The error bound for the mean would decrease because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean.

2,185

6,765

595

1. 1. 8629
   2. 6944
   3. 35
   4. 34
2. t34
   ${�}_{34}$ 
3. 1. CI: (6244, 11,014)

   2. 

      Figure 8.19
4. It will become smaller

1. 1. x−
      $\overline{�}$ 

      = 2.51

   2. sx
      ${�}_{�}$ 

      = 0.318

   3. n = 9
   4. n – 1 = 8
2. the effective length of time for a tranquilizer
3. the mean effective length of time of tranquilizers from a sample of nine patients
4. We need to use a Student’s-t distribution, because we do not know the population standard deviation.
5. 1. CI: (2.27, 2.76)
   2. Check student’s solution.
6. If we were to sample many groups of nine patients, 95% of the samples would contain the true population mean length of time.

x¯=$251,854.23

$\overline{�}=\$251,854.23$ 

s= $521,130.41

$�=\$521,130.41$ 

Note that we are not given the population standard deviation, only the standard deviation of the sample.

There are 30 measures in the sample, so n = 30, and df = 30 – 1 = 29

CL = 0.96, so α = 1 – CL = 1 – 0.96 = 0.04

α2=0.02tα2=t0.02

$\frac{�}{2}=0.02{�}_{\frac{�}{2}}={�}_{0.02}$ = 2.150

EBM=tα2(sn√)=2.150(521,130.4130√) ~ $204,561.66

$���={�}_{\frac{�}{2}}\left(\frac{�}{\sqrt{�}}\right)=2.150\left(\frac{521,130.41}{\sqrt{30}}\right) ~ \$204,561.66$ 

x−−

$\overline{�}$ – EBM = $251,854.23 – $204,561.66 = $47,292.57

x−−

$\overline{�}$ + EBM = $251,854.23+ $204,561.66 = $456,415.89

We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89.

1. 1. x¯
      $\overline{�}$ 

      = 11.6

   2. sx
      ${�}_{�}$ 

      = 4.1

   3. n = 225
   4. n – 1 = 224
2. X is the number of unoccupied seats on a single flight. X−
   $\overline{�}$ 

   is the mean number of unoccupied seats from a sample of 225 flights.

3. We will use a Student’s-t distribution, because we do not know the population standard deviation.
4. 1. CI: (11.12 , 12.08)
   2. Check student’s solution.

1. 1. CI: (7.64 , 9.36)

   2. 

      Figure 8.20
2. The sample should have been increased.
3. Answers will vary.
4. Answers will vary.
5. Answers will vary.

b

1. 1,068
2. The sample size would need to be increased since the critical value increases as the confidence level increases.

1. X = the number of people who feel that the president is doing an acceptable job;

   P′ = the proportion of people in a sample who feel that the president is doing an acceptable job.

2. N(0.61,(0.61)(0.39)1200−−−−−−−−√)
   $�\left(0.61,\sqrt{\frac{(0.61)(0.39)}{1200}}\right)$ 
3. 1. CI: (0.59, 0.63)
   2. Check student’s solution

1. 1. (0.72, 0.82)
   2. (0.65, 0.76)
   3. (0.60, 0.72)
2. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.
3. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families.
4. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Black person into their families.

1. X = the number of adult Americans who feel that crime is the main problem; P′ = the proportion of adult Americans who feel that crime is the main problem
2. Since we are estimating a proportion, given P′ = 0.2 and n = 1000, the distribution we should use is N(0.2,(0.2)(0.8)1000−−−−−−√)
   $�\left(0.2,\sqrt{\frac{(0.2)(0.8)}{1000}}\right)$ 

   .

3. 1. CI: (0.18, 0.22)
   2. Check student’s solution.
4. One way to lower the sampling error is to increase the sample size.
5. The stated “± 3%” represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3%. Thus, they estimate the percentage of adult Americans who feel that crime is the main problem to be between 18% and 22%.

c

a

1. p′ = (0.55 + 0.49)2
   $\frac{\text{(0}\text{.55 + 0}\text{.49)}}{\text{2}}$ 

   = 0.52; EBP = 0.55 – 0.52 = 0.03

2. No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this.
3. CL = 0.75, so α = 1 – 0.75 = 0.25 and α2=0.125 zα2=1.150
   $\frac{�}{2}=0.125 {�}_{\frac{�}{2}}=1.150$ 

   . (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.)
   EBP=(1.150)0.52(0.48)1,000−−−−−−−√≈0.018

   $���=(1.150)\sqrt{\frac{0.52(0.48)}{1,000}}\approx 0.018$ 

   
   (p′ – EBP, p′ + EBP) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538)

4. Yes – this interval does not fall less than 0.50 so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75% confidence.

1. 1. 71
   2. 2.8
   3. 48
2. X is the height of a male Swede, and x––
   $\stackrel{\_}{�}$ 

   is the mean height from a sample of 48 male Swedes.

3. Normal. We know the standard deviation for the population, and the sample size is greater than 30.
4. 1. CI: (70.151, 71.85)

   2. 

      Figure 8.21
5. The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean.

1. 1. x¯
      $\overline{�}$ 

      = 23.6

   2. σ
      $�$ 

      = 7

   3. n = 100
2. X is the time needed to complete an individual tax form. X¯¯¯
   $\overline{�}$ 

   is the mean time to complete tax forms from a sample of 100 customers.

3. N(23.6,7100√)
   $�\left(23.6,\frac{7}{\sqrt{100}}\right)$ 

   because we know sigma.

4. 1. (22.228, 24.972)

   2. 

      Figure 8.22
5. It will need to change the sample size. The firm needs to determine what the confidence level should be, then apply the error bound formula to determine the necessary sample size.
6. The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval.
7. According to the error bound formula, the firm needs to survey 206 people. Since we increase the confidence level, we need to increase either our error bound or the sample size.

1. 1. 7.9
   2. 2.5
   3. 20
2. X is the number of letters a single camper will send home. X¯¯¯
   $\overline{�}$ 

   is the mean number of letters sent home from a sample of 20 campers.

3. N 7.9(2.520√)

   $7.9\left(\frac{2.5}{\sqrt{20}}\right)$ 

4. 1. CI: (6.98, 8.82)

   2. 

      Figure 8.23
5. The error bound and confidence interval will decrease.

1. x−
   $\overline{�}$ 

   = $568,873

2. CL = 0.95 α = 1 – 0.95 = 0.05 zα2
   ${�}_{\frac{�}{2}}$ 

   = 1.96
   EBM = z0.025σn√

   ${�}_{0.025}\frac{�}{\sqrt{�}}$ 

   = 1.96 90920040√

   $\frac{909200}{\sqrt{40}}$ 

   = $281,764

3. x−
   $\overline{�}$ 

   − EBM = 568,873 − 281,764 = 287,109
   x−

   $\overline{�}$ 

   + EBM = 568,873 + 281,764 = 850,637

4. We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637.

Higher

It would increase to four times the prior value.

No, It could have no affect if it were to change to 1 – p, for example. If it gets closer to 0.5 the minimum sample size would increase.

Yes

1. No
2. No