Only D and e need help ### Zinc Content Analysis in Pet FoodA pet food company is concerned that their pet food must have a certain percentage of the mineral zinc. Insufficient zinc levels can make dogs very sick. If the machinery is working correctly, the average zinc content is supposed to be $ \mu = 0.375 $ grams per pound of food. They sample 25 bags of food and find:- Sample mean ($ \bar{x} $): $ 0.359 $ grams- Standard deviation ($ \sigma $): $ 0.05 $ gramsThe firm wants to determine if their pet food is healthy by conducting a Z-test at a 95% level of significance.#### Analysis Steps:a) **State the Hypotheses:**- **Null Hypothesis ($ H_0 $):** The average zinc content is $ 0.375 $ grams per pound ($ \mu = 0.375 $).- **Alternative Hypothesis ($ H_a $):** The average zinc content is different from $ 0.375 $ grams per pound ($ \mu \neq 0.375 $).b) **Draw a Picture/Graph:** - Although not depicted here, a graph involves plotting a normal distribution curve with the mean of $ 0.375 $ grams at the center. Highlight the rejection regions on both tails based on the Z-test critical values.c) **Find the Critical Value:** - For a 95% confidence level, the Z critical values are approximately $ \pm 1.96 $.d) **Compute the Test Statistic:** - The Z-test formula is: \[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] - Substituting the values, \[ Z = \frac{0.359 - 0.375}{\frac{0.05}{\sqrt{25}}} = \frac{-0.016}{0.01} = -1.6 \]e) **Conclusion:** - Since $-1.6$ is within the range of $-1.96$ to $1.96$, we fail to reject the null hypothesis. Thus, there is not enough statistical evidence to suggest the average zinc content is different from \( 0

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Description: Only D and e need help ### Zinc Content Analysis in Pet FoodA pet food company is concerned that their pet food must have a certain percentage of the mineral zinc. Insufficient zinc levels can make dogs very sick. If the machinery is working correctl

The provided information are: Population meanμ=0.375Sample meanx¯=0.379Population standard…

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A pet food company is concerned that their pet food must have a certain % of the mineral zinc in the food. Too little and dogs get very sick. If the machinery is working correctly, the average zinc is u = .375 gram/lb of food. They sample 25 bags of food and find: x = .359 grams o = .05 grams The firm wants to know if their pet food is healthy. They decide on a 95% level of significance. Do the Z test. They ask you to perform the analysis. a) State the null and alternative hypotheses. Но: На: b) Draw a picture/graph of what you are trying to find. c) Find the critical value you will use. d) Compute the test statistic (from the data) e) What do you conclude? Why?

A pet food company is concerned that their pet food must have a certain % of the mineral zinc in the food. Too little and dogs get very sick. If the machinery is working correctly, the average zinc is u = .375 gram/lb of food. They sample 25 bags of food and find: x = .359 grams o = .05 grams The firm wants to know if their pet food is healthy. They decide on a 95% level of significance. Do the Z test. They ask you to perform the analysis. a) State the null and alternative hypotheses. Но: На: b) Draw a picture/graph of what you are trying to find. c) Find the critical value you will use. d) Compute the test statistic (from the data) e) What do you conclude? Why?

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A percentage is a number indicated as a fraction of 100. It is a dimensionless number often expressed using the symbol %.

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Only D and e need help

### Zinc Content Analysis in Pet Food

A pet food company is concerned that their pet food must have a certain percentage of the mineral zinc. Insufficient zinc levels can make dogs very sick. If the machinery is working correctly, the average zinc content is supposed to be $ \mu = 0.375 $ grams per pound of food.

They sample 25 bags of food and find:
- Sample mean ($ \bar{x} $): $ 0.359 $ grams
- Standard deviation ($ \sigma $): $ 0.05 $ grams

The firm wants to determine if their pet food is healthy by conducting a Z-test at a 95% level of significance.

#### Analysis Steps:

a) **State the Hypotheses:**

- **Null Hypothesis ($ H_0 $):** The average zinc content is $ 0.375 $ grams per pound ($ \mu = 0.375 $).
- **Alternative Hypothesis ($ H_a $):** The average zinc content is different from $ 0.375 $ grams per pound ($ \mu \neq 0.375 $).

b) **Draw a Picture/Graph:**
- Although not depicted here, a graph involves plotting a normal distribution curve with the mean of $ 0.375 $ grams at the center. Highlight the rejection regions on both tails based on the Z-test critical values.

c) **Find the Critical Value:**
- For a 95% confidence level, the Z critical values are approximately $ \pm 1.96 $.

d) **Compute the Test Statistic:**
- The Z-test formula is:
\[
Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]
- Substituting the values,
\[
Z = \frac{0.359 - 0.375}{\frac{0.05}{\sqrt{25}}} = \frac{-0.016}{0.01} = -1.6
\]

e) **Conclusion:**
- Since $-1.6$ is within the range of $-1.96$ to $1.96$, we fail to reject the null hypothesis. Thus, there is not enough statistical evidence to suggest the average zinc content is different from \( 0

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