5.31 Chicken diet and weight, Part I.

a) Describe the distributions of weights of chickens that were fed linseed and horsebean.

By oberving the box-and-whisker plot of linseed and horsebean, linseed has an approximately normal distribution because it appears to be symmetric, and the horsebean has a skewd left distibution since it has a longer tail on the left.

b) Do these data provide strong evidence that the average weights of chickens that were fed linseed and horsebean are different? Use a 5% significance level.

Let’s state the hypothesis test first.

\[H_0: mean_l - mean_h = 0\] \[H_A: mean_l - mean_h != 0\] Here, we will use 2-tailed sample t test.

mean.diff <- 218.75 - 160.20
df <- 12 + 10 - 2
pooled.var <-(52.24^2 * 11 + 38.63^2 * 9) / df
se.diff <- sqrt(pooled.var/11 + pooled.var/9)
t.obt <- mean.diff / se.diff
t.obt
## [1] 2.794803
p.value <- 2 * pt(abs(t.obt), df = df, lower.tail = F)
p.value
## [1] 0.01118485

Since the test statistics is 2.79 which is greater than the critical value of T is 2.08, and P-value is 0.01112 which is less than alpha/2 = 0.025, we can reject \[H_0\] and accept \[H_A\]. It means that these data provide strong evidence that average weights of chickens that were fed linseed and horsebean are different.

c) What type of error might we have committed? Explain.

Type I error: We may incorrect reject a true null hypotheis. It means that the average weights of chickens that were fed linseed and horsebean are not different but we rejected it.

D) Would your conclusion change if we used alpha = 0.01?

If we use alpha = 0.01, The critical t is 2.85 which is greater than t statistics and P-value is also greater than alpha/2 = 0.005. We fail to reject \[H_0\].