Find the t dcore you would use for an 95% confidence interval. 80 students,72 average, 4 standard deviation.
To find the t-score for a 95% confidence interval, you need to know the sample size, mean, and standard deviation. From your question, I can see that the sample size is 80, the average is 72, and the standard deviation is 4.
To calculate the t-score, you can use the formula:
t = (X - μ) / (s / √n)
Where:
- t is the t-score
- X is the average
- μ is the population mean (unknown in this case)
- s is the standard deviation
- n is the sample size
Since we want to construct a 95% confidence interval, we need to find the critical value for a two-tailed test with a confidence level of 95%. This means that we need to find the t-value associated with a 2.5% (1 - 0.95) tail on each side of the distribution.
The degrees of freedom (df) for this case will be n - 1, where n is the sample size. So, in this case, df = 80 - 1 = 79.
Now, we can use a table or a statistical software to find the t-value for a 2.5% tail with 79 degrees of freedom. However, I'll demonstrate how to find it using a Python code snippet:
```python
import scipy.stats as stats
confidence_level = 0.95
sample_size = 80
degrees_of_freedom = sample_size - 1
t_value = stats.t.ppf((1 + confidence_level) / 2, degrees_of_freedom)
print(t_value)
```
Running the code will give you the t-value of approximately 1.990.
Therefore, the t-score for a 95% confidence interval, given a sample size of 80, an average of 72, and a standard deviation of 4, is approximately 1.990.