The national average for the number of students per teacher for all U. S. public schools in 15.9. A random sample of 12 school districts from a moderately populated area showed that the mean number of students per teacher was 19.2 with a variance of 4.41. Estimate the true mean number of students per teacher with 95% confidence. How does your estimate compare with the national average

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.

answer are

step-by-step explanation:

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

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