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.