Answer :
Certainly! Let's go through the steps to solve this problem.
a. Finding the Test Statistic
First, we will determine the test statistic for the given samples. The question involves testing if the means of two populations are different. This type of test is known as a two-sample t-test for independent samples.
1. Collect Sample Data:
- Sample #1: 90.3, 54.4, 57.3, 79.1, 64.6, 47.7, 68.3, 49.8, 66.4, 68.3
- Sample #2: 69.3, 82.3, 44.7, 65.2, 62.9, 67.7, 62.6, 41.6, 66.6, 80.7, 51.7
2. Calculate the Means for Each Sample:
- Mean of Sample #1: 64.62
- Mean of Sample #2: 63.36
3. Calculate the Variances for Each Sample:
- Variance of Sample #1: [tex]\(s_1^2 = 198.84\)[/tex]
- Variance of Sample #2: [tex]\(s_2^2 = 150.11\)[/tex]
4. Calculate Sample Sizes:
- [tex]\(n_1 = 10\)[/tex] (Sample #1)
- [tex]\(n_2 = 11\)[/tex] (Sample #2)
5. Calculate the Pooled Standard Deviation (pooled_std):
[tex]\[
\text{pooled\_std} = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}}
\][/tex]
6. Calculate the Test Statistic:
[tex]\[
\text{test statistic} = \frac{\text{mean1} - \text{mean2}}{\text{pooled\_std} \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
\][/tex]
The calculated test statistic is approximately 0.2469.
b. Finding the p-value
To find the p-value for the test statistic:
1. Determine the Degrees of Freedom:
[tex]\[
\text{df} = n_1 + n_2 - 2 = 19
\][/tex]
2. Calculate the p-value:
Since our test is two-tailed (based on the alternative hypothesis [tex]\(\mu_1 \neq \mu_2\)[/tex]), the p-value is calculated as:
[tex]\[
\text{p-value} = 2 \times (1 - \text{CDF of } t \text{ at } | \text{test statistic} | \text{ with } \text{df} = 19)
\][/tex]
After performing these calculations, the p-value is approximately 0.8076.
Conclusion:
- The test statistic is 0.2469.
- The p-value is approximately 0.8076.
Given a significance level of [tex]\(\alpha = 0.002\)[/tex], since the p-value is much larger than [tex]\(\alpha\)[/tex], we do not reject the null hypothesis. Thus, we do not have sufficient evidence to support the claim that the first population mean is different from the second population mean.
a. Finding the Test Statistic
First, we will determine the test statistic for the given samples. The question involves testing if the means of two populations are different. This type of test is known as a two-sample t-test for independent samples.
1. Collect Sample Data:
- Sample #1: 90.3, 54.4, 57.3, 79.1, 64.6, 47.7, 68.3, 49.8, 66.4, 68.3
- Sample #2: 69.3, 82.3, 44.7, 65.2, 62.9, 67.7, 62.6, 41.6, 66.6, 80.7, 51.7
2. Calculate the Means for Each Sample:
- Mean of Sample #1: 64.62
- Mean of Sample #2: 63.36
3. Calculate the Variances for Each Sample:
- Variance of Sample #1: [tex]\(s_1^2 = 198.84\)[/tex]
- Variance of Sample #2: [tex]\(s_2^2 = 150.11\)[/tex]
4. Calculate Sample Sizes:
- [tex]\(n_1 = 10\)[/tex] (Sample #1)
- [tex]\(n_2 = 11\)[/tex] (Sample #2)
5. Calculate the Pooled Standard Deviation (pooled_std):
[tex]\[
\text{pooled\_std} = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}}
\][/tex]
6. Calculate the Test Statistic:
[tex]\[
\text{test statistic} = \frac{\text{mean1} - \text{mean2}}{\text{pooled\_std} \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
\][/tex]
The calculated test statistic is approximately 0.2469.
b. Finding the p-value
To find the p-value for the test statistic:
1. Determine the Degrees of Freedom:
[tex]\[
\text{df} = n_1 + n_2 - 2 = 19
\][/tex]
2. Calculate the p-value:
Since our test is two-tailed (based on the alternative hypothesis [tex]\(\mu_1 \neq \mu_2\)[/tex]), the p-value is calculated as:
[tex]\[
\text{p-value} = 2 \times (1 - \text{CDF of } t \text{ at } | \text{test statistic} | \text{ with } \text{df} = 19)
\][/tex]
After performing these calculations, the p-value is approximately 0.8076.
Conclusion:
- The test statistic is 0.2469.
- The p-value is approximately 0.8076.
Given a significance level of [tex]\(\alpha = 0.002\)[/tex], since the p-value is much larger than [tex]\(\alpha\)[/tex], we do not reject the null hypothesis. Thus, we do not have sufficient evidence to support the claim that the first population mean is different from the second population mean.
