Question 1167904
This problem requires a hypothesis test to compare the means of two independent samples. Since the population standard deviations are unknown and the sample sizes are small (n < 30), a **two-sample t-test** is appropriate. Specifically, since we are testing if the mean of Orno is *greater than* the mean of Edne, it will be a **one-tailed test**.

Given the small sample sizes, we should ideally check for approximate normality of the data or assume that the underlying populations are normally distributed. We also need to decide whether to assume equal variances or unequal variances between the two populations. A common approach is to perform an F-test for equality of variances first, or simply assume unequal variances (Welch's t-test) which is more robust when the assumption of equal variances might be violated. For simplicity, and typical textbook problems of this nature, if variances are not drastically different, one might proceed with the pooled-variance t-test. However, if no specific instruction, assuming unequal variances is safer. Let's calculate the standard deviations for each group.

**Data:**

* **Edne (Sample 1):** $n_1 = 6$
    Scores: $X_1 = \{5, 8, 7, 6, 9, 7\}$
* **Orno (Sample 2):** $n_2 = 8$
    Scores: $X_2 = \{8, 10, 7, 11, 9, 12, 14, 9\}$

**1. Calculate Sample Statistics:**

* **Edne (Sample 1):**
    * Sum: $\sum X_1 = 5+8+7+6+9+7 = 42$
    * Mean: $\bar{X}_1 = \frac{42}{6} = 7$
    * Sum of Squares: $\sum X_1^2 = 5^2+8^2+7^2+6^2+9^2+7^2 = 25+64+49+36+81+49 = 304$
    * Sample Variance: $s_1^2 = \frac{\sum X_1^2 - n_1(\bar{X}_1)^2}{n_1-1} = \frac{304 - 6(7)^2}{6-1} = \frac{304 - 6(49)}{5} = \frac{304 - 294}{5} = \frac{10}{5} = 2$
    * Sample Standard Deviation: $s_1 = \sqrt{2} \approx 1.414$

* **Orno (Sample 2):**
    * Sum: $\sum X_2 = 8+10+7+11+9+12+14+9 = 80$
    * Mean: $\bar{X}_2 = \frac{80}{8} = 10$
    * Sum of Squares: $\sum X_2^2 = 8^2+10^2+7^2+11^2+9^2+12^2+14^2+9^2 = 64+100+49+121+81+144+196+81 = 836$
    * Sample Variance: $s_2^2 = \frac{\sum X_2^2 - n_2(\bar{X}_2)^2}{n_2-1} = \frac{836 - 8(10)^2}{8-1} = \frac{836 - 8(100)}{7} = \frac{836 - 800}{7} = \frac{36}{7} \approx 5.143$
    * Sample Standard Deviation: $s_2 = \sqrt{\frac{36}{7}} = \frac{6}{\sqrt{7}} \approx 2.268$

**2. Formulate Hypotheses:**

* **Null Hypothesis ($H_0$):** $\mu_{Orno} \le \mu_{Edne}$ (The mean weight of bottles from Orno is not greater than Edne)
* **Alternative Hypothesis ($H_1$):** $\mu_{Orno} > \mu_{Edne}$ (The mean weight of bottles from Orno is greater than Edne)

This is a **one-tailed (right-tailed) t-test**.

**3. Choose Significance Level:**

* $\alpha = 0.05$

**4. Choose Test Statistic:**

Given that sample sizes are small and population standard deviations are unknown, a t-test is used. Since the sample variances ($s_1^2 = 2$ and $s_2^2 = 36/7 \approx 5.143$) are not drastically different, we can proceed with a pooled-variance t-test. (Alternatively, a Welch's t-test for unequal variances could be used, leading to a slightly different calculation for degrees of freedom.)

**Pooled Variance ($s_p^2$):**
$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}$
$s_p^2 = \frac{(6-1)(2) + (8-1)(\frac{36}{7})}{6+8-2}$
$s_p^2 = \frac{5(2) + 7(\frac{36}{7})}{12}$
$s_p^2 = \frac{10 + 36}{12} = \frac{46}{12} = \frac{23}{6} \approx 3.833$

**t-statistic for pooled variance:**
$t = \frac{(\bar{X}_2 - \bar{X}_1) - (\mu_2 - \mu_1)}{\sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$
Under $H_0$, $\mu_2 - \mu_1 = 0$.
$t = \frac{(10 - 7) - 0}{\sqrt{\frac{23}{6} \left(\frac{1}{6} + \frac{1}{8}\right)}}$
$t = \frac{3}{\sqrt{\frac{23}{6} \left(\frac{4+3}{24}\right)}}$
$t = \frac{3}{\sqrt{\frac{23}{6} \left(\frac{7}{24}\right)}}$
$t = \frac{3}{\sqrt{\frac{161}{144}}}$
$t = \frac{3}{\frac{\sqrt{161}}{12}} = \frac{3 \times 12}{\sqrt{161}} = \frac{36}{\sqrt{161}}$
$t \approx \frac{36}{12.688} \approx 2.837$

**5. Determine Degrees of Freedom and Critical Value:**

* Degrees of Freedom ($df$) = $n_1 + n_2 - 2 = 6 + 8 - 2 = 12$
* For a one-tailed (right-tailed) t-test with $\alpha = 0.05$ and $df = 12$, the critical t-value ($t_{crit}$) from a t-distribution table is approximately $1.782$.

**6. Make a Decision:**

* Compare the calculated t-statistic to the critical t-value.
* Calculated $t = 2.837$
* Critical $t_{crit} = 1.782$

Since $2.837 > 1.782$, the calculated t-statistic falls into the rejection region.

**7. Conclusion:**

At the 0.05 significance level, we **reject the null hypothesis ($H_0$)**.

There is sufficient statistical evidence to conclude that the mean weight of the bottles filled by the Orno Machine is significantly greater than the mean weight of the bottles filled by the Edne Machine.