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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![### 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9ff3616-361e-41ff-bd24-ad1fe99bbb43%2Fd0ae1ffe-8676-457a-a8a8-1d3b1cc7256c%2Fjqee11h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### 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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