The principal randomly selected six students to take an aptitude test. their scores were: 87.4 86.9 89.9 78.3 75.1 70.6 determine a 90% confidence interval for the mean score for all students. assume the population is normally distributed. give the lower limit of the interval as your answer below. round to two decimal places. (hint from class: note that you will need to calculate the sample mean (xbar) and sample standard deviation (s) for this sample data in order to find the confidence interval endpoints. in class i referred to the formula for that -- and erroneously gave the formula for the sample variance instead of the sample standard deviation. the sample standard deviation would be the square root of the sample variance. formulas for both can be found on the test formula sheets posted on ilearn among other places.)

79.5688 < µ < 83.1646

Step-by-step explanation:

Sample mean is the sum of all scores, divided the the total number of test takers.  In this case, the sample mean is:

(87.4 + 86.9 + 89.9 + 78.3 + 75.1 + 70.6)/6 =  488.2/2 = 81.3667

The sample standard deviation is the square root of the sample variance.  See attached photo 1 for calculation of these values...

The sample standard deviation is 3.1856

We need to make a 90% confidence interval for this data.  Since n < 30, we will use a t-value.  The degrees of freedom is always one less than the sample size so on the t-distribution chart, look under the column for Area under the curve = 0.10, and the row for 5.  The t-value you should see is t = 2.015

See attached photo 2 for the construction of the confidence interval

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation:

i’m here for points.