6-5/8= (5/8 is just like that but slanted)
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Answer:
The 95% confidence interval for the true mean number of students per teacher is (17.9, 20.5). The lower bound of the interval is above the national average, which means that this estimate is higher than the national average.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 12 - 1 = 11
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 11 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.201
The margin of error is:
In which s is the standard deviation of the sample(square root of the variance) and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 19.2 - 1.3 = 17.9 students.
The upper end of the interval is the sample mean added to M. So it is 19.2 + 1.3 = 20.5 students
The 95% confidence interval for the true mean number of students per teacher is (17.9, 20.5). The lower bound of the interval is above the national average, which means that this estimate is higher than the national average.