5.6 Working Backwards, Part II

n <- 25

# margin of error is (b-a)/2
ME <- ((77-65)/2)
ME
## [1] 6
# sample mean is calculated as (a+b)/2 
se <- ((77+65)/2)
se
## [1] 71
# using t(.05)*s/sqrt(n) to calculate the sample standard devation
df <- 25-1
t.value <- qt(.95, df)
t.value
## [1] 1.710882
sd <- (ME/t.value)*5
sd
## [1] 17.53481

5.14 SAT Scores

a) Raina wants to use a 90% confidence interval. How large of a sample should she collect?

me1 <- 25
s.d <- 250
z <- qnorm(0.95)
z
## [1] 1.644854
n <- ((z * s.d)/ me1)^2
n
## [1] 270.5543

b) Luke wants to use a 99% confidence interval. Without calculating the actual sample size, determine whether his sample should be larger or smaller than Raina’s, and explain your reasoning.

Luke's sample size should be larger than Raina's.

c) Calculate the minimum required sample size for Luke.

me2 <- 25
sd <- 250
z <- qnorm(0.99)
z
## [1] 2.326348
n <- ((z * sd)/ me2)^2
n
## [1] 541.1894

5.20 High School and Beyond, Part I

a) Is there a clear difference in the average reading and writing scores?

No. The difference in schores is approximately normal.

b) Are the reading and writing scores of each student independent of each other?

Yes, each student's readinga and writing scores are independent of other student's scores.

c) Create hypotheses appropriate for the following research question: is there an evident difference in the average score of students in the reading and writing exam?

H0: There is no difference in the average of reading and writing score.
Ha: There is difference in the average of reading and writing score.

d) Check the conditions required to complete this test.

Independence of observations and the observations should be aprroximately normal distribution.

e) The average observed difference in scores iss x{read-write}=-0.545, and the standard deviation of the differences is 8.887 points.

mean_diff <- -0.545
sd <- 8.887
n <- 200

# Find the standard erro
sde <- sd / sqrt(n)
# T stat
t <- (mean_diff - 0)/ sde
df1 <- n - 1
p <- pt(t, df=df1)
p
## [1] 0.1934182
P is more than 0.05 we fail to reject the null hypothesis.

f) What type of error might we have made? Explain what the error means in the context of the application.

It is Type II error, we incorrectly reject the alternative hypothesis.

g) Based on the results of this hypothesis test, would you expect a confidence interval for the average difference between the reading and writing scores to include 0? Explain your reasoning.

There is no difference between writing and reading, because we fail to reject null hypothesis,

5.32 Fuel efficiency of manual and automatic cares, Part I

H0: There is no difference between automatic car’s MPG and manual car’s MPG. Ha: There is differnce between automatic car’s MPG and manual car’s MPG.

n <- 26
mean_diff1 <- 16.12 - 19.85
standard_erro_diff <- sqrt((3.58^2/n)+(4.51^2/n))
standard_erro_diff
## [1] 1.12927
t1 <- (mean_diff1 - 0)/ standard_erro_diff
t1 
## [1] -3.30302
P_value <- pt(t1, df=n-1)
P_value
## [1] 0.001441807
We reject the null hypothesis and accept alternative hypothesis because the p-value is less than 0.05. In this case, there is strong evidence to support the difference between manual and automatic transmissions.

5.48 Work hours and education

a) Write the hypotheses for evaluating whether the average number of hours worked varies across the five groups.

H0:There is no difference in average hours worked across the 5 groups.
Ha: At least 1 group has a difference in the average hours worked with the other groups.

b) Check conditions and describe any assumptions you must make to proceed with the test.

The observations are independent and approximately normal, and the variablity across the groups is about the same.

c) Below is part of the output associated with this test. Fill in the empty cells.

n <- 1172
k <- 5
df1 <- k -1
df2 <- n-k
mean_total <- 40.45
group <- data.frame(n=c(121,546,97,253,155),
                    sd=c(15.81,14.97,18.1,13.62,15.51),
                    mean=c(38.67, 39.6, 41.39,42.55,40.85))
sg <- sum (group$n * (group$mean - mean_total)^2)
sg
## [1] 2004.101
Msg <- (1/df1)*sg
sse <- 267382
mse <- sse / df2
mse
## [1] 229.1191
F <- Msg / mse
F
## [1] 2.186745
Df Sum Sq Mean Sq F value Pr (>F)
degree 4 2004.1 501.54 2.186745 0.0682
Residuals 1167 267382 229.12
Total 1171 269386.1

d) What is the conclusion of the test?

Since P-value = 0.0682, that is greater than 0.05, we fail to reject the null hypothesis. There is no difference between groups.