1. A local sandwich shop claims to put 6 ounces of meat on each sandwich the You take a random sample of 19 sandwiches, and find that the average amount of meat is 5.6 ounces, with a sample standard deviation of 1.3 ounces. Can we say that the amount of meat on the sandwiches is significantly lower than the shop's claim? (Use a = .10) %3D Type of Test: Null Hypothesis: Alternate Hypothesis: SE: Test Statistic: P-value: Decision:

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### Statistical Hypothesis Testing Example

#### Problem Statement
A local sandwich shop claims to put 6 ounces of meat on each sandwich they make. You take a random sample of 19 sandwiches and find that the average amount of meat is 5.6 ounces, with a sample standard deviation of 1.3 ounces. Can we say that the amount of meat on the sandwiches is significantly lower than the shop's claim? (Use α = .10)

#### Required Information:
- **Type of Test:**
  
- **Null Hypothesis (H₀):**

- **Alternate Hypothesis (H₁):**

- **Standard Error (SE):**

- **Test Statistic:**

- **P-value:**

- **Decision:**

- **Conclusion Sentence:**

### Explanation
To determine whether the sandwich shop's claim is valid or the amount of meat is significantly lower, we will conduct a hypothesis test. Here is the step-by-step process:

1. **Type of Test:** 
   We will use a one-sample t-test because the sample size is small (n < 30), and we do not know the population standard deviation.

2. **Null Hypothesis (H₀):**
   The null hypothesis states that the true mean amount of meat is 6 ounces.
   \[ H₀: \mu = 6 \]

3. **Alternate Hypothesis (H₁):**
   The alternate hypothesis states that the true mean amount of meat is less than 6 ounces.
   \[ H₁: \mu < 6 \]

4. **Standard Error (SE):**
   The standard error of the mean can be calculated using the formula:
   \[ SE = \frac{s}{\sqrt{n}} \]
   where \( s \) is the sample standard deviation and \( n \) is the sample size.
   \[ SE = \frac{1.3}{\sqrt{19}} \]

5. **Test Statistic:**
   The test statistic for a one-sample t-test is calculated as:
   \[ t = \frac{\bar{x} - \mu}{SE} \]
   where \( \bar{x} \) is the sample mean.
   \[ t = \frac{5.6 - 6}{SE} \]

6. **P-value:**
   Using the t-distribution table (or a statistical software), we compare the calculated t
Transcribed Image Text:### Statistical Hypothesis Testing Example #### Problem Statement A local sandwich shop claims to put 6 ounces of meat on each sandwich they make. You take a random sample of 19 sandwiches and find that the average amount of meat is 5.6 ounces, with a sample standard deviation of 1.3 ounces. Can we say that the amount of meat on the sandwiches is significantly lower than the shop's claim? (Use α = .10) #### Required Information: - **Type of Test:** - **Null Hypothesis (H₀):** - **Alternate Hypothesis (H₁):** - **Standard Error (SE):** - **Test Statistic:** - **P-value:** - **Decision:** - **Conclusion Sentence:** ### Explanation To determine whether the sandwich shop's claim is valid or the amount of meat is significantly lower, we will conduct a hypothesis test. Here is the step-by-step process: 1. **Type of Test:** We will use a one-sample t-test because the sample size is small (n < 30), and we do not know the population standard deviation. 2. **Null Hypothesis (H₀):** The null hypothesis states that the true mean amount of meat is 6 ounces. \[ H₀: \mu = 6 \] 3. **Alternate Hypothesis (H₁):** The alternate hypothesis states that the true mean amount of meat is less than 6 ounces. \[ H₁: \mu < 6 \] 4. **Standard Error (SE):** The standard error of the mean can be calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} \] where \( s \) is the sample standard deviation and \( n \) is the sample size. \[ SE = \frac{1.3}{\sqrt{19}} \] 5. **Test Statistic:** The test statistic for a one-sample t-test is calculated as: \[ t = \frac{\bar{x} - \mu}{SE} \] where \( \bar{x} \) is the sample mean. \[ t = \frac{5.6 - 6}{SE} \] 6. **P-value:** Using the t-distribution table (or a statistical software), we compare the calculated t
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