Answer :
Certainly! Let's tackle each of these parts step-by-step.
### Part (a)
We need to construct a 90% confidence interval for the mean test score in the population.
#### Step 1: Identify the given data
- Sample mean, [tex]\( \bar{Y} = 712.1 \)[/tex]
- Sample standard deviation, [tex]\( s_Y = 23.2 \)[/tex]
- Sample size, [tex]\( n = 400 \)[/tex]
- Confidence level, [tex]\( 90\% \)[/tex]
#### Step 2: Determine the critical value
For a 90% confidence interval, we use the z-distribution since the sample size is large ([tex]\( n > 30 \)[/tex]). The critical z-value [tex]\( z_{\alpha/2} \)[/tex] for a 90% confidence level is approximately 1.645.
#### Step 3: Calculate the standard error of the mean
The standard error of the mean (SEM) is given by:
[tex]\[ \text{SEM} = \frac{s_Y}{\sqrt{n}} = \frac{23.2}{\sqrt{400}} = 1.16 \][/tex]
#### Step 4: Calculate the margin of error
The margin of error (MOE) is given by:
[tex]\[ \text{MOE} = z_{\alpha/2} \times \text{SEM} = 1.645 \times 1.16 \approx 1.908 \][/tex]
#### Step 5: Construct the confidence interval
The confidence interval is calculated as:
[tex]\[ \left( \bar{Y} - \text{MOE}, \bar{Y} + \text{MOE} \right) = \left( 712.1 - 1.908, 712.1 + 1.908 \right) \][/tex]
[tex]\[ \text{Confidence Interval} = (710.192, 714.008) \][/tex]
### Part (b)
We need to determine if there is statistically significant evidence that districts with smaller classes have higher average test scores.
#### Step 1: Identify the given data
For small classes:
- Sample mean, [tex]\( \bar{Y}_{\text{small}} = 721.8 \)[/tex]
- Sample standard deviation, [tex]\( s_{Y,\text{small}} = 24.4 \)[/tex]
- Sample size, [tex]\( n_{\text{small}} = 150 \)[/tex]
For large classes:
- Sample mean, [tex]\( \bar{Y}_{\text{large}} = 710.9 \)[/tex]
- Sample standard deviation, [tex]\( s_{Y,\text{large}} = 20.6 \)[/tex]
- Sample size, [tex]\( n_{\text{large}} = 250 \)[/tex]
#### Step 2: Calculate the standard error of the difference in means
The standard error of the difference in means (SED) is calculated as:
[tex]\[ \text{SED} = \sqrt{ \left( \frac{s_{Y,\text{small}}^2}{n_{\text{small}}} \right) + \left( \frac{s_{Y,\text{large}}^2}{n_{\text{large}}} \right) } \][/tex]
[tex]\[ \text{SED} = \sqrt{ \left( \frac{24.4^2}{150} \right) + \left( \frac{20.6^2}{250} \right) } \approx 2.388 \][/tex]
#### Step 3: Calculate the test statistic (z-value)
The test statistic [tex]\( z \)[/tex] is calculated as:
[tex]\[ z = \frac{\left( \bar{Y}_{\text{small}} - \bar{Y}_{\text{large}} \right)}{\text{SED}} = \frac{(721.8 - 710.9)}{2.388} \approx 4.579 \][/tex]
#### Step 4: Calculate the p-value
For the calculated z-value, we find the corresponding p-value. Since we are performing a two-tailed test:
[tex]\[ p = 2 \times (1 - \Phi(|z|)) \][/tex]
where [tex]\( \Phi \)[/tex] is the cumulative distribution function of the standard normal distribution.
Given the z-value obtained, the p-value is very small ([tex]\( 4.672 \times 10^{-6} \)[/tex]).
#### Step 5: Determine statistical significance
Compare the p-value with the significance level [tex]\( \alpha = 0.10 \)[/tex]. Since the p-value is much smaller than [tex]\( \alpha \)[/tex], we reject the null hypothesis. This indicates:
There is statistically significant evidence that districts with smaller classes have higher average test scores.
### Part (a)
We need to construct a 90% confidence interval for the mean test score in the population.
#### Step 1: Identify the given data
- Sample mean, [tex]\( \bar{Y} = 712.1 \)[/tex]
- Sample standard deviation, [tex]\( s_Y = 23.2 \)[/tex]
- Sample size, [tex]\( n = 400 \)[/tex]
- Confidence level, [tex]\( 90\% \)[/tex]
#### Step 2: Determine the critical value
For a 90% confidence interval, we use the z-distribution since the sample size is large ([tex]\( n > 30 \)[/tex]). The critical z-value [tex]\( z_{\alpha/2} \)[/tex] for a 90% confidence level is approximately 1.645.
#### Step 3: Calculate the standard error of the mean
The standard error of the mean (SEM) is given by:
[tex]\[ \text{SEM} = \frac{s_Y}{\sqrt{n}} = \frac{23.2}{\sqrt{400}} = 1.16 \][/tex]
#### Step 4: Calculate the margin of error
The margin of error (MOE) is given by:
[tex]\[ \text{MOE} = z_{\alpha/2} \times \text{SEM} = 1.645 \times 1.16 \approx 1.908 \][/tex]
#### Step 5: Construct the confidence interval
The confidence interval is calculated as:
[tex]\[ \left( \bar{Y} - \text{MOE}, \bar{Y} + \text{MOE} \right) = \left( 712.1 - 1.908, 712.1 + 1.908 \right) \][/tex]
[tex]\[ \text{Confidence Interval} = (710.192, 714.008) \][/tex]
### Part (b)
We need to determine if there is statistically significant evidence that districts with smaller classes have higher average test scores.
#### Step 1: Identify the given data
For small classes:
- Sample mean, [tex]\( \bar{Y}_{\text{small}} = 721.8 \)[/tex]
- Sample standard deviation, [tex]\( s_{Y,\text{small}} = 24.4 \)[/tex]
- Sample size, [tex]\( n_{\text{small}} = 150 \)[/tex]
For large classes:
- Sample mean, [tex]\( \bar{Y}_{\text{large}} = 710.9 \)[/tex]
- Sample standard deviation, [tex]\( s_{Y,\text{large}} = 20.6 \)[/tex]
- Sample size, [tex]\( n_{\text{large}} = 250 \)[/tex]
#### Step 2: Calculate the standard error of the difference in means
The standard error of the difference in means (SED) is calculated as:
[tex]\[ \text{SED} = \sqrt{ \left( \frac{s_{Y,\text{small}}^2}{n_{\text{small}}} \right) + \left( \frac{s_{Y,\text{large}}^2}{n_{\text{large}}} \right) } \][/tex]
[tex]\[ \text{SED} = \sqrt{ \left( \frac{24.4^2}{150} \right) + \left( \frac{20.6^2}{250} \right) } \approx 2.388 \][/tex]
#### Step 3: Calculate the test statistic (z-value)
The test statistic [tex]\( z \)[/tex] is calculated as:
[tex]\[ z = \frac{\left( \bar{Y}_{\text{small}} - \bar{Y}_{\text{large}} \right)}{\text{SED}} = \frac{(721.8 - 710.9)}{2.388} \approx 4.579 \][/tex]
#### Step 4: Calculate the p-value
For the calculated z-value, we find the corresponding p-value. Since we are performing a two-tailed test:
[tex]\[ p = 2 \times (1 - \Phi(|z|)) \][/tex]
where [tex]\( \Phi \)[/tex] is the cumulative distribution function of the standard normal distribution.
Given the z-value obtained, the p-value is very small ([tex]\( 4.672 \times 10^{-6} \)[/tex]).
#### Step 5: Determine statistical significance
Compare the p-value with the significance level [tex]\( \alpha = 0.10 \)[/tex]. Since the p-value is much smaller than [tex]\( \alpha \)[/tex], we reject the null hypothesis. This indicates:
There is statistically significant evidence that districts with smaller classes have higher average test scores.
