Hypothesis Testing: Mean Chicken Weight

In Graph Form:

Raising chickens in commercial chicken farms is a growing industry, as there is a shift from red meat to more white meat. New Jersey Red Chickens are a favorite chicken and the Feed is Us company has a new chicken food that they claim is excellent for increasing weight and naturally it costs more. You have heard this line before and therefore are skeptical but the potential pay off is huge. You buy a small amount and feed it to ten chickens, which you choose at random.

Chicken Weights

4.41

4.37

4.33

4.35

4.30

4.39

4.36

4.38

4.40

4.39

At the .01 level of significance is there evidence that the chicken weight exceeds the average chicken weight of 4.35 pounds.

The first step of the hypothesis testing procedure is to determine the alternate and the null hypothesis. Based on our problem statement we are trying to decide if the New Jersey Red Chickens which ate the new feed on average weigh more than the chickens that ate regular food which weighs 4.35 pounds. In our case, the alternate hypothesis (H1) states the chickens that ate the feed weigh more than the chickens that didn't. Therefore, the null hypothesis (H0) states that the chickens who ate the feed weigh less than or equal to the chickens that did not eat the food.
The second step of the hypothesis testing procedure is to state the decision rule. Using an alpha of .01 means we want to be 99% sure of our results. Since the sample size is small we will be using a Student's t Test. The degrees of freedom we will be using for this t Test is n-1, which in our case equal 9. Using the Student's t Chart the critical value is 2.821. Therefore, the decision rule is "reject the null hypothesis if t > 2.821.
The third step of the hypothesis testing procedure is to calculate the actual t value. By using the t formula we get a t value of 1.678. This value is less than the critical value in step two of 2.821.
In step four, we will fail to reject the null hypothesis because 1.678 < 2.821. This leads us to step five and our conclusion that the average weight of the chickens that have been fed the special feed do not exceed the average weight of the chickens that did not eat the special feed.
State the problem:
H0:µ<=4.35
H1:µ>4.35
Decision Rule: Reject the H0 if t>2.821 with 9 degrees of freedom and an &#945;=.01
Student's t Test: t = (Xbar-&#956;)/(s/&#8730;n)=1.678
Decision: Fail to reject the H0
Conclusion: The average weight of the chickens with the special feed does not exceed the average weight of the chickens without the special feed.