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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 5.1 eggs/week.

Sarah has 37 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 5.5 eggs/day.

Sarah wants to determine whether the mean laying rate [tex]$\mu$[/tex] for her hens is higher than the mean rate for all Golden Comets. State the appropriate null and alternate hypotheses.

A) [tex]$H_0: \mu\ \textless \ 5.5, H_1: \mu=5.5$[/tex]

B) [tex]$H_0: \mu=5.5, H_1: \mu \neq 5.5$[/tex]

C) [tex]$H_0: \mu=5.1, H_1: \mu\ \textgreater \ 5.1$[/tex]

D) [tex]$H_0: \mu\ \textgreater \ 5.1, H_1: \mu=5.1$[/tex]



Answer :

GinnyAnswer
To determine whether Sarah's hens have a higher mean egg-laying rate than the general mean for Golden Comets, we can set up the following hypotheses:

1. Null Hypothesis (H0): This is the hypothesis that there is no effect or difference. Since Sarah wants to find out if her hens' mean laying rate is higher than the general mean, we start with the assumption that her hens' mean laying rate is equal to the general mean laying rate. Hence, [tex]\( H_0: \mu = 5.1 \)[/tex].

2. Alternative Hypothesis (H1): This is the hypothesis that indicates the presence of an effect or difference. Since Sarah suspects that her hens might be laying more eggs, we are interested in testing if the mean laying rate for her hens is greater than the general mean. Therefore, the alternative hypothesis would be [tex]\( H_1: \mu > 5.1 \)[/tex].

In hypothesis notation for this scenario:
- The null hypothesis ([tex]\( H_0 \)[/tex]) is that the mean laying rate of Sarah's hens is equal to the general mean laying rate: [tex]\( H_0: \mu = 5.1 \)[/tex].
- The alternative hypothesis ([tex]\( H_1 \)[/tex]) is that the mean laying rate of Sarah's hens is higher than the general mean laying rate: [tex]\( H_1: \mu > 5.1 \)[/tex].

This matches option C:
[tex]$ \underline{H_0}: \mu = 5.1, \underline{H_1}: \underline{\mu} > 5.1 $[/tex]

So, the correct answer is option C.

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