A data set includes data from student evaluations of courses. The summary statistics are nequals89, x overbarequals3.58, sequals0.53. Use a 0.01 significance level to test the claim that the population of student course evaluations has a mean equal to 3.75. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.

|t| =| - 3.0303| = 3.03

Hence we rejected null hypothesis.

The claim that the population of student course evaluations has a mean is not equal to 3.75

Step-by-step explanation:

Step(i)

A data set includes data from student evaluations of courses.

Given n= 89

sample mean x⁻ = 3.58

Sample Standard deviation (S) = 0.53

Given data population of student course evaluations has a mean equal to 3.75.

μ = 3.75

Step(ii)

Null hypothesis:-H₀: the population of student course evaluations has a mean equal to 3.75

μ = 3.75

Alternate hypothesis :- H₁: μ ≠ 3.75

Level of significance :-∝=0.01

The test statistic

Step(iii):

The degrees of freedom γ=n-1 = 89-1=88

tabulated value t = 2.3733

Calculated value

       t =  - 3.0303

|t| =| - 3.0303| = 3.03

The tabulated value of 't' for 88 degree of freedom at 1% of level of significance is 2.3733

since calculated value t=3.03 is greater than tabulated value t = 2.3733 of t

Hence we rejected null hypothesis.

the claim that the population of student course evaluations has a mean is not equal to 3.75

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: