5.6

Sample mean = (77+65)/2 = 71

Margin of error = (77-65)/2 = 6

Sample standard deviation = 17.54 (see steps below)

Margin of error = T score x SE

T score = 1.71 based on 90% confidence interval and df = 24

So 6 = 1.71 x SE

SE = 3.51

SE = Sample standard deviation / sqrt(n)

3.51 = s / sqrt(25)

s = 17.54

5.12

(a)

Ho: Police officer average blood level concentration = 35 μg/l

Ha: Police officer average blood level concentration > 35 μg/l

(b)

Independence is reasonable because I would think that 52 police officers is less than 10% of urban police officers. The normality assumption should be good given that the sample size is greater than 30.

(c)

x <- rnorm(52, mean = 124.32, sd = 37.74)
t.test(x, mu = 35, conf.level = 0.95, alternative = "greater")
## 
##  One Sample t-test
## 
## data:  x
## t = 17.808, df = 51, p-value < 2.2e-16
## alternative hypothesis: true mean is greater than 35
## 95 percent confidence interval:
##  115.8797      Inf
## sample estimates:
## mean of x 
##  124.2786

P-value is less than 0.05 significance level so reject the null hypothesis and conclude that downtown police officers have a higher lead exposure

(d)

The result would not change because the p-value is still lower than a 0.01 significance level.

5.18

(a)

Yes, because each day has a value for Intel and a corresponding value for Southwest.

(b)

Yes, because each item has a value for Target and a corresponding value for Walmart.

(c)

No, because each would only have one corresponding value, which would be their own SAT score.

5.24

False, because there needs to be a connection between the two groups. If not, you cannot use a paired test.

5.30

Point estimate = 56.81 - 44.51 = 12.30

SE = sqrt((13.322/23)+(16.132/23)) = 4.36

df = 22 so the T statistic for 95% CI is 2.07

CI = 12.30 +- (2.07 * 4.36)

(3.275,21.325)

5.36

Ho: Average number of items recalled by the patients eating = average number of items recalled by the patients in the control group

Average number of items recalled by the patients eating ≠ average number of items recalled by the patients in the control group

Point estimate = 6.1 - 4.9 = 1.2

SE = sqrt((1.82/22)+(1.82/22)) = 0.543

T statistic = (1.2 - 0)/0.543 = 2.21

Using df = 21, the distribution table tells us that the p-value is between 0.05 and 0.02 so p-value is less than 0.05 significance level and we can reject the null. The data provide strong evidence that the average number of food items recalled by the patients in the treatment and control groups are different

5.42

ANOVA test would be most appropriate because we want to test the sample means across more than two groups.

5.48

(a)

Ho: Mu1 = Mu2 = Mu3 = Mu4 = Mu5

Ha: At least one mean is different

(b)

The observations are independent within and across groups, because each sample represents less than 10% of the population for that group and we can assume that groups are not impacted by each other.

The data within each group are nearly normal. Outliers are less of a concern due to the large sample size of the groups with the most outliers.

The variability across the groups is about equal based on the box plots provided.

(c)

degree DF = 4 (5-1)

residuals DF = 1,167 (1,172-5)

total DF = 1,171

degree Sum Sq = 2006 (degree DF x degree Mean Sq)

total Sum Sq = 269,388

residuals Mean Sq = 229.12 (residuals Sum Sq / residuals DF)

F value = 2.189 (degree Mean Sq / residuals Mean Sq)

(d)

The p-value is larger than 0.05, indicating the evidence is not strong enough to reject the null hypothesis at a significance level of 0.05.