statistical math question?

Question

Experience raising New Jersey Red chickens revealed the mean weight of the chickens at five months is 4.35 pounds. The weights follow the normal distribution. In an effort to increase their weight, a special additive is added to the chicken feed. The subsequent weights of a sample of five-month-old chickens were (in pounds):

4.41 4.37 4.33 4.35 4.30 4.39 4.36 4.38 4.40 4.39
At the .01 level, has the special additive increased the mean weight of the chickens? Estimate the p-value.

Answer 1

I have the notes below on how to do this by hand. the quick solution is to use software like R.

> data <- c(4.41, 4.37, 4.33, 4.35, 4.30, 4.39, 4.36, 4.38, 4.40, 4.39)
> t.test(data, NULL, "greater", 4.35)

One Sample t-test

data: data
t = 1.6777, df = 9, p-value = 0.06386
alternative hypothesis: true mean is greater than 4.35
95 percent confidence interval:
4.348333 Inf
sample estimates:
mean of x
4.368

since the p-value is 0.06386, larger than the significance level we conclude the null hypothesis is plausible; that the mean is the same or smaller.


/// ==== notes === \\\

Hypothesis Test for mean:

assuming you have a large enough sample such that the central limit theorem holds and the mean is normally distributed then to test the null hypothesis H0: μ = Δ

find the test statistic z = (xBar - Δ) / (sx / sqrt(n))

where xbar is the sample average
sx is the sample standard deviation
n is the sample size

The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.

H1: μ > Δ; p-value is the area to the right of z
H1: μ < Δ; p-value is the area to the left of z
H1: μ ≠ Δ; p-value is the area in the tails greater than |z|

for a small sample test for the mean every thing is the same save the test statistic is a t statistic with n - 1 degrees of freedom. However, in this case the underlying distribution must be normal for this test to be valid.

Answer 2

Standard deviation is not given, so it must be computed from the sample.
sample mean=4.37
sample sd=0.03 (approximately)
t=(xbar-mean)/s/sqrt(n)
=(4.37-4.35)/0.03/sqrt(10)=0.02/0.0095=2.105
Look at the t-table with 9 degrees of freedom (n-1), n = sample size(10)
t critical value at alpha=0.01(9 df)=2.821
No evidence to support the claim that the additive has increased the mean weight of chickens at 0.01 level.
Please check all of my computation.
Compute the standard deviation.

Answer 3

mean is just average so,
4.41+4.37 +4.33 +4.35 +4.30+ 4.39 +4.36+ 4.38+ 4.40+ 4.39=43.68
43.68/10=4.368
4.368-4.35=.018

you should have it from here