A survey measures the motivation, attitudes, and study habits of freshman in college. Freshmans’ scores on this survey range from 0 to 200 and follow (approximately) a Normal distribution, with mean of 110 and standard deviation σ = 20. You suspect that the next group of freshman will have a mean μ, which is different from 110 because they are often excited yet anxious about leaving college. To verify your suspicion, you test the hypotheses H0: μ = 110, Ha: μ ≠ 110 You give the survey to 50 high school seniors who are incoming freshman and find their mean score. Suppose you observed the same sample mean of 115.35, but based on a sample of 100 freshman students. What would the corresponding P-value be?

a. 0.0074
b. 0.9926
c. None of the answers are correct
d. 0.0037

Option A) 0.0074    

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 110

Sample mean, = 20

Sample size, n = 100

Alpha, α = 0.05

Population standard deviation, σ = 115.35

First, we design the null and the alternate hypothesis

We use two-tailed z test to perform this hypothesis.

Formula:

Putting all the values, we have

Now, we calculate the p-value from the standard normal table.

P-value  = 0.0074

Thus, the correct answer is

Option A) 0.0074

a 360 degree rotation

step-by-step explanation:

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b