5.6

(sample_mean <- (65 + 77) / 2)

## [1] 71

n <- 25
df <- n - 1
(ME <- 77 - sample_mean)

## [1] 6

t <- abs(qt(.1/2, 24))
SE <- ME / t
(SD <- SE * sqrt(n))

## [1] 17.53481

5.14

(a)

SD <- 250 
ME <- 25
z <- 1.65
n <- z^2 * SD^2 / ME^2 

sample size should be greater than 

ceiling(n)

## [1] 273

(b)
since sample size is directly proportional to the square of z score, Luke
needs a sample size larger than Raina.

(c)

SD <- 250 
ME <- 25
z <- 2.58
n <- z^2 * SD^2 / ME^2 

sample size should be greater than 

ceiling(n)

## [1] 666

5.20

(a) There is no clear difference in the average reading and writing scores
as the distribution is fairly normal around 0.

(b) They are dependent.

(c) H0: u(reading) - u(writing) = 0  
    HA: u(reading) - u(writing) not equal 0
    
(d) Students are selected simple random sample and represent less than 10%
of all students. The sample size is atleast 30. The box plots shows no obvious
outliers so data is not skewed.

(e) 

x = -0.545
u = 8.887
n = 200
(SE = u / sqrt(n))

## [1] 0.6284058

(T <- (x - 0) / SE)

## [1] -0.867274

(p <- pt(T, n - 1))

## [1] 0.1934182

Since p > 0.05, we cannot reject null hypothesis. The data doesn't provide
convincing evidence for difference.

(f) Type 2 Error, since we may have incorrectly failed to reject H0. 
There may be difference, but we were unable to detect it.

(g) Yes, since we failed to reject H0, which had a null value of 0.

5.32
u = u(automatic) - u(manual)
H0: u = 0
HA: u not equal to 0

x_auto <- 16.12
x_manual <- 19.85
sd_auto <- 3.58
sd_manual <- 4.51
n_auto <- n_manual <- 26
(SE <- sqrt(sd_auto^2 / n_auto + sd_manual^2 / n_manual))

## [1] 1.12927

(t <- ((x_auto - x_manual) - 0) / SE)

## [1] -3.30302

df <- n_auto - 1
(p <- pt(t, df))

## [1] 0.001441807

Since the p-value is less than significance level (0.05), we reject the null 
hypothesis H0 and conclude that there is strong evidence of a difference in 
fuel efficiency between manual and automatic transmissions.

5.48
a. H0: Average hours worked is the same for all educational attainments
   HA: Atlease one pair of means are different
   
b. Independence: We don’t have any information on how the data were collected, 
so we cannot assess independence. To proceed, we must assume the subjects in 
each group are independent. In practice, we would inquire for more details. 

Approx.normal: Fairly symmetrical with some categories like HS and Bachelor's
have visible outliers so some skew. Since the sample size is fairly large,
skew is acceptable.

Constant variance: This condition is sufficiently met, as the standard 
deviations are reasonably consistent across groups.

            Df    Sum Sq    Mean Sq   F value   Pr(>F)
degree      4     2006.16    501.54     2.19    0.0682
Residuals  1167   267,382    229.12
Total      1171   269388.16

Calculation
-----------
Sum sq = Mean Sq * df = 501.54 * 4 = 2006.16
Total Sum Sq = 2006.16 + 267382 = 269388.16
Residuals Mean Sq = 267382 / 1167 = 229.12
F value = 501.54 / 229.12 = 2.19

(d) since p value (0.682) greater than 0.05, test fail to reject H0. 
The data do not provide convincing evidence of a difference between the average
hours worked for all educational attainments.