Let's solve this problem step by step.
### Step 1: Calculate the 95% Confidence Interval for the Mean GPA
Given Data:
- Sample size ([tex]\( n \)[/tex]) = 30
- Sample mean ([tex]\( \bar{x} \)[/tex]) = 3.01
- Sample standard deviation ([tex]\( s \)[/tex]) = 0.24
- Critical value for a 95% confidence interval with 29 degrees of freedom (df = [tex]\( n - 1 = 29 \)[/tex]) = 2.045
Formula for Confidence Interval:
[tex]\[ \text{Confidence Interval} = \left( \bar{x} - t_\alpha \frac{s}{\sqrt{n}}, \bar{x} + t_\alpha \frac{s}{\sqrt{n}} \right) \][/tex]
Where:
- [tex]\( \bar{x} \)[/tex] is the sample mean
- [tex]\( t_\alpha \)[/tex] is the critical value (t-score)
- [tex]\( s \)[/tex] is the sample standard deviation
- [tex]\( n \)[/tex] is the sample size
Calculation of the Margin of Error:
[tex]\[ \text{Margin of Error} = t_\alpha \times \frac{s}{\sqrt{n}} \][/tex]
[tex]\[ \text{Margin of Error} = 2.045 \times \frac{0.24}{\sqrt{30}} = 0.09 \][/tex]
Confidence Interval:
[tex]\[ \bar{x} - \text{Margin of Error} = 3.01 - 0.09 = 2.92 \][/tex]
[tex]\[ \bar{x} + \text{Margin of Error} = 3.01 + 0.09 = 3.10 \][/tex]
So, the 95% confidence interval for the mean GPA is:
[tex]\[ (2.92, 3.10) \][/tex]
### Step 2: Determine the Null and Alternative Hypotheses for a Two-Sided Test
The null and alternative hypotheses for a two-sided test at a significance level of 0.05 are:
[tex]\( H_0: \mu = 2.87 \)[/tex]
[tex]\( H_a: \mu \neq 2.87 \)[/tex]
Therefore, the correct choice is:
A. [tex]\( H_0: \mu = 2.87 \)[/tex]
[tex]\( H_a: \mu \neq 2.87 \)[/tex]
### Step 3: Calculate the Test Statistic
Formula for Test Statistic:
[tex]\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]
Where:
- [tex]\( \bar{x} \)[/tex] is the sample mean
- [tex]\( \mu \)[/tex] is the hypothesized population mean
- [tex]\( s \)[/tex] is the sample standard deviation
- [tex]\( n \)[/tex] is the sample size
Calculation:
[tex]\[ t = \frac{3.01 - 2.87}{\frac{0.24}{\sqrt{30}}} = 3.18 \][/tex]
### Step 4: Calculate the p-value
The p-value for a two-sided test is calculated as:
[tex]\[ p = 2 \times (1 - \text{CDF}(|t|)) \][/tex]
Given our test statistic [tex]\( t = 3.18 \)[/tex] and degrees of freedom [tex]\( df = 29 \)[/tex]:
[tex]\[ p = 0.0035 \][/tex]
### Conclusion
- 95% Confidence Interval: [tex]\( (2.92, 3.10) \)[/tex]
- Test Statistic: [tex]\( 3.18 \)[/tex]
- p-value: [tex]\( 0.0035 \)[/tex]
Because the p-value (0.0035) is less than the significance level (0.05), we reject the null hypothesis [tex]\( H_0: \mu = 2.87 \)[/tex].
So, if a two-sided alternative had been used with a significance level of 0.05, the hypothesized mean of 2.87 would have been rejected.
