Answer :
Sure! Let's test the difference between the performances of Part 1 and Part 2 of the quiz using a significance level of 0.01. Given that the population variances are unknown and not equal, we will use Welch's t-test for our analysis.
Here are the given values from the problem:
[tex]\[ \begin{array}{|l|c|c|} \hline & \textbf{Part 1} & \textbf{Part 2} \\ \hline \text{Sample size} & n_1 = 10 & n_2 = 10 \\ \hline \text{Sample mean} & \text{mean}_1 = 35 & \text{mean}_2 = 24 \\ \hline \text{Sample standard deviation} & \text{std}_1 = 2.5 & \text{std}_2 = 2.1 \\ \hline \end{array} \][/tex]
We will follow these steps:
1. State the null hypothesis (H0) and the alternative hypothesis (H1):
- H0: [tex]\(\mu_1 = \mu_2\)[/tex] (There is no difference in performance between Part 1 and Part 2)
- H1: [tex]\(\mu_1 \neq \mu_2\)[/tex] (There is a difference in performance between Part 1 and Part 2)
2. Calculate the pooled standard deviation for unequal variances:
[tex]\[ s_1^2 = (2.5)^2 = 6.25 \][/tex]
[tex]\[ s_2^2 = (2.1)^2 = 4.41 \][/tex]
3. Compute the test statistic:
[tex]\[ t_{\text{stat}} = \frac{\text{mean}_1 - \text{mean}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \][/tex]
[tex]\[ t_{\text{stat}} = \frac{35 - 24}{\sqrt{\frac{6.25}{10} + \frac{4.41}{10}}} \][/tex]
[tex]\[ t_{\text{stat}} = \frac{11}{\sqrt{0.625 + 0.441}} = \frac{11}{\sqrt{1.066}} \approx 10.65 \][/tex]
4. Determine the degrees of freedom for Welch's t-test:
[tex]\[ \text{df} = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \][/tex]
[tex]\[ \text{df} = \frac{(0.625 + 0.441)^2}{\frac{(0.625)^2}{9} + \frac{(0.441)^2}{9}} \approx 17.48 \][/tex]
5. Determine the critical value from the t-distribution for a two-tailed test at the 0.01 significance level:
[tex]\[ t_{\text{crit}} \approx 2.89 \][/tex]
6. Compute the p-value:
[tex]\[ \text{p-value} \approx 4.55 \times 10^{-9} \][/tex]
7. Compare the p-value to the significance level (α = 0.01) and make a decision:
[tex]\[ \text{p-value} < \alpha \][/tex]
Since the p-value is much smaller than the significance level of 0.01, we reject the null hypothesis.
### Conclusion:
There is significant evidence to conclude that there is a difference in performance between Part 1 and Part 2 of the quiz at the 0.01 significance level.
Here are the given values from the problem:
[tex]\[ \begin{array}{|l|c|c|} \hline & \textbf{Part 1} & \textbf{Part 2} \\ \hline \text{Sample size} & n_1 = 10 & n_2 = 10 \\ \hline \text{Sample mean} & \text{mean}_1 = 35 & \text{mean}_2 = 24 \\ \hline \text{Sample standard deviation} & \text{std}_1 = 2.5 & \text{std}_2 = 2.1 \\ \hline \end{array} \][/tex]
We will follow these steps:
1. State the null hypothesis (H0) and the alternative hypothesis (H1):
- H0: [tex]\(\mu_1 = \mu_2\)[/tex] (There is no difference in performance between Part 1 and Part 2)
- H1: [tex]\(\mu_1 \neq \mu_2\)[/tex] (There is a difference in performance between Part 1 and Part 2)
2. Calculate the pooled standard deviation for unequal variances:
[tex]\[ s_1^2 = (2.5)^2 = 6.25 \][/tex]
[tex]\[ s_2^2 = (2.1)^2 = 4.41 \][/tex]
3. Compute the test statistic:
[tex]\[ t_{\text{stat}} = \frac{\text{mean}_1 - \text{mean}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \][/tex]
[tex]\[ t_{\text{stat}} = \frac{35 - 24}{\sqrt{\frac{6.25}{10} + \frac{4.41}{10}}} \][/tex]
[tex]\[ t_{\text{stat}} = \frac{11}{\sqrt{0.625 + 0.441}} = \frac{11}{\sqrt{1.066}} \approx 10.65 \][/tex]
4. Determine the degrees of freedom for Welch's t-test:
[tex]\[ \text{df} = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \][/tex]
[tex]\[ \text{df} = \frac{(0.625 + 0.441)^2}{\frac{(0.625)^2}{9} + \frac{(0.441)^2}{9}} \approx 17.48 \][/tex]
5. Determine the critical value from the t-distribution for a two-tailed test at the 0.01 significance level:
[tex]\[ t_{\text{crit}} \approx 2.89 \][/tex]
6. Compute the p-value:
[tex]\[ \text{p-value} \approx 4.55 \times 10^{-9} \][/tex]
7. Compare the p-value to the significance level (α = 0.01) and make a decision:
[tex]\[ \text{p-value} < \alpha \][/tex]
Since the p-value is much smaller than the significance level of 0.01, we reject the null hypothesis.
### Conclusion:
There is significant evidence to conclude that there is a difference in performance between Part 1 and Part 2 of the quiz at the 0.01 significance level.
