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. State the null and alternative hypotheses. o: a: Draw a picture/graph of what you are trying to find. Find the critical value you will use. Compute the test statistic (from the data) What do you conclude? Why?

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**Title: Conducting a Z Test for Mineral Content in Pet Food**

**Introduction:**
A pet food company is concerned about maintaining an adequate percentage of the mineral zinc in their products. Insufficient zinc levels can result in health issues for pets. The standard zinc content, if the machinery is functioning properly, is \( \mu = 0.375 \) grams per pound of food. Upon sampling 25 bags of food, the company finds:

- Sample mean \( \bar{x} = 0.359 \) grams
- Standard deviation \( \sigma = 0.05 \) grams

The firm aims to determine if the pet food is healthy by checking if the zinc content is on target. A significance level of 95% is chosen for this test.

**Steps for Analysis:**

**a) State the Null and Alternative Hypotheses:**

- **Null Hypothesis (\( H_0 \)):** The average zinc content is equal to the target value. \( \mu = 0.375 \) grams.
  
- **Alternative Hypothesis (\( H_a \)):** The average zinc content is not equal to the target value. \( \mu \neq 0.375 \) grams.

**b) Visual Representation:**
Create a graph of a normal distribution curve to illustrate the standard zinc content with the sample mean highlighted. This will help visualize where the sample mean falls relative to the expected mean.

**c) Find the Critical Value:**
Using a Z-table, determine the critical Z-value corresponding to a 95% confidence level. For a two-tailed test, the critical Z-values are approximately \( \pm 1.96 \).

**d) Compute the Test Statistic:**
The Z-score is calculated using the formula:

\[
Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]

Where:
- \( \bar{x} = 0.359 \) grams
- \( \mu = 0.375 \) grams
- \( \sigma = 0.05 \) grams
- \( n = 25 \) bags

\[ 
Z = \frac{0.359 - 0.375}{\frac{0.05}{\sqrt{25}}} 
\]

This calculation gives a Z-score which will be used to determine if the sample mean is significantly different from the hypothesized population mean.

**e
Transcribed Image Text:**Title: Conducting a Z Test for Mineral Content in Pet Food** **Introduction:** A pet food company is concerned about maintaining an adequate percentage of the mineral zinc in their products. Insufficient zinc levels can result in health issues for pets. The standard zinc content, if the machinery is functioning properly, is \( \mu = 0.375 \) grams per pound of food. Upon sampling 25 bags of food, the company finds: - Sample mean \( \bar{x} = 0.359 \) grams - Standard deviation \( \sigma = 0.05 \) grams The firm aims to determine if the pet food is healthy by checking if the zinc content is on target. A significance level of 95% is chosen for this test. **Steps for Analysis:** **a) State the Null and Alternative Hypotheses:** - **Null Hypothesis (\( H_0 \)):** The average zinc content is equal to the target value. \( \mu = 0.375 \) grams. - **Alternative Hypothesis (\( H_a \)):** The average zinc content is not equal to the target value. \( \mu \neq 0.375 \) grams. **b) Visual Representation:** Create a graph of a normal distribution curve to illustrate the standard zinc content with the sample mean highlighted. This will help visualize where the sample mean falls relative to the expected mean. **c) Find the Critical Value:** Using a Z-table, determine the critical Z-value corresponding to a 95% confidence level. For a two-tailed test, the critical Z-values are approximately \( \pm 1.96 \). **d) Compute the Test Statistic:** The Z-score is calculated using the formula: \[ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] Where: - \( \bar{x} = 0.359 \) grams - \( \mu = 0.375 \) grams - \( \sigma = 0.05 \) grams - \( n = 25 \) bags \[ Z = \frac{0.359 - 0.375}{\frac{0.05}{\sqrt{25}}} \] This calculation gives a Z-score which will be used to determine if the sample mean is significantly different from the hypothesized population mean. **e
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