To better understand the financial burden students are faced with each term, the statistics department would like to know how much their ST201 students are spending on school materials on average. Let’s use our class data to calculate a 95% confidence interval to estimate the average amount ST201 students spend on materials each term. The average from our student survey is $248 and the number of students sampled is 90. Assume . State the question of interest. On average, how much do ST201 students spend on school materials each term? a. (1 point) Identify the parameter. b. Check the conditions. a. (2 points) Does the data come from a random sample? What are some potential biases about the way the data was collected? (1 point) Is the sample size large enough for distribution of the sample mean to be normal according to the rules for Central Limit Theorem?

a).

The amount spent on school materials for each term of all ST201students

b).

a).

It is not a random sample. This looks like a convenience sampling and there is sampling bias. This sample is not representative of the entire population. Since it is not a random sample it is not appropriate to generalize the results to all students.

b).

The sample size is 80 which is greater than 30. It is large enough to assume normal distribution according to central limit theorem.

c).

mean: $617

z critical value at 95%: 1.96

standard error = σ/sqrt(n) =500/sqrt(80) = 55.9017

lower limit= mean-1.96*se = 617-1.96*55.9017=507.43

upper limit= mean+1.96*se = 617+1.96*55.9017=726.57

d).

The amount spent on school materials for each term for the 80 ST201students is $617. We are 95% confident that amount spent on school materials for each term of all ST201students falls in the interval ($507.43, $726.57).

Step-by-step explanation:

hard

step-by-step explanation:

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)