Aprofessor is interested in comparing the average amount spent on textbooks for freshmen and sophomores. a random sample of 10 freshmen yielded a sample mean amount of $1023 and a sample standard deviation of $68. a random sample of 14 sophomores yielded a sample mean amount of $1257 and a sample standard deviation of $338. construct a 99% confidence interval for the difference between the mean amount spent on textbooks by freshmen and the mean amount spent by sophomores (ie. do freshmen minus sophomores). since the sample standard deviations are wildly different, use a method which does not assume the populations have the same variance. assume normality.

-332.86 < µ1 - µ2 < -141.14

Step-by-step explanation:

We will be constructing a 99% confidence interval for a difference of means.  

We have 2 samples sizes

Sample 1: (freshmen)

n = 10

x = 1023

s = 68

Sample 2: (sophomores)

n = 14

x = 1257

s = 338

We are to assume normality, but the equality in the population variances.  This determines the formula we will use.  For this situation, we use t-distribution to find the error, and our degrees of freedom are one less than the size of the smallest sample.  In this case, 9 degrees of freedom.  

Our t-value is: 3.25

*See attached photo for the construction of the confidence interval

let's solve your equation step-by-step.

24−3x=−27

step 1: simplify both sides of the equation.

24−3x=−27

24+−3x=−27

−3x+24=−27

step 2: subtract 24 from both sides.

−3x+24−24=−27−24

−3x=−51

step 3: divide both sides by -3.

−3x

−3

=

−51

−3

x=17

answer:

x=17

step-by-step explanation:

mark brainliest and have a great day!