Each student was asked to record and report the amount of money they spent on text books in a semester. The sample of 130 students resulted in the average of $422 and a standard deviation of $57. Find a 99% confidence interval for the mean amount of money spent by collage students on textbooks

(409.36, 434.64)

Step-by-step explanation:

When population standard deviation is unknown and sample is not so large , then the formula to find the confidence interval for population mean is given by :-

, where = sample mean , n = sample size , s= sample population standard deviation, t*= two tailed critical value.

As , per given , , s=$57, n=130

For 99% confidence ,

By t-distribution table , t-value for   (two tailed) and df =129 [∵df=n-1] would be

t*=2.6145

Now , the  99% confidence interval for the mean amount of money spent by collage students on textbooks will be :

Hence, a 99% confidence interval for the mean amount of money spent by collage students on textbooks will be (409.36,  434.64).

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).