Answer :
To solve this problem, we want to test the hypothesis that the mean from Sample #1 ([tex]\(\mu_1\)[/tex]) is greater than the mean from Sample #2 ([tex]\(\mu_2\)[/tex]). We're working with a significance level of [tex]\(\alpha = 0.10\)[/tex].
Step-by-Step Solution:
1. State the Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu_1 = \mu_2\)[/tex]
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu_1 > \mu_2\)[/tex]
2. Set up the Data:
- Sample #1: Gather the data points provided for Sample #1.
- Sample #2: Gather the data points provided for Sample #2.
3. Calculate the Sample Statistics:
- Calculate the mean ([tex]\(\bar{x}_1\)[/tex]) and standard deviation ([tex]\(s_1\)[/tex]) for Sample #1.
- Calculate the mean ([tex]\(\bar{x}_2\)[/tex]) and standard deviation ([tex]\(s_2\)[/tex]) for Sample #2.
- Determine the sample sizes: [tex]\(n_1\)[/tex] for Sample #1 and [tex]\(n_2\)[/tex] for Sample #2.
4. Calculate the Pooled Standard Deviation:
- Use the formula for pooled standard deviation:
[tex]\[
s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}
\][/tex]
5. Compute the Test Statistic:
- Use the formula for the two-sample t-test statistic:
[tex]\[
t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
\][/tex]
6. Determine Degrees of Freedom:
- The degrees of freedom for the t-distribution in this test is [tex]\(n_1 + n_2 - 2\)[/tex].
7. Find the p-value:
- Use the cumulative distribution function (CDF) for the t-distribution to find the p-value associated with the calculated test statistic. This p-value reflects the probability of observing a test statistic as extreme as, or more extreme than, the one calculated, assuming the null hypothesis is true.
8. Decision:
- Compare the p-value to the significance level [tex]\(\alpha = 0.10\)[/tex]. If the p-value is less than [tex]\(\alpha\)[/tex], reject the null hypothesis in favor of the alternative hypothesis. Otherwise, do not reject the null hypothesis.
Results:
- The test statistic calculated is approximately [tex]\(1.002\)[/tex].
- The p-value calculated is approximately [tex]\(0.1592\)[/tex].
Since the p-value ([tex]\(0.1592\)[/tex]) is greater than the significance level of [tex]\(0.10\)[/tex], we do not have enough evidence to reject the null hypothesis. Therefore, we conclude that there is not sufficient evidence to support the claim that [tex]\(\mu_1 > \mu_2\)[/tex] at the 0.10 significance level.
Step-by-Step Solution:
1. State the Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu_1 = \mu_2\)[/tex]
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu_1 > \mu_2\)[/tex]
2. Set up the Data:
- Sample #1: Gather the data points provided for Sample #1.
- Sample #2: Gather the data points provided for Sample #2.
3. Calculate the Sample Statistics:
- Calculate the mean ([tex]\(\bar{x}_1\)[/tex]) and standard deviation ([tex]\(s_1\)[/tex]) for Sample #1.
- Calculate the mean ([tex]\(\bar{x}_2\)[/tex]) and standard deviation ([tex]\(s_2\)[/tex]) for Sample #2.
- Determine the sample sizes: [tex]\(n_1\)[/tex] for Sample #1 and [tex]\(n_2\)[/tex] for Sample #2.
4. Calculate the Pooled Standard Deviation:
- Use the formula for pooled standard deviation:
[tex]\[
s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}
\][/tex]
5. Compute the Test Statistic:
- Use the formula for the two-sample t-test statistic:
[tex]\[
t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
\][/tex]
6. Determine Degrees of Freedom:
- The degrees of freedom for the t-distribution in this test is [tex]\(n_1 + n_2 - 2\)[/tex].
7. Find the p-value:
- Use the cumulative distribution function (CDF) for the t-distribution to find the p-value associated with the calculated test statistic. This p-value reflects the probability of observing a test statistic as extreme as, or more extreme than, the one calculated, assuming the null hypothesis is true.
8. Decision:
- Compare the p-value to the significance level [tex]\(\alpha = 0.10\)[/tex]. If the p-value is less than [tex]\(\alpha\)[/tex], reject the null hypothesis in favor of the alternative hypothesis. Otherwise, do not reject the null hypothesis.
Results:
- The test statistic calculated is approximately [tex]\(1.002\)[/tex].
- The p-value calculated is approximately [tex]\(0.1592\)[/tex].
Since the p-value ([tex]\(0.1592\)[/tex]) is greater than the significance level of [tex]\(0.10\)[/tex], we do not have enough evidence to reject the null hypothesis. Therefore, we conclude that there is not sufficient evidence to support the claim that [tex]\(\mu_1 > \mu_2\)[/tex] at the 0.10 significance level.
