Answered

2) 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 5.2 eggs/week.

Sarah has 46 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 5.4 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. Assume the population standard deviation to be [tex]\sigma = 1.1[/tex] eggs/day. Compute the value of the test statistic.

A) 1.29
B) 0.18
C) 1.23
D) 0.89



Answer :

GinnyAnswer
To determine whether the mean laying rate for Sarah's hens is statistically higher than the mean rate for all Golden Comets, we need to calculate a test statistic. We'll use the Z-test formula for the population mean with the known population standard deviation. Here is a detailed, step-by-step solution to the problem:

Step 1: Identify the given values.
- Population mean (μ): 5.2 eggs/day
- Sample mean ([tex]\( \bar{x} \)[/tex]): 5.4 eggs/day
- Population standard deviation (σ): 1.1 eggs/day
- Sample size (n): 46 hens

Step 2: Write down the Z-test formula for the population mean:
[tex]\[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \][/tex]

Step 3: Substitute the given values into the formula.
[tex]\[ Z = \frac{5.4 - 5.2}{\frac{1.1}{\sqrt{46}}} \][/tex]

Step 4: Simplify the denominator by calculating [tex]\( \frac{1.1}{\sqrt{46}} \)[/tex].
[tex]\[ \sqrt{46} \approx 6.782 \][/tex]
[tex]\[ \frac{1.1}{6.782} \approx 0.1622 \][/tex]

Step 5: Plug the simplified value back into the formula.
[tex]\[ Z = \frac{0.2}{0.1622} \][/tex]

Step 6: Perform the division.
[tex]\[ Z \approx 1.233150906022777 \][/tex]

So, the computed value for the test statistic is approximately 1.23.

Answer:
C) 1.23