Question

The Golden Comet is a hybrid chicken that is prized for its high egg production rate and gentle disposition. According to recent studies, the mean rate of egg production for 1-year-old Golden Comets is 5.0 eggs/week. Sarah has 39 1-year-old hens that are fed exclusively on natural scratch feed: insects, seeds, and plants that the hens obtain as they range freely around the farm. Her hens exhibit a mean egg-laying rate of 5.2 eggs/day. Sarah wants to determine whether the mean laying rate μ for her hens is higher than the mean rate for all Golden Comets. Assume the population standard deviation to be eggs/day.
i. State the appropriate null and alternate hypotheses.
ii. Compute the value of the test statistic.
iii. State a conclusion. Use the level of significance.

Answer 1

Answer:

There is not enough evidence to state that mean laying rate μ for her hens is higher than the mean rate for all Golden Comets.      

Step-by-step explanation:

We are given the following in the question:

Population mean, μ =  5.0 eggs/week

Sample mean, $\bar{x}$ = 5.2 eggs/day

Sample size, n = 39

Alpha, α = 0.05

Population standard deviation, σ = 1.1 eggs/Day

a) First, we design the null and the alternate hypothesis

$H_{0}: \mu = 5\\H_A: \mu > 5$

We use one-tailed z test to perform this hypothesis.

b) Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{5.2 - 5}{\frac{1.1}{\sqrt{39}} } = 1.135$

c) Now, $z_{critical} \text{ at 0.05 level of significance } = 1.64$

Since,  

$z_{stat} < z_{critical}$

We fail to reject the null hypothesis and accept the null hypothesis. Thus, mean laying rate μ for Sarah's hens is same as the mean rate for all Golden Comets.  There is not enough evidence to state that mean laying rate μ for her hens is higher than the mean rate for all Golden Comets.