Munchie's Potato Chip Company claims that the weight of the contents of a 10-ounce bag of chips is normally distributed, with mean 10 ounces. A consumer group, Snack Munchers for Truth (SMFT), says that the average weight is less than this: SMFT weighs the contents of 7 randomly selected bags of potato chips and gets the weights shown below. Test SMFT's claim.
10.1, 10, 9.7, 9.9, 9.7, 9.8, 9.7

Fill in the blank with the test value.

H_0:\mu=

What sign should appear in the alternative hypothesis?
 A. <
 B. >
 C. \neq

The test statistic for this test is 
.

The p-value for this test is 
.

Select the appropriate conclusion for this test at a significance level of \alpha=0.05.
 A. Reject H_0. We have significant evidence that the average weight is less than 10.
 B. Fail to reject H_0. We do not have significant evidence that the average weight is less than 10

To test SMFT's claim about the average weight of the chips, we can conduct a one-sample t-test. Here’s how we proceed:

1. **Formulating the Hypotheses:**
   - Null Hypothesis (\(H_0\)): The true mean weight of the chips is 10 ounces.  
     \[
     H_0: \mu = 10
     \]
   - Alternative Hypothesis (\(H_a\)): The true mean weight of the chips is less than 10 ounces.
     - Answer to alternative hypothesis sign: \( A. < \)

2. **Calculating the Sample Mean (\(\bar{x}\)) and Standard Deviation (s):**

   Given weights: \(10.1, 10, 9.7, 9.9, 9.7, 9.8, 9.7\)

   The sample mean is:
   \[
   \bar{x} = \frac{10.1 + 10 + 9.7 + 9.9 + 9.7 + 9.8 + 9.7}{7} \approx 9.8414
   \]

   The sample standard deviation (\(s\)) can be calculated as follows:
   \[
   s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
   \]

3. **Test Statistic Calculation:**
   
   Using the formula for the t-test statistic:
   \[
   t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
   \]
   where \(\mu_0 = 10\) (the hypothesized mean), \(n = 7\) (sample size), and \(\bar{x}\) and \(s\) are as calculated above.

4. **P-Value Calculation:**

   With \( t \)-value and degrees of freedom (\(df = n - 1 = 6\)), we find the p-value for a one-tailed test.

5. **Conclusion (Significance Level \(\alpha = 0.05\)):**

   If the p-value is less than 0.05, we reject \(H_0\); otherwise, we fail to reject \(H_0\).

Let's compute the exact values for the test statistic and p-value.The results are as follows:

- **Sample Mean (\(\bar{x}\))**: 9.8429
- **Sample Standard Deviation (s)**: 0.1618
- **Test Statistic (t)**: -2.5690
- **P-Value**: 0.0212

### Conclusion at \(\alpha = 0.05\):
Since the p-value (0.0212) is less than the significance level of 0.05, we **reject the null hypothesis** \( H_0 \).

### Final Answer:
- **Alternative Hypothesis Sign**: A. <
- **Conclusion**: A. Reject \( H_0 \). We have significant evidence that the average weight is less than 10 ounces.

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Would you like more details or have any questions?

### Related Questions:
1. How would the test change if the significance level was set to 0.01?
2. What if we had a larger sample size? How would it affect the test results?
3. What are the assumptions of a one-sample t-test?
4. How can we interpret the p-value in practical terms?
5. What if we wanted to test if the average weight differs from 10 (two-tailed)?

### Tip:
Remember, a smaller p-value indicates stronger evidence against the null hypothesis in hypothesis testing.

Munchie's Potato Chip Company claims that the weight of the contents of a 10-ounce bag of chips is normally distributed, with mean 10 ounces. A consumer group, Snack Munchers for Truth (SMFT), says that the average weight is less than this: SMFT weighs the contents of 7 randomly selected bags of potato chips and gets the weights shown below. Test SMFT's claim. 10.1, 10, 9.7, 9.9, 9.7, 9.8, 9.7

Explore how to perform a one-sample t-test to evaluate the claim made by Snack Munchers for Truth (SMFT) about the mean weight of Munchie's potato chips. This problem demonstrates key statistical concepts like hypothesis testing, the t-distribution, and p-value interpretation. Suitable for college-level statistics students.

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