To conduct a hypothesis test for the two independent samples provided, we need to follow these steps:

1. State the Hypotheses:

   • Null hypothesis $H_0: \mu_1 = \mu_2$ (the means of the two samples are equal).
   • Alternative hypothesis $H_a: \mu_1 \neq \mu_2$ (the means of the two samples are not equal, indicating a two-tailed test).

2. Identify the Test Type:

   • Since the alternative hypothesis states that the means are not equal, this is a two-tailed test.

3. Calculate the Sample Means and Sample Standard Deviations:

   • Sample 1: $907, 349, 154$
   • Sample 2: $754, 274, 175$

   Calculating the means and standard deviations:

   • Mean of Sample 1: $\bar{x}_1 = \frac{907 + 349 + 154}{3} = \frac{1410}{3} = 470$

   • Mean of Sample 2: $\bar{x}_2 = \frac{754 + 274 + 175}{3} = \frac{1203}{3} = 401$

   • Sample standard deviation of Sample 1: $s_1 = \sqrt{\frac{(907-470)^2 + (349-470)^2 + (154-470)^2}{n_1 - 1}} = \sqrt{\frac{(437)^2 + (-121)^2 + (-316)^2}{2}}$ $= \sqrt{\frac{190969 + 14641 + 99856}{2}} = \sqrt{\frac{299466}{2}} = \sqrt{149733} \approx 386.5$

   • Sample standard deviation of Sample 2: $s_2 = \sqrt{\frac{(754-401)^2 + (274-401)^2 + (175-401)^2}{n_2 - 1}} = \sqrt{\frac{(353)^2 + (-127)^2 + (-226)^2}{2}}$ $= \sqrt{\frac{124609 + 16129 + 51076}{2}} = \sqrt{\frac{191814}{2}} = \sqrt{95907} \approx 309.7$

4. Calculate the Test Statistic: The test statistic for comparing two means is given by: $t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$ Here, $n_1 = n_2 = 3$.

   Plugging in the values: $t = \frac{470 - 401}{\sqrt{\frac{(386.5)^2}{3} + \frac{(309.7)^2}{3}}}$ $= \frac{69}{\sqrt{\frac{149733}{3} + \frac{95907}{3}}} = \frac{69}{\sqrt{49911 + 31969}} = \frac{69}{\sqrt{81980}} \approx \frac{69}{286.8} \approx 0.240$

5. Calculate the Degrees of Freedom: Using the formula for degrees of freedom for two independent samples: $df = n_1 + n_2 - 2 = 3 + 3 - 2 = 4$

6. Find the p-value: For a two-tailed test, we look up the critical value in the t-distribution table. The test statistic $t \approx 0.240$ and degrees of freedom $df = 4$.

   Using a t-table or calculator, find the p-value for $t = 0.240$ with $df = 4$. This yields a two-tailed p-value of approximately $0.8108$.

Summary of Results

• Test Statistic: $t \approx 0.2400$
• p-value: $\approx 0.8108$

Conclusion

Since the p-value is significantly greater than any common alpha level (e.g., 0.05), we fail to reject the null hypothesis.

Do you want any details or have any questions? Here are some related questions you might consider:

1. How do you interpret the results of this hypothesis test?
2. What assumptions are necessary for this test?
3. How would the results change if we had larger sample sizes?
4. What are the implications of failing to reject the null hypothesis?
5. How would you conduct this test using software like R or Python?

Tip: Always check the assumptions of normality and equal variances before performing a t-test.