Question

A producer manufactures batteries by using techniques I and II. The capacities (in ampere hours) of 12 randomly selected batteries manufactured by using technique I are: 140, 132, 136, 142, 138, 150, 154, 150, 152, 136, 144, and 142. Moreover, the capacities (in ampere hours) of 14 randomly selected batteries manufactured by using technique II are: 144, 134, 132, 130, 136, 146, 140, 128, 131, 128, 150, 137, 130, and 135. Suppose that battery capacities manufactured by using techniques I and II are normally distributed, with unknown means, \( \mu_I \) and \( \mu_{II} \) respectively, and an unknown common variance \( \sigma^2 \). Consider the null hypothesis \( H_0: (\mu_I - \mu_{II}) = \gamma \) ampere hours. For which of the following values of \( \gamma \) should the null hypothesis \( H_0 \) be accepted against the alternative hypothesis \( H_1: (\mu_I - \mu_{II}) \neq \gamma \) ampere hours, at 10% level of significance?

• A. \( \gamma = 4 \)
• B. \( \gamma = 7 \)
• C. \( \gamma = 10 \)
• D. \( \gamma = 13 \)

Hint:

In hypothesis testing, critical values for the t-distribution depend on the degrees of freedom and the significance level. Compare your test statistic to the critical value to determine whether to accept or reject the null hypothesis.

Answer 1

The Correct Option is A, B, C

Step 1: Hypothesis Testing and Confidence Interval
The problem is testing whether the difference in means \( (\mu_I - \mu_{II}) \) is equal to some value \( \gamma \). We are using a two-sample t-test. The null hypothesis is that the difference in means is \( \gamma \), and the alternative is that it is not. We are also provided with the following information: - 12 and 14 randomly selected samples from techniques I and II, respectively.
- The standard normal and t-distribution values for significance levels.
Step 2: Test Statistic Calculation
To perform this hypothesis test, we need the sample means and standard deviations for the two groups. The test statistic for a two-sample t-test is: \[ t = \frac{\bar{X_1} - \bar{X_2} - \gamma}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Where:
- \( \bar{X_1}, \bar{X_2} \) are the sample means,
- \( s_1, s_2 \) are the sample standard deviations,
- \( n_1, n_2 \) are the sample sizes.
The degree of freedom for the t-distribution will be calculated using: \[ df = \min(n_1 - 1, n_2 - 1) \] Step 3: Applying the Critical Values for \( t \)-Distribution
Given that the significance level is 10%, we can refer to the critical t-values for \( \alpha = 0.10 \) and the appropriate degrees of freedom \( df \). Based on the provided t-values, we compare the calculated test statistic with the critical values to determine whether to accept or reject the null hypothesis. Step 4: Conclusion
After calculating and comparing the test statistics, we find that the values of \( \gamma = 4 \), \( \gamma = 7 \), and \( \gamma = 10 \) provide acceptable conditions for not rejecting the null hypothesis at the 10% significance level. Therefore, the correct answers are \( \gamma = 4 \), \( \gamma = 7 \), and \( \gamma = 10 \).