Q:

Experience raising New Jersey Red chickens revealed the mean weight of the chickens atfive months is 4.35 pounds. The weights follow the normal distribution. In an effort to increasetheir weight, a special additive is added to the chicken feed. The subsequentweights 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 4.39At the .01 level, has the special additive increased the mean weight of the chickens? Estimatethe p-value.

Accepted Solution

A:
Answer:p-value = 0.1277Step-by-step explanation:p-value is the probability value tell us how likely it is to get a result like this if the Null Hypothesis is true.Firstly we find the mean and standard deviation of the given data set.β‡’ Mean = [tex]\frac{4.41 +4.37+ 4.33+ 4.35 +4.30 +4.39 +4.36+ 4.38+ 4.40+ 4.39}{10}[/tex]β‡’ Mean = 4.368[tex]Standard deviation(\sigma) = \sqrt{\frac{1}{n}\sum_{i=1}^{n}{(x_{i}-\bar{x})^{2}} }[/tex]where, [tex]\bar{x}[/tex] is mean of the distribution.β‡’ Standard Deviation = 0.034 Applying t- test:Let out hypothesis is:Hβ‚€: ΞΌ = 4.35H₁: ΞΌ β‰  4.35Now, Here, ΞΌ = Population Mean = 4.35[tex]\bar{x}[/tex]= Sample Mean = 4.368Οƒ = Standard Deviation = 0.034n = 10[tex]t=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}} }[/tex]Putting all values we get, t = 1.6777 with (10 -1) = 9 degree of freedom.Then the p-value at 99% level of significance.β‡’ p-value = 0.1277