(a) Is there a clear di↵erence in the average reading and writing scores?

No, there does not appear to be a significant difference because the (read-write) histogram was normally distributed and grouped around zero.

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

I would guess that the two samples are not independent because one students reading would impact the same students writing. We can assume that they are independent between students.

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

\[H_o:\mu_r-\mu_w = 0 \] \[H_a:\mu_r-\mu_w \ne 0 \]

(d) Check the conditions required to complete this test.

The conditions required to perform the hypothesis test are independence and normalcy.

1. Independence

• Because they are paired they are not independent.

2. Normally distributed

• They do look normally distributed. It is still possible to do a t test on these.

(e) The average observed difference in scores is \(\bar x_{read} − \bar x_{write} = −0.545\) , and the standard deviation of the differences is 8.887 points. Do these data provide convincing evidence of a difference between the average scores on the two exams?

sd <- 8.887
xbar <- 0.545
tVal <- xbar/sd
n <- 200
p <- pt(tVal, n-1)

Since the p value for the test is only 0.5244192 we cannot reject the null hypothesis that the means are different.

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

There are two types of error:

1. Type 1 error: *Reject the null when the true value is not actually different.
2. Type 2 error: *Do not reject the null when there is a difference between the means.

Since we did not reject the null, we might be suffering from Type 2 error.

(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.

Yes, because we tested whethere or not the means were different from each other, to reject the null would require the 95% CI to not include zero, because we did not reject the null, we know that the 95% CI does include zero.

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

\[H_O: \mu_l=\mu_{hs}=\mu_{Jr Col}=\mu_{barhalors}=\mu_{grad} \] \[H_a: \exists \mu_i \ne \mu_j \]

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

1. The observations are indipendant within and between groups.

• This is most likely true.

2. The data within groups is normal.

• This is also most likely true because we can see some individual points on the bar charts that don’t look clumped.

3. The cross group variance is relitively equal.

• This also looks like it is satisfied.

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

xbar <- c(38.67, 39.6, 41.39, 42.55, 40.85)
sd <- c(15.81, 14.97, 18.1, 13.62, 15.51)
n <- c(121, 546, 97, 253, 155)
df <- data.frame (xbar, sd, n)
df %>% knitr::kable()

n <- sum(df$n)
k <- nrow(df)
degFre <- n - k - 1
res <- n - k

fStat <- qf(1 - 0.0682, degFre, res)

MSE <- 501.54/fStat

SSD <- degFre * 501.54
SSE <- 267382 # Ambiguous naming. 
SST <- SSD + SSE

(d) What is the conclusion of the test?

Since 0.5244192 is > 0.05, we fail to reject that any of the means is different.