(M\right)=\frac{1731}{2201}$.
• (b) We are looking for $P\left(N\right)$. Of the 2201 passengers onboard, 1490 of them did not survive. Therefore $P\left(N\right)=\frac{1490}{2201}$.
• (c) Those who were both male and did not survive represent ($M$ and $N$). From Your Turn #9, p. 262, there are 1364 out of the 2201 passengers who were male and did not survive. Therefore $P\left(M\hspace{0.17em}\text{and}N\right)=\frac{1364}{2201}$.
• (d) Here we seek $P\left(M\hspace{0.17em}\text{or}N\right)$. From the Addition Rule, $P\left(M\hspace{0.17em}\text{or}N\right)=P\left(M\right)+P\left(N\right)-P\left(M\hspace{0.17em}\text{and}\hspace{0.17em}N\right)=\frac{1731}{2201}+\frac{1490}{2201}-\frac{1364}{2201}=\frac{1857}{2201}$.

• (a) The event did not respond to the marketing campaign is the complement of the event responded to the marketing campaign. Therefore $R^C$ denotes the event did not respond to the marketing campaign. $P\left(R^C\right)=\frac{N\left(R^C\right)}{N\left(S\right)}=\frac{240}{288}=0.8333$. We could have also used the formula for the probability of a complement. This gives us $P\left(R^C\right)=1-P\left(R\right)=1-0.1667=0.8333$.
• (b) The event did not have a credit card on file is the complement of the event has a credit card on file. Therefore $C^C$ denotes the event does not have a credit card on file. Then $P\left(R^C\hspace{0.17em}\text{given}\hspace{0.17em}{C}^C\right)=P\left(R^C|C^C\right)=\frac{N\left(R^C\hspace{0.17em}\text{and}{C}^C\right)}{N\left(C^C\right)}=\frac{161}{178}=0.9043$.

• $M$: Person is male.
• $N$: Person did not survive.

• $F$: Roll a 6 on the first toss.
• $D$: Roll a 6 on the second toss.

• $G$: Observe a heart on the first draw.
• $H$: Observe a heart on the second draw.

• $G$: Observe a heart on the first draw.
• $H$: Observe a heart on the second draw.

$\begin{array}{l}P\left(S_1\hspace{0.17em}\text{and}{S}_2\hspace{0.17em}\text{and}{S}_3\hspace{0.17em}\text{and}{S}_4\hspace{0.17em}\text{and}{S}_5\hspace{0.17em}\text{and}{S}_6\hspace{0.17em}\text{and}{S}_7\hspace{0.17em}\text{and}{S}_8\hspace{0.17em}\text{and}{S}_9\hspace{0.17em}\text{and}{S}_{10}\right)=P\left(S_1\right)·\hfill \\ \hspace{1em}P\left(S_2\right)·P\left(S_3\right)·P\left(S_3\right)·P\left(S_4\right)·P\left(S_6\right)·P\left(S_7\right)·P\left(S_8\right)·P\left(S_9\right)·P\left(S_{10}\right)\hfill \\ \begin{array}{l}=\left(0.24\right)\left(0.24\right)\left(0.24\right)\left(0.24\right)\left(0.24\right)\left(0.24\right)\left(0.24\right)\left(0.24\right)\left(0.24\right)\left(0.24\right)\\ ={\left(0.24\right)}^{10}\approx \mathrm{0.0000006340.}\end{array}\hfill \end{array}$

$\begin{array}{c}=1-P\left(\text{None of the 4 Americans smoke}\right)=1-P\left(\text{First one doesn}\text{'}\text{t smoke and Second one doesn}\text{'}\text{t smoke and Third one doesn}\text{'}\text{t smoke and Fourth one doesn}\text{'}\text{t smoke}\right)\\ =1-P\left(\text{First one doesn}\text{'}\text{t smoke}\right)·P\left(\text{Second one doesn}\text{'}\text{t smoke}\right)=1-{\left(0.76\right)}^4\\ \approx \mathrm{0.6664.}\end{array}$

$\begin{array}{c}P\left(H\hspace{0.17em}\text{given}\hspace{0.17em}C\right)=\frac{P\left(H\right)·P\left(C\hspace{0.17em}\text{given}H\right)}{P\left(H\right)·P\left(C\hspace{0.17em}\text{given}H\right)+P\left(H^C\right)·P\left(C\hspace{0.17em}\text{given}{H}^C\right)}\\ =\frac{\left(\frac{84}{1413}\right)\left(\frac{39}{84}\right)}{\left(\frac{84}{1413}\right)\left(\frac{39}{84}\right)+\left(\frac{1329}{1413}\right)\left(\frac{430}{1329}\right)}=\frac{\frac{39}{1413}}{\frac{39}{1413}+\frac{430}{1413}}=\frac{39}{469}\approx 0.08\end{array}$

• (a) 9! = 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 362,880
• (b) 10! = 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 3,628,800, as in Your Turn #22.

• (a) ${\hspace{0.17em}}_5{P}_3\hspace{0.17em}=\frac{5!}{\left(5-3\right)!}=\frac{5·4·3·2!}{2!}=5·4·3=60$
• (b) ${\hspace{0.17em}}_6{P}_4\hspace{0.17em}=\frac{6!}{\left(6-4\right)!}=\frac{6·5·4·3·2!}{2!}=6·5·4·3=360$
• (c) ${\hspace{0.17em}}_2{P}_2\hspace{0.17em}=\frac{2!}{\left(2-2\right)!}=\frac{2·1!}{0!}=2·1=2$

${\hspace{0.17em}}_{10}{C}_2\hspace{0.17em}=\frac{10!}{2!\left(10-2\right)!}=\frac{10!}{2!8!}=\frac{10·9·8!}{2·1·8!}=\frac{90}{2}=45$

• (a) ${\hspace{0.17em}}_5{C}_4\hspace{0.17em}=\frac{5!}{4!\left(5-4\right)!}=\frac{5!}{4!1!}=\frac{5·4!}{4·1!}=\frac{5}{1}=5$
• (b) ${\hspace{0.17em}}_5{C}_3\hspace{0.17em}=\frac{5!}{3!\left(5-3\right)!}=\frac{5!}{3!2!}=\frac{5·4·3!}{3·2·1!}=\frac{20}{2}=10$
• (c) ${\hspace{0.17em}}_5{C}_2\hspace{0.17em}=\frac{5!}{2!\left(5-2\right)!}=\frac{5!}{2!3!}=\frac{5·4·3!}{2·1·3!}=\frac{20}{2}=10$

Chapter 6

• (a) Your best friend's height is something that is measured, not counted. Therefore your best friend's height is a continuous variable. Possible values for your best friend's height in feet are $3\le H\le 7$.
• (b) The number of cats you have is something you count. Therefore the number of cats you have is a discrete variable. Possible values are {0,1,2,3,…}.

• (a) $P\left(X=1\hspace{0.17em}\text{point}or\hspace{0.17em}X=3\hspace{0.17em}\text{points}\right)=P\left(X=1\right)+P\left(X=3\right)=0.24+0.5=\mathrm{0.74.}$
• (b) The outcomes $X=1$ and $X=3$ mutually exclusive. Therefore, $P\left(X=1\hspace{0.17em}\text{point}\hspace{0.17em}and\hspace{0.17em}X=3\hspace{0.17em}\text{points}\right)=0$.
• (c) The phrase at most means “that many or fewer.” Thus, $P\left(X\le 3\right)=P\left(X=3\hspace{0.17em}\text{points}\hspace{0.17em}or\hspace{0.17em}X=1\hspace{0.17em}\text{point}or\hspace{0.17em}X=0\hspace{0.17em}\text{points}\right)=P\left(X=3\right)+P\left(X=1\right)+P\left(X=0\right)=0.5+0.24+0.26=1$.
• (d) The phrase at least means “that many or more.” Thus, $P\left(X\ge 3\right)=P\left(X=3\right)=0.5$.

$n=3,\hspace{0.17em}p=0.5,\hspace{0.17em}q=1-p=1-0.5=0.5$

• (a) $P\left(X=0\right)=\left({\hspace{0.17em}}_3{C}_0\right){\left(0.5\right)}^0{\left(0.5\right)}^{3-0}=\left(1\right)\left(1\right)\left(0.125\right)=0.125$
• (b) $P\left(X=0\right)=\left({\hspace{0.17em}}_3{C}_1\right){\left(0.5\right)}^1{\left(0.5\right)}^{3-1}=\left(3\right)\left(1\right)\left(0.25\right)=0.375$
• (c) $P\left(X\le 1\right)=P\left(X=0\right)+P\left(X=1\right)=0.500$

$n=10,\hspace{0.17em}p=0.5$

• (a) $\begin{array}{l}P\left(X\le 5\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)+\\ P\left(X=5\right)=0.0010+0.0098+0.0439+0.1172+0.2051+0.2461=\\ \mathrm{0.6231.}\end{array}$

  From the TI-84 Plus: 0.623046875

• (b) $\begin{array}{l}P\left(X<6\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+\\ P\left(X=4\right)+P\left(X=5\right)=0.0010+0.0098+0.0439+0.1172+\\ 0.2051+0.2461=\mathrm{0.6231.}\end{array}$

  From the TI-84 Plus: 0.623046875

• (c) From the binomial table:

  $\begin{array}{l}P\left(0\le X\le 5\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+\\ P\left(X=4\right)+P\left(X=5\right)=0.0010+0.0098+0.0439+0.1172+\\ 0.2051+0.2461=\mathrm{0.6231.}\end{array}$

  From the TI-84 Plus: 0.623046875

• (d) From the binomial table:

  $\begin{array}{l}P\left(0<X<5\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)\\ =0.0098+0.0439+0.1172+0.2051=\mathrm{0.376.}\end{array}$

  From the TI-84 Plus: 0.3759765625

$n=50,\hspace{0.17em}p=0.6,\hspace{0.17em}q=1-p=1\sigma$

• (a) $\mu =n·p=\left(50\right)\left(0.6\right)=30$
• (b) $\begin{array}{l}{\sigma }^2=n·p·q=\left(50\right)\left(0.6\right)\left(0.4\right)=12\\ \sigma =\sqrt{{\sigma }^2}=\sqrt{12}=3.46101615\approx 3.4641\end{array}$
• (c) We use the $Z$-score method. The $Z$-score for $X=36$ is $z=\frac{X-\mu }{\sigma }=\frac{36-30}{3.4641}\approx 1.73\hspace{0.17em}\text{since}-2<Z<2,\hspace{0.17em}X=36$ is neither an outlier nor unusual.

• (a) Here, $X=10$, so the probability that $X=10$ is

  $P\left(10\right)=\frac{{10}^{10}·e^{-10}}{10!}\approx \mathrm{0.1251.}$

• (b) Here, $X=12$, so the probability that $X=12$ is

  $P\left(12\right)=\frac{{10}^{12}·e^{-10}}{12!}\approx \mathrm{0.0948.}$

• (a) Mean = μ = 10, Variance = ${\sigma }^2$ = 10, Standard deviation = $\sigma =\sqrt{\mu }=\sqrt{10}\approx 3.1623$.

• (b) A data value farther than 2 standard deviations from the mean is considered moderately unusual. The number of customers to a boutique shop that lie 2 standard deviations above and below the mean are calculated as follows:

  $\begin{array}{c}\mu +2\sigma \approx 10+2\left(3.1623\right)=16.3246\\ \mu -2\sigma \approx 10-2\left(3.1623\right)=3.6754\end{array}$

  Thus if there were 3 or less customers to the boutique shop this would be considered moderately unusual. Similarly, if there were 17 or more customers to the boutique shop this would also be considered moderately unusual.

$\text{Area}=\text{base}\times \text{height}=\left(8-4\right)\times 0.1=4\times 0.1=0.4$

$1-0.8997=\mathrm{0.1003.}$

$1-0.975=\mathrm{0.025.}$

$Z=-1.645\hspace{0.17em}\text{and}\hspace{0.17em}Z=\mathrm{1.645.}$

$1-0.7673=\mathrm{0.2327.}$

$\begin{array}{c}P\left(305<X<605\right)=P\left(\frac{305-514}{118}<\frac{X-514}{118}<\frac{605-514}{118}\right)\\ \approx P\left(-1.77<Z<0.77\right)=0.7794-0.0384=\mathrm{0.741.}\end{array}$

$\begin{array}{l}P\left(X_{\text{binomial}}<12\right)\approx P\left(Y_{\text{normal}}<11.5\right)\approx P\left(\frac{Y_{\text{normal}}-12.8}{3.2}<\frac{11.5-12.8}{3.2}\right)\\ \hspace{1em}\approx P\left(Z<-0.41\right)=\mathrm{0.3409.}\end{array}$

Chapter 7

$Z=\frac{18.5-20}{1}=-\mathrm{1.50.}$

$\hat{p}=\frac{X}{n}=\frac{1009}{3058}\approx \mathrm{0.33.}$

${\sigma }_{\hat{p}}=\sqrt{\frac{p·q}{n}}=\sqrt{\frac{0.043·\left(1-0.043\right)}{225}}\approx \sqrt{0.0001828933333}\approx \mathrm{0.01352.}$

$P\left(\hat{p}<0.04\right)=\mathrm{0.4122.}$

$Z=\frac{0.04-{\mu }_{\hat{p}}}{{\sigma }_{\hat{p}}}=\frac{0.04-0.043}{0.01352}\approx -0.22$

Chapter 8

• (a) The sample mean yield is calculated as

  $\bar{x}=\frac{\sum x}{n}=\frac{36+68+65+70+63}{5}=\mathrm{60.4.}$

• (b) The point estimate of μ, the unknown nationwide mean winter wheat yield for all 50 states, is 60.4 bushels per acre.

• (a) Since the population is not normally distributed and the sample size is less than 30, the $Z$ interval may not be used.
• (b) Since the sample size is greater than or equal to 30 and σ is known, the $Z$ interval may be used.

• (a) Table 1 gives us $Z_{\alpha /2}=2.576$.
• (b) Table 1 gives us $Z_{\alpha /2}=1.645$.

$\text{lower}\hspace{0.17em}\text{bound}=\bar{x}-Z_{\alpha /2}\left(\sigma /\sqrt{n}\right)\hspace{1em}\hspace{1em}\hspace{1em}\text{upper}\hspace{0.17em}\text{bound}=\bar{x}+Z_{\alpha /2}\left(\sigma /\sqrt{n}\right)$

$\text{lower}\hspace{0.17em}\text{bound}=510-1.96\left(118/\sqrt{25}\right)=463.74$

$\text{upper}\hspace{0.17em}\text{bound}=510+1.96\left(118/\sqrt{25}\right)=556.26$

$\text{lower}\hspace{0.17em}\text{bound}=\bar{x}-Z_{\alpha /2}\left(\sigma /\sqrt{n}\right)$

$\text{upper}\hspace{0.17em}\text{bound}=\bar{x}+Z_{\alpha /2}\left(\sigma /\sqrt{n}\right)$

$\text{lower}\hspace{0.17em}\text{bound}=20.71-2.576\left(5.637/\sqrt{100}\right)\approx 20.71-1.45=19.26$

$\text{upper}\hspace{0.17em}\text{bound}=20.71+2.576\left(5.637/\sqrt{100}\right)\approx 20.71+1.45=22.16$

$n=\left(\frac{Z_{\alpha /2}·\sigma }{E}\right)={\left(\frac{1.96·\$5000}{\$100}\right)}^2=9604$

• (a) The sample size is not large ($n$ is not ≥ 30) and we are not told that the population is normal. Therefore, the conditions are not met for the $t$ interval for μ. It is not okay to construct the $t$ interval.
• (b) We are not told that the population is normal. However, the sample size is large ($n$ is ≥ 30). Therefore, the conditions are met for the $t$ interval for μ. It is okay to construct the $t$ interval.

$\text{lower}\hspace{0.17em}\text{bound}=\bar{x}-t_{\alpha /2}\left(s/\sqrt{n}\right),\hspace{0.17em}\text{upper}\hspace{0.17em}\text{bound}=\bar{x}+t_{\alpha /2}\left(s/\sqrt{n}\right)$

$\text{lower}\hspace{0.17em}\text{bound}=185.9-\left(2.131\right)\left(56.8/\sqrt{16}\right)=185.9-30.3=159.6$

$\text{upper}\hspace{0.17em}\text{bound}=185.9+\left(2.131\right)\left(56.8/\sqrt{16}\right)=185.9+30.3=216.2$

$E=\left(2.131\right)\left(56.8/\sqrt{16}\right)=30.3$

• (a) $\hat{p}=\frac{x}{n}=\frac{50}{100}=0.5$

  The point estimate of the proportion $p$ is 0.5.

• (b) $\hat{p}=\frac{x}{n}=\frac{90}{160}=0.5625$

  The point estimate of the proportion $p$ is 0.5625.

$\begin{array}{c}\text{lower}\hspace{0.17em}\text{bound}=\hat{p}-Z_{\alpha /2}\sqrt{\frac{\hat{p}·\hat{q}}{n}}=0.5-1.96\sqrt{\frac{0.5\left(0.5\right)}{100}}\\ =0.5-1.96\left(0.05\right)=0.5-0.098=0.402\\ \text{upper}\hspace{0.17em}\text{bound}\hspace{0.17em}=\hat{p}+Z_{\alpha /2}\sqrt{\frac{\hat{p}·\hat{q}}{n}}=0.5+1.96\sqrt{\frac{0.5\left(0.5\right)}{100}}\\ =0.5+1.96\left(0.05\right)=0.5+0.098=0.598\end{array}$

$\begin{array}{c}\text{lower}\hspace{0.17em}\text{bound}=\hat{p}-Z_{\alpha /2}\sqrt{\frac{\hat{p}·\hat{q}}{n}}=0.5625-1.96\sqrt{\frac{0.5625\left(0.4375\right)}{160}}\\ =0.5625-1.96\left(0.0392184387\right)=0.5625-0.0768681399=0.4856318601\approx 0.4856\\ \text{upper}\hspace{0.17em}\text{bound}\hspace{0.17em}=\hat{p}+Z_{\alpha /2}\sqrt{\frac{\hat{p}·\hat{q}}{n}}=0.5625+1.96\sqrt{\frac{0.5625\left(0.4375\right)}{160}}\\ =0.5625+1.96\left(0.0392184387\right)=0.5625+0.0768681399=0.6393681399\approx 0.6394\end{array}$

• (a) The margin of error is

  $E=Z_{\alpha /2}\sqrt{\frac{\hat{p}·\hat{q}}{n}}=1.96\sqrt{\frac{0.5\left(0.5\right)}{100}}=1.96\left(0.05\right)=0.098$

  We can estimate the population proportion to within 0.098 with 95% confidence.

• (b) The margin of error is

  $E=Z_{\alpha /2}\sqrt{\frac{\hat{p}·\hat{q}}{n}}=1.96\sqrt{\frac{0.5625\left(0.4375\right)}{160}}=0.0768681399\approx 0.0769$

  We can estimate the population proportion to within 0.0769 with 95% confidence.

$n=\hat{p}·\hat{q}{\left(\frac{Z_{\alpha /2}}{E}\right)}^2=0.37\left(0.63\right){\left(\frac{2.576}{0.03}\right)}^2\approx 1718.665984$

$n={\left[\frac{0.5Z_{\alpha /2}}{E}\right]}^2={\left[\frac{\left(0.5\right)\left(1.96\right)}{0.05}\right]}^2=384.16$

$\left(1-\alpha \right)=0.95\hspace{1em}\hspace{1em}\frac{\alpha }{2}=\frac{0.05}{2}=0.025\hspace{1em}\hspace{1em}1-\frac{\alpha }{2}=1-0.025=0.975$

Chapter 9

• Step 1 Search the word problem for certain key English words and select the appropriate symbol.

  The problem uses the word “increased” which means “is greater than.” Thus, we will use a hypothesis that contains the > symbol.

• Step 2 Determine the form of the hypothesis.

  From Table 1, we see that the > symbol means that we use a right-tail test:

  $H_0:\mu ={\mu }_0\hspace{1em}\hspace{1em}\text{versus}\hspace{1em}\hspace{1em}{H}_a:\mu >{\mu }_0$

• Step 3 Find the value for ${\mu }_0$ and write your hypotheses.

  The alternative hypothesis $H_a$ states that the mean monthly amount of time that iPhone and Android users spend using the apps on their devices has increased from some value ${\mu }_0$. Increased from what? 30 hours per month. Write the two hypotheses with ${\mu }_0=30$.

  $H_0:\mu =30\hspace{1em}\hspace{1em}\text{versus}\hspace{1em}\hspace{1em}{H}_a:\mu >30$

• (a) A Type I error would be to reject $H_0$ when $H_0$ is true. This would be to conclude that μ is less than 7 when, in reality, $\mu =7$. In other words, a Type I error would be to conclude that the population mean of the pair of dice tossed in Example 1 is less than 7 when, in reality, it is equal to 7.

  A Type II error occurs when we do not reject $H_0$ when $H_0$ is false. This would be to conclude that μ is equal to 7 when, in reality, it is less than 7.

• (b) A Type I error would be to reject $H_0$ when $H_0$ is true. This would be to conclude that μ had decreased when, in reality, it had stayed the same. In other words, a Type I error would be to conclude that the population mean rainfall in Arizona had decreased when, in reality, it had not decreased.

  A Type II error occurs when we do not reject $H_0$ when $H_0$ is false. This would be to conclude that μ had stayed the same when, in reality, it had decreased.

$z_{\text{data}}=\frac{\bar{x}-{\mu }_0}{\sigma /\sqrt{n}}=\frac{480-413}{670/\sqrt{36}}=0.6$

• (a) We have a right-tailed test and level of significance $\alpha =0.10$, so Table 4 tells us that the critical value is $Z_{\text{crit}}=1.28$.
• (b)
  

• (a) We have a right-tailed test, so that the $p$-value is the area in the right tail:

  $p\text{<mo>-</mo>value}=P\left(Z>Z_{\text{data}}\right)=P\left(Z>0.5\right)$

  The $Z$ table gives the probability for $P\left(Z<0.5\right)$. Thus,

  $p\text{<mo>-</mo>value}=P\left(Z>0.5\right)=1-P\left(Z<0.5\right)=1-0.6915=0.3085$

  

• (b) We have a left-tailed test, so that the $p$-value is the area in the left tail:

  $p\text{<mo>-</mo>value}=P\left(Z<Z_{\text{data}}\right)=P\left(Z<-1.2\right)=0.1151$

  
  Page A-16

• (c) Here, we have a two-tailed test, so that the $p$-value is the sum of the area in the two tails:

  $\begin{array}{c}p\text{<mo>-</mo>value}=P\left(Z<-\left|Z_{\text{data}}\right|\right)+P\left(Z>\left|Z_{\text{data}}\right|\right)=P\left(Z<-\left|-0.1\right|\right)+P\left(Z>\left|-0.1\right|\right)\\ =P\left(Z<-0.1\right)+P\left(Z>0.1\right)=0.4602+0.4602=0.9204\end{array}$

  

• (a) The $p$-value of 0.1587 implies that there is no evidence against the null hypothesis that the population mean equals 3.0.
• (b) The $p$-value of 0.0735 implies that there is moderate evidence against the null hypothesis that the population mean equals 10.
• (c) The $p$-value of 0.0456 implies that there is solid evidence against the null hypothesis that the population mean equals 100.

• • (a) Step 1 State the hypotheses.

    From Example 18 the hypotheses are

    $H_0:\mu =15\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu <15$

    where $\mu$ refers to the population mean age at onset.

  • Step 2 Find $t_{\text{crit}}$ and state the rejection rule.

    We still have a left-tailed test and df = 19, but now $\alpha =0.01$. From the table $t_{\text{crit}}=2.539$. Therefore we will reject $H_0$ if $t_{\text{data}}\le -2.539$.

  • Step 3 Calculate $t_{\text{data}}$.

    From Example 18 $t_{\text{data}}=-2.2183$.

  • Step 4 State the conclusion and interpretation.

    The rejection rule from Step 2 says to reject $H_0$ if $t_{\text{data}}\le -2.539$. From Step 3, we have $t_{\text{data}}=-2.2183$. Because −2.2183 is not less than −2.539, our conclusion is to not reject $H_0$. There is insufficient evidence at level of significance $\alpha =0.01$ that the population mean age of onset has decreased from its previous level of 15 years.

• (b) In Example 18, $\alpha =0.05$ and the conclusion is to reject $H_0$. In (a), $\alpha =0.01$ and the conclusion is to not reject $H_0$. This is because the cutoff for rejecting the null hypothesis at $\alpha =0.01=-2.539<t_{\text{data}}=-2.2183\le -1.729=$ the cutoff for rejecting the null hypothesis at $\alpha =0.05$.

  From the “Developing Your Statistical Sense” box on p. 515, there are two alternatives available in situations like this.

  1. Turn to a direct assessment of the strength of evidence against the null hypothesis.
  2. Obtain more data, perhaps through a call for further research.

• (a) The hypotheses do not depend on α. Therefore, a change in α will not affect the hypotheses.
• (b) The test statistic $t_{\text{data}}$ does not depend on α. Therefore, a change in α will not affect $t_{\text{data}}$.
• (c) The $p$-value does not depend on α. Therefore, a change in α will not affect the $p$-value.
• (d) The conclusion does depend on α. Decreasing α from $\alpha =0.05$ to $\alpha =0.01$ changes the rejection rule from reject $H_0$ if the $p-\text{value}\le \hspace{0.17em}\alpha =0.05$ to reject $H_0$ if the $p-\text{value}\le \alpha =0.01$. The $p-\text{value}=0.0228$ is not ≤ 0.01. Therefore we do not reject $H_0$. Thus the conclusion has changed from reject $H_0$ if $\alpha =0.05$ to do not reject $H_0$ if $\alpha =0.01$.

$H_0:\mu =15\hspace{1em}\text{versus}\hspace{1em}{H}_a:\mu <15.$

$\hat{p}=\frac{x}{n}=\frac{\text{number in sample that are Chromebooks}}{\text{sample size}}=\frac{50}{400}=\mathrm{0.125.}$

$Z_{\text{data}}=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0·q_0}{n}}}=\frac{0.125-0.20}{\sqrt{\frac{0.20\left(0.80\right)}{400}}}=\frac{-0.075}{0.02}=-3.75$

• Step 1 State the hypotheses and the rejection rule.

  The hypotheses are

  $H_0:p=0.12\hspace{1em}\hspace{1em}\text{versus}\hspace{1em}\hspace{1em}{H}_a:p>0.12$

  where $p$ represents the population proportion of young people ages 18–24 who had an accident. We reject $H_0$ if the $p$-value $\le \alpha =0.05$.

• Step 2 Calculate $Z_{\text{data}}$.

  Our sample proportion is $\hat{p}=150/1000=0.15$.

  Page A-17

  Thus, our test statistic is

  $Z_{\text{data}}=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0·q_0}{n}}}=\frac{0.15-0.12}{\sqrt{\frac{0.12\left(0.88\right)}{1000}}}\approx 2.92$

• Step 3 Find the $p$-value.

  We have a right-tailed test, so the $p$-value is

  $\begin{array}{c}p\text{<mo>-</mo>value}=P\left(Z>Z_{\text{data}}\right)=P\left(Z>2.92\right)=1-P\left(Z<2.92\right)\\ =1-0.9982=0.0018\end{array}$

• Step 4 State the conclusion and the interpretation.

  The $p$-value 0.0018 is $\le \alpha =0.05$, so we reject $H_0$. There is evidence that the population proportion of young people ages 18–24 who had an accident has increased.

Chapter 10

$\begin{array}{ll}\text{Henrik}:\hfill & 90-92=-2\hfill \\ \text{Ivana}:\hfill & 70-70=0\hfill \\ \text{Jen}:\hfill & 76-75=1\hfill \\ \text{Kayla}:\hfill & 61-60=1\hfill \\ \text{Luisa}:\hfill & 60-58=2\hfill \\ \text{Manuel}:\hfill & 90-80=4\hfill \end{array}$

• Step 1 State the hypotheses.

  The hypotheses are

  $H_0:{\mu }_d=0\hspace{1em}\text{versus}\hspace{1em}{H}_a:{\mu }_d>0$

  where ${\mu }_d$ represents the population mean difference in English quiz scores.

• Step 2 Find the critical value $t_{\text{crit}}$ and state the rejection rule.

  We have a right-tailed test with area in one tail equal to $\alpha =0.10$ and degrees of freedom equal to $\text{df}=n-1=6-1=5$. Therefore our critical value from the table is $t_{\text{crit}}=1.476$. Therefore our rejection rule is that we will reject $H_0$ when $t_{\text{data}}$ is greater than or equal to 1.476.

• Step 3 Find $t_{\text{data}}$.

  From Your Turn #1, ${\bar{x}}_d=1\hspace{0.17em}\text{and}\hspace{0.17em}{s}_d=2$. Therefore

  $t_{\text{data}}=\frac{{\bar{x}}_d}{s_d/\sqrt{n}}=\frac{1}{2/\sqrt{6}}\approx 1.22$

• Step 4 State the conclusion and the interpretation.

  Since $t_{\text{data}}\approx 1.22$ is not greater than or equal to 1.476, we do not reject $H_0$. There is insufficient evidence at level of significance $\alpha =0.10$ that the population mean difference of English quiz scores is greater than 0.

$\begin{array}{c}\text{lower bound}={\bar{x}}_d-t_{\alpha /2}\left(s_d/\sqrt{n}\right)\\ =1-\left(2.015\right)\left(2/\sqrt{6}\right)\\ \approx 1-1.6452=-0.6452\\ \text{upper}\hspace{0.17em}\text{bound}={\bar{x}}_d+t_{\alpha /2}\left(s_d/\sqrt{n}\right)\\ =1+\left(2.015\right)\left(2/\sqrt{6}\right)\\ \approx 1+1.6452=2.6452\end{array}$

• (a) $H_0:{\mu }_d=2\hspace{1em}\text{vs}\text{.}\hspace{1em}{H}_a:{\mu }_d\ne 2$

  ${\mu }_0=2$ lies inside the interval, so we do not reject $H_0$.

• (b) $H_0:{\mu }_d=5.6\hspace{1em}\text{vs}\text{.}\hspace{1em}{H}_a:{\mu }_d\ne 5.6$

  ${\mu }_0=5.6$ lies outside the interval, so we reject $H_0$.

• (c) $H_0:{\mu }_d=5.7\hspace{1em}\text{vs}\text{.}\hspace{1em}{H}_a:{\mu }_d\ne 5.7$

  ${\mu }_0=5.7$ lies outside the interval, so we reject $H_0$.

Chapter 11

• (a) There are $k=4$ possible outcomes: paperbacks, hardcovers, e-Books, and all other formats. Assigning probabilities using the relative frequency method, we have the following hypothesized proportions for each book format:

  $p_{\text{paperback}}=0.41,\hspace{0.17em}{p}_{\text{hardcover}}=0.34,\hspace{0.17em}{p}_{\text{e}-\text{Book}}=0.13,\hspace{0.17em}{p}_{\text{all other formats}}=0.12$

  and

  $\sum p_i=0.41+0.34+0.13+0.12=1.$

  Therefore, $X$ = book format is a valid multinomial random variable.

• (b) We have $n=2000$ books (sample size = 2000), so the expected frequencies are as provided in the following table.

  Category	$Expected\hspace{0.17em}frequenc{y}_i=E_i=n·p_i$
  Paperback	$E_{\text{paperback}}=2000·0.41=820$
  Hardcover	$E_{\text{hardcover}}=2000·0.34=680$
  e-Book	$E_{\text{e<mo>-</mo>Book}}=2000·0.13=260$
  All other formats	$E_{\text{all other formats}}=2000·0.12=240$

${\chi }_{\text{data}}^2=\sum \frac{{\left(O_i-E_i\right)}^2}{E_i}\approx 0.122+0+1.538+0.417=2.077$

• Step 1 State the hypotheses and check the conditions. The hypotheses are:

  $H_0:p_{\text{paperback}}=0.41,\hspace{0.17em}\hspace{0.17em}{p}_{\text{hardcover}}=0.34,\hspace{0.17em}{p}_{\text{e}-\text{Book}}=0.13,\hspace{0.17em}{p}_{\text{all}\hspace{0.17em}\text{other}\hspace{0.17em}\text{formats}}=0.12$

  $H_a$ : The random variable does not follow the distribution specified in $H_0$.

  Checking the conditions, the expected frequencies from Your Turn #1 are

  $E_{\text{paperback}}=820,\hspace{0.17em}\hspace{0.17em}{E}_{\text{hardcover}}=680,\hspace{0.17em}{E}_{\text{e}-\text{Book}}=260,\hspace{0.17em}{E}_{\text{all}\hspace{0.17em}\text{other}\hspace{0.17em}\text{formats}}=240$

  Because none of these expected frequencies is less than one, and none of the expected frequencies is less than five, the conditions for performing the goodness of fit test are met.

• Step 2 Find the ${\chi }^2$ critical value, ${\chi }_{\text{crit}}^2$ and state the rejection rule.

  We have degrees of freedom $k-1=4-1=3$ and $\alpha =0.05$ Then, from the table, ${\chi }_{\text{crit}}^2={\chi }_{0.05}^2=7.815$. The rejection rule is “Reject$H_0$ if ${\chi }_{\text{data}}^2\ge 7.815$.”

• Step 3 Find ${\chi }_{\text{data}}^2$.

  From Your Turn #2, ${\chi }_{\text{data}}^2\approx 2.077$

• Step 4 State the conclusion and interpretation.

  Compare ${\chi }_{\text{data}}^2$ with ${\chi }_{\text{crit}}^2·{\chi }_{\text{data}}^2\approx 2.077$ is not $\ge \hspace{0.17em}\text{to}\hspace{0.17em}{\chi }_{\text{crit}}^2=7.815$. Therefore, we do not reject $H_0$. There is insufficient evidence that the variable book format does not follow the distribution specified in $H_0$. In other words, there is insufficient evidence that the book format market shares have changed.

Chapter 12

Chapter 13

• (a)

  

  As age increases, the score on the video game tends to increase.

• (b) As calculated by the TI-84, $r=0.894427191\approx 0.8944$. Age and score are positively correlated. An increase in age is associated with an increase in the score on the video game.
• (c) As calculated by the TI-84, $\hat{y}=2x+54$. The slope $b_1=2$ means there is an estimated increase of 2 in the score on the video game for each one year increase in age. The $y$-intercept $b_0=54$ is the estimated score on the video game of someone aged $x=0$.

• (d) For a 22-year-old person, the estimated score on the video game is $\hat{y}=2x+54=2\left(22\right)+54=98$.

  The actual score of the 22-year-old person in the sample is $y=100$. Thus, our prediction error is $y-\hat{y}=100-98=2$. The 22-year-old person scored slightly higher than expected.

• (a)

  $x=\text{Age}$	$y$ = score on video game	Fitted (predicted) values $\hat{y}=2\left(\text{age}\right)+54$	Residuals $y-\hat{y}$
  14	80	82	−2
  16	90	86	4
  18	90	90	0
  20	90	94	−4
  22	100	98	2

• (b)

  

  The scatterplot of the residuals versus the fitted values shows no evidence of the unhealthy patterns shown in Figure 4. Thus, the independence assumption, the constant variance assumption, and the zero-mean assumption are verified.

  

  Also, the normal probability plot of the residuals indicates no evidence of departures from normality in the residuals. Therefore, we conclude that the regression assumptions are verified.

• Step 1 State the hypotheses and the rejection rule.

  $H_0:{\beta }_1=0$ No linear relationship exists between age and score.

  $H_a:{\beta }_1\ne 0$ A linear relationship exists between age and score.

  Reject $H_0$ if the $p-\text{value}\le \alpha =0.05$.

  Page A-19

• Step 2 Calculate $t_{\text{data}}$.

  $t_{\text{data}}=\frac{b_1}{s/\sqrt{{\sum \left(x-\bar{x}\right)}^2}}$

  From Your Turn #1, $b_1=2$. From the TI-84, $s_x^2=10$. Therefore,

  $\sum {\left(x-\bar{x}\right)}^2=\left(n-1\right)s_x^2=\left(5-1\right)10=40.$

  $\text{SSE}=\sum {\left(y-\hat{y}\right)}^2$. The residuals $y-\hat{y}$ were calculated in Your Turn #2 (a). They are −2, 4, 0, −4, and 2. Therefore,

  $\begin{array}{c}\text{SSE}=\sum {\left(y-\hat{y}\right)}^2={\left(-2\right)}^2+{\left(4\right)}^2+{\left(0\right)}^2+{\left(-4\right)}^2+{\left(2\right)}^2\\ =4+16+0+16+4=40.\hspace{0.17em}\text{Then}\\ s=\sqrt{\frac{\text{SSE}}{n-2}}=\sqrt{\frac{40}{5-2}}=\mathrm{3.651483717.}\end{array}$

  Thus

  $t_{\text{data}}=\frac{b_1}{s/\sqrt{\sum {\left(x-\bar{x}\right)}^2}}=\frac{2}{3.65148317/\sqrt{40}}=3.464101615\approx \mathrm{3.4641.}$

• Step 3 Find the $p$-value.

  Using the TI-84 we get $p-\text{value}=0.0405193264$.

• Step 4 State the conclusion and the interpretation.

  The $p$-value = 0.0405193264 is $\le \alpha =0.05$, so we reject $H_0$. Evidence exists, at level of significance $\alpha =0.05$, for a linear relationship between age and score.

• Step 1 State the hypotheses $H_0:{\beta }_1=0$ No linear relationship exists between age and score.

  $H_a:{\beta }_1\ne 0$ A linear relationship exists between age and score.

• Step 2 Find the critical value $t_{\text{crit}}$ and the rejection rule.

  The degrees of freedom are $\text{df}=n-2=5-2=3$. We have a two-tailed test with $\alpha =0.05$. From the table, $t_{\text{crit}}=3.182$. Therefore the rejection rule is Reject $H_0$ if $t_{\text{data}}\le -3.182\hspace{0.17em}\text{or}\hspace{0.17em}{t}_{\text{data}}\ge 3.182$.

• Step 3 Calculate $t_{\text{data}}$.

  $t_{\text{data}}=\frac{b_1}{s/\sqrt{\sum {\left(x-\bar{x}\right)}^2}}$

  From Your Turn #1, $b_1=2$. From the TI-84, $s_x^2=10$. Therefore,

  $\sum {\left(x-\bar{x}\right)}^2=\left(n-1\right)s_x^2=\left(5-1\right)10=40.$

  $\text{SSE}=\sum {\left(y-\hat{y}\right)}^2$. The residuals $y-\hat{y}$ were calculated in Your Turn #2 (a). They are −2, 4, 0, −4, and 2. Therefore,

  $\begin{array}{c}\text{SSE}=\sum {\left(y-\hat{y}\right)}^2={\left(-2\right)}^2+{\left(4\right)}^2+{\left(0\right)}^2+{\left(-4\right)}^2+{\left(2\right)}^2\\ =4+16+0+16+4=40.\hspace{0.17em}\text{Then}\\ s=\sqrt{\frac{\text{SSE}}{n-2}}=\sqrt{\frac{40}{5-2}}=\mathrm{3.651583717.}\end{array}$

  Thus

  $t_{\text{data}}=\frac{b_1}{s/\sqrt{\sum {\left(x-\bar{x}\right)}^2}}=\frac{2}{3.65148317/\sqrt{40}}=3.464101615\approx \mathrm{3.4641.}$

• Step 4 State the conclusion and the interpretation.

  $t_{\text{data}}\approx 3.4641$ is ≥ 3.182, so we reject $H_0$. Evidence exists, at level of significance $\alpha =0.05$, for a linear relationship between age and score.

  Page A-20