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, [tex]\bar{x}[/tex] = 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
[tex]H_{0}: \mu = 5\\H_A: \mu > 5[/tex]
We use one-tailed z test to perform this hypothesis.
b) Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{5.2 - 5}{\frac{1.1}{\sqrt{39}} } = 1.135[/tex]
c) Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]
Since,
[tex]z_{stat} < z_{critical}[/tex]
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.
