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### Step 1: Find the 95% confidence interval for the mean GPA.
Given data from the sample:
- Mean GPA ([tex]\(\bar{x}\)[/tex]): We calculated this to be 2.93 (from previously summarized data).
- Standard Deviation (s): We calculated this to be 0.18 (sample standard deviation).
- Sample Size (n): 30 (as stated).
The formula for the 95% confidence interval for the mean, when the standard deviation of the population is unknown, uses the t-distribution and is given by:
[tex]\[ \bar{x} \pm t^{*} \frac{s}{\sqrt{n}} \][/tex]
Where:
- [tex]\(t^{}\)[/tex] is the critical value from the t-distribution for 95% confidence and [tex]\(n-1\)[/tex] degrees of freedom ([tex]\(df = 29\)[/tex]).
Using the data and calculations provided:
- Critical value [tex]\(t^{}\)[/tex]: 2.045 (typical critical value for 95% confidence with 29 df)
- Margin of error [tex]\(ME\)[/tex]: [tex]\(t^{*} \frac{s}{\sqrt{n}} = 2.045 \times \frac{0.18}{\sqrt{30}} \approx 0.07\)[/tex]
Thus, the confidence interval is:
[tex]\[ (2.93 - 0.07,\ 2.93 + 0.07) = (2.84,\ 3.01) \][/tex]
So, the 95% confidence interval is: [tex]\( \boxed{(2.84, 3.01)} \)[/tex]
### Step 2: Test the two-sided alternative with a significance level of 0.05.
Determine the null and alternative hypotheses. The correct formulation for a two-tailed test is:
- [tex]\( H_0: \mu = 2.86 \)[/tex]
- [tex]\( H_a: \mu \neq 2.86 \)[/tex]
Calculate the test statistic:
[tex]\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{2.93 - 2.86}{\frac{0.18}{\sqrt{30}}} = 1.58 \][/tex]
The test statistic value is [tex]\( \boxed{1.58} \)[/tex].
### Step 3: Determine the p-value.
For a two-tailed test, the p-value is calculated as:
- p-value = [tex]\( 2 \times P(T > |t|) \)[/tex] where T follows a t-distribution with [tex]\( df = 29 \)[/tex].
From the summarized results:
- p-value = 0.124
Thus, the p-value is [tex]\( \boxed{0.124} \)[/tex].
### Step 4: Interpret the results of the test.
Since the p-value is [tex]\(0.124\)[/tex], which is greater than the significance level ([tex]\( \alpha = 0.05 \)[/tex]), we do not reject the null hypothesis.
[tex]\( \boxed{\text{Interpretation: Since the p-value is greater than the significance level of 0.05, we do not reject } H_0. \text{ There is insufficient evidence to conclude that the mean GPA for statistics students is different from 2.86.}} \)[/tex]
### Step 1: Find the 95% confidence interval for the mean GPA.
Given data from the sample:
- Mean GPA ([tex]\(\bar{x}\)[/tex]): We calculated this to be 2.93 (from previously summarized data).
- Standard Deviation (s): We calculated this to be 0.18 (sample standard deviation).
- Sample Size (n): 30 (as stated).
The formula for the 95% confidence interval for the mean, when the standard deviation of the population is unknown, uses the t-distribution and is given by:
[tex]\[ \bar{x} \pm t^{*} \frac{s}{\sqrt{n}} \][/tex]
Where:
- [tex]\(t^{}\)[/tex] is the critical value from the t-distribution for 95% confidence and [tex]\(n-1\)[/tex] degrees of freedom ([tex]\(df = 29\)[/tex]).
Using the data and calculations provided:
- Critical value [tex]\(t^{}\)[/tex]: 2.045 (typical critical value for 95% confidence with 29 df)
- Margin of error [tex]\(ME\)[/tex]: [tex]\(t^{*} \frac{s}{\sqrt{n}} = 2.045 \times \frac{0.18}{\sqrt{30}} \approx 0.07\)[/tex]
Thus, the confidence interval is:
[tex]\[ (2.93 - 0.07,\ 2.93 + 0.07) = (2.84,\ 3.01) \][/tex]
So, the 95% confidence interval is: [tex]\( \boxed{(2.84, 3.01)} \)[/tex]
### Step 2: Test the two-sided alternative with a significance level of 0.05.
Determine the null and alternative hypotheses. The correct formulation for a two-tailed test is:
- [tex]\( H_0: \mu = 2.86 \)[/tex]
- [tex]\( H_a: \mu \neq 2.86 \)[/tex]
Calculate the test statistic:
[tex]\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{2.93 - 2.86}{\frac{0.18}{\sqrt{30}}} = 1.58 \][/tex]
The test statistic value is [tex]\( \boxed{1.58} \)[/tex].
### Step 3: Determine the p-value.
For a two-tailed test, the p-value is calculated as:
- p-value = [tex]\( 2 \times P(T > |t|) \)[/tex] where T follows a t-distribution with [tex]\( df = 29 \)[/tex].
From the summarized results:
- p-value = 0.124
Thus, the p-value is [tex]\( \boxed{0.124} \)[/tex].
### Step 4: Interpret the results of the test.
Since the p-value is [tex]\(0.124\)[/tex], which is greater than the significance level ([tex]\( \alpha = 0.05 \)[/tex]), we do not reject the null hypothesis.
[tex]\( \boxed{\text{Interpretation: Since the p-value is greater than the significance level of 0.05, we do not reject } H_0. \text{ There is insufficient evidence to conclude that the mean GPA for statistics students is different from 2.86.}} \)[/tex]
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